H C Verma solutions, MAGNETIC FIELD, Chapter-34, Questions for Short Answer, Concepts of Physics, Part-II

Magnetic Field

Questions for Short Answer

    1.  Suppose a charged particle moves with a velocity v near a wire carrying an electric current. A magnetic force, therefore, acts on it. If the same particle is seen from a frame moving with velocity v in the same direction, the charge will be found at rest. Will the magnetic force become zero in this frame? Will the magnetic field become zero in this frame?  

ANSWER: Since the charge is at rest in the second frame, the magnetic force is zero in this frame.    

    In the second frame, the magnetic field appears to be zero, but it will appear as an electric field or a combination of both because electric and magnetic fields are not basically independent.

    2.  Can a charged particle be accelerated by a magnetic field? Can its speed be increased?

ANSWER: A uniform magnetic field exerts a centripetal force on a charged particle, hence it can accelerate centripetally. 

     Since the magnetic force on the charged particle is always perpendicular to its direction, its speed can not be increased.   

    3.  Will a current loop placed in a magnetic field always experience a zero force?  

ANSWER: The current loop will always experience a zero force if placed in a uniform magnetic field. But not if the magnetic field is non-uniform. 

    4.  The free electrons in a conducting wire are in constant thermal motion. If such a wire carrying no current is placed in a magnetic field, is there a magnetic force on each free electron? On the wire?  

ANSWER: Yes, there will be a magnetic force on each free electron. Since the electrons are in random motions, the net magnetic force on the wire is zero.  

    5.  Assume that the magnetic field is uniform in a cubical region and is zero outside. Can you project a charged particle from outside into the field so that the particle describes a complete circle in the field?  

ANSWER: No. Suppose the charged particle is projected in a plane perpendicular to the uniform magnetic field to describe a circle. As soon as the charged particle enters the cubical area, a magnetic force perpendicular to the direction of velocity starts acting on it, and it moves in a circular path. But the path can not be a complete circle, as the direction of velocity at the entry point is tangent to the circular path being described, and the particle can not return to this point inside the cube. See the diagram below.  

Diagram for Q-5

    6.  An electron beam projected along the positive X-axis deflects along the positive Y-axis. If this deflection is caused by a magnetic field, what is the direction of the field? Can we conclude that the field is parallel to the Z-axis?  

ANSWER: The direction of the component of the magnetic field causing the deflection is along the Z-axis. Another component of the magnetic field may be along the X-axis, which will have no effect on the charged particle.  

      It can not be concluded that the resultant field is parallel tothe  Z-axis. 

    7.  Is it possible for a current loop to stay without rotating in a uniform magnetic field? If yes, what should be the orientation of the loop?  

ANSWER: Yes, a current loop can stay without rotating in a uniform field. 

    The plane of the loop should be perpendicular to the direction of the magnetic field.   

    8.  The net charge in a current-carrying wire is zero. Then, why does a magnetic field exert a force on it? 

ANSWER: The net charge in a current-carrying wire is zero, but the moving charge in it is only electrons, which are also in the same direction. Hence, a magnetic field exerts a force on a current-carrying wire through the moving electrons. 

    9.  The torque on a current loop is zero if the angle between the positive normal and the magnetic field is either θ =0 or θ =180°. In which of the two orientations is the equilibrium stable? 

ANSWER: The equilibrium is stable in the case of minimum potential energy. Let us calculate the potential energy of a loop when deflected from θ =0° to θ.  

   Torque on the loop,

Γ =iABsinθ

   Where i = current in the loop, A =area of the loop, and B = magnitude of the uniform magnetic field. Let us rotate the loop by a very small angle dθ against the torque very slowly. 

    The work done on the loop against the magnetic field = The potential energy stored =Γ dθ.

→dU = iAB sinθdθ

On integration,

U =iAB [-cosθ ]

    Putting the limit of integration between 0° and θ, we get,

U =iAB(-cosθ +1)

→U =iAB(1 -cosθ)

   Now, for the two zero torque positions, let us check the potential energy.

   For θ = 0°, U = 0. 

But for θ =180°, U =2iAB.

   Clearly, the potential energy is minimum for θ =0°, hence the equilibrium is stable for θ =0°.   

    10.  Verify that the units weber and volt-second are the same.   

ANSWER: When a loop of one turn is rotated in a magnetic field, the electromotive force produced in it is, 

v =dφ/dt, i.e. rate of change of magnetic flux. Thus, 

dφ =v dt 

Unit of φ, magnetic flux is weber. On the right-hand side, the same is volt-second. So both are the same.   

OBJECTIVE - I

    1.  A positively charged particle projected towards the east is deflected towards the north by a magnetic field. The field maybe 

(a) towards west

(b) towards south

(c) upwards

(d) downwards  

ANSWER: (d)        

EXPLANATION: Since the deflection of the positively charged particle is towards the north, the magnetic field causing this deflection may be perpendicular to both directions, i.e., in the vertical direction.

Diagram for Q-1

    The force by the magnetic field is given as, 

F = q vxB                

From this vector cross-product relation, we can conclude that the direction of the magnetic field is downwards. Option (d) is correct.   

   

    2.  A charged particle is whirled in a horizontal circle on a frictionless table by attaching it to a string fixed at one point. If a magnetic field is switched on in the vertical direction, the tension in the string 

(a) will increase

(b) will decrease

(c) will remain the same

(d) may increase or decrease.       

ANSWER: (d)        

EXPLANATION: Neither the direction of the movement of the particle is known (clockwise or anticlockwise), nor the direction of the magnetic field is given (upwards or downwards). The nature of the charge on the particle is also not given. Depending on the nature of the charge and these two directions, the force on the charged particle may be either towards the fixed point (center) or away from it. So the tension in the string may increase or decrease. Option (d) is correct.      

   

    3.  Which of the following particles will experience a maximum magnetic force (magnitude) when projected with the same velocity perpendicular to a magnetic field?    

(a) electron

(b) proton

(c) He⁺

(d) Li⁺⁺ 

ANSWER: (d)        

EXPLANATION: The maximum magnitude of the force is 

F = qvB   

Given that v and B are fixed, the particle having the maximum charge will experience the maximum magnetic force. Out of the four particles, electron, proton, and He⁺ have the same magnitude of charge, but Li⁺⁺ has double that magnitude. Hence, Li⁺⁺ will experience the maximum magnetic force. Option (d) is correct.

  

    4.  Which of the following particles will describe the smallest circle when projected with the same velocity perpendicular to a magnetic field?

(a) electron

(b) proton

(c) He⁺

(d) Li⁺ 

   

ANSWER: (a)        

EXPLANATION: The magnitude of the charge on all of the given particles is the same. Hence, the magnetic force F will have equal magnitudes. This magnetic force provides the centripetal force for the circular motion of the particles, hence 

F = mv²/r 

→r =mv²/F 

    v²/F is fixed, so the radius r will be minimum for the particle having the least mass. Here it is, the electron. Thus, option (a) is correct.  

       

    5.  Which of the following particles will have a minimum frequency of revolution when projected with the same velocity perpendicular to a magnetic field?   

(a) electron

(b) proton

(c) He⁺

(d) Li⁺ 

ANSWER: (d)        

EXPLANATION: The time period T of a circular motion of a particle with speed v is, 

T =2πr/v

  So the frequency, 

f =1/T =v/2πr 

Also, the magnetic force,

qvB =mv²/r 

→v =qrB/m   

Now frequency,

f = qrB/2πrm =qB/2πm 

    Here, for the given particles, the magnitude of charge q is the same, B is constant, thus the frequency f is inversely proportional to the mass m of the particle. Since Li has the greatest mass among the four particles, it will have the minimum frequency. Option (d) is correct.  

               

    6.  A circular loop of area 1 cm², carrying a current of 10 A, is placed in a magnetic field of 0·1 T perpendicular to the plane of the loop. The torque on the loop due to the magnetic field is 

(a) zero

(b) 10⁻⁴ N-m 

(c) 10⁻² N-m  

(d) 1 N-m.   

     

ANSWER: (a)        

EXPLANATION: Since the direction of the magnetic field is perpendicular to the plane of the loop, the torque on the loop will always be zero, whatever be the current in the wire, the area of the loop, or the magnitude of the magnetic field. Option (a) is correct.     

            

    7.  A beam consisting of protons and electrons moving at the same speed goes through a thin region in which there is a magnetic field perpendicular to the beam. The protons and electrons  

(a) will go un-deviated 

(b) will be deviated by the same angle and will not separate 

(c) will be deviated by different angles and hence separate

(d) will be deviated by the same angle, but will separate.       

ANSWER: (c)        

EXPLANATION: The magnitude of the charge on a proton or an electron is the same. Hence, the force on them in the thin magnetic region will be the same but opposite in direction because the nature of the charge on them is opposite. 

  Due to the force, the acceleration in the direction of the force =F/m. Mass is different for them, hence the different accelerations. Otherwise, we may think that in the thin magnetic field, the radius of the arc on which they move =mv²/F.

    Due to the difference in mass, they will deviate by different angles and separate. Option (c) is correct.  

        

    8.  A charged particle in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be 

(a) a straight line

(b) a circle

(c) helix with uniform pitch

(d) a helix with a non-uniform pitch.     

ANSWER: (c)        

EXPLANATION: We can resolve the velocity along the magnetic field and perpendicular to the magnetic field. Due to the perpendicular component of the velocity, the charged particle will move in a circle in a plane perpendicular to the direction of the magnetic field. On the other hand, the component of the velocity along the magnetic field will be unaffected, and due to this, the particle will have a uniform straight motion along the field. When we combine these two motions for the charged particle, the path will be a helix with a uniform pitch. Option (c) is correct.  

      

    9.  A particle moves in a region having a uniform magnetic field and a parallel, uniform electric field. At some instant, the velocity of the particle is perpendicular to the field direction. The path of the particle will be

(a) a straight line

(b) a circle

(c) a helix with uniform pitch

(d) a helix with a non-uniform pitch.   

ANSWER: (d)        

EXPLANATION: The effect of the magnetic field will be to move the charged particle in a circular path, while the effect of the electric field on the particle will be to accelerate it towards either of the directions of the field, depending upon the nature of the charge. So, combining a uniform circular motion and an accelerated straight motion perpendicular to the circular motion will give a path of a helix with a non-uniform pitch. Option (d) is correct. 

 

    10.  An electric current 'i' enters and leaves a uniform circular wire of radius 'a' through diametrically opposite points. A charged particle q moving along the axis of the circular wire passes through its center at speed v. The magnetic field acting on the particle when it passes through the center has a magnitude  

(a) qv(µ₀i/2a) 

(b) qv(µ₀i/2πa)

(c) qv(µ₀i/a)

(d) zero.       

ANSWER: (d)        

EXPLANATION: Since the current 'i' enters and leaves the wire loop at diametrically opposite points, the magnitude of the current in each half of the wire will be equal to i/2. The direction of the current in one half will be clockwise, while in the other it will be anti-clockwise. So whatever magnetic field they produce at the center will be equal and opposite in direction. Thus, the net magnetic field at the center = zero. Option (d) is correct.

  

OBJECTIVE - II

    1.  If a charged particle at rest experiences no electromagnetic force, 

(a) the electric field must be zero.

(b) The magnetic field must be zero.

(c) The electric field may or may not be zero.

(d) The magnetic field may or may not be zero.     

ANSWER: (a), (d).   

EXPLANATION: There will not be a magnetic force on a charged particle at rest, even if there is a magnetic field present. Option (d) is correct, not option (b). 

   But if there is an electric field, the charged particle can not remain at rest in the absence of a magnetic force on it. So the electric field here must be zero. Option (a) is correct, and option (c) is not correct.    

    2.  If a charged particle kept at rest experiences an electromagnetic force, 

(a) the electric field must not be zero.

(b) The magnetic field must not be zero.

(c) The electric field may or may not be zero.

(d) The magnetic field may or may not be zero.     

ANSWER: (a), (d).  

EXPLANATION: A magnetic field does not exert a force on a charged particle kept at rest. So if a charged particle is at rest, there may or may not exist a magnetic field. Even if the particle experiences an electromagnetic force, there must be the presence of an electric field. Hence, only options (a) and (d) are correct.  

    3.  If a charged particle projected in a gravity-free room deflects, 

(a) there must be an electric field. 

(b) There must be a magnetic field

(c) Both fields cannot be zero.

(d) Both fields can be nonzero.        

ANSWER: (c), (d).   

EXPLANATION: If a charged particle projected in a gravity-free room deflects, it may be due to either of the two fields. Hence, options (a) and (b) are not correct.  

   If both fields are zero, then the force on the particle is zero, and it can not deflect. So both fields can not be zero. Option (c) is correct. 

     If both fields are present with nonzero magnitudes, even then the particle can deflect. Option (d) is correct.      

    4.  A charged particle moves in a gravity-free space without a change in velocity. Which of the following is/are possible? 

(a) E =0, B =0. 

(b) E =0, B ≠0.

(c) E ≠0, B =0.

(d) E ≠0, B ≠0. 

ANSWER: (a), (b), (d).  

EXPLANATION: If both fields are zero, there will be no force on the charged particle. Thus, it can move without a change in velocity. Option (a) is correct. 

     If there is only the magnetic field present, even then the charged particle can move with a constant velocity if the direction of the magnetic field is parallel to the direction of the velocity. Option (b) is correct.    

      If the only electric field is present, then the charged particle will always experience a force due to i,t and it can not move with a constant velocity. Option (c) is not correct.    

      The magnetic force on a moving charged particle is perpendicular to the magnetic field and the velocity. Now, if the direction of the electric field is such that it exerts a force equal and opposite to the magnetic force, then the resultant force on the particle is zero, and it can move with a constant velocity. Option (d) is correct. See the picture below:- 

The direction of F is perpendicular to both v and B

 

    5.  A charged particle moves along a circle under the action of possible constant electric and magnetic fields. Which of the following are possible? 

(a) E =0, B =0. 

(b) E =0, B ≠0.

(c) E ≠0, B =0.

(d) E ≠0, B ≠0. 

     

ANSWER: (b).   

EXPLANATION: Without a centripetal force, a particle can not move along a circle. In the absence of both fields, the force on the particle is zero. So both fields can not be zero in this case. Option (a) is incorrect.  

   If the electric field is zero but the magnetic field is non-zero, and the charge moves perpendicular to the magnetic field, a magnetic force perpendicular to the velocity will act. It will provide a centripetal force to the particle, and it will move in a circle in a plane perpendicular to the magnetic field. Option (b) is correct.  

      If the magnetic field is zero but the electric field is non-zero, then the charged particle can not move along a circle because it can not provide a constant magnitude of force always perpendicular to the velocity in any condition. Option (c) is incorrect.

       If both the fields are non-zero, the charged particle will have an additional straight-line velocity, and due to the combined effects of these velocities, the particle cannot move in a complete circle. Option (d) is incorrect.         

             

    6.  A charged particle goes undeflected in a region containing electric and magnetic fields. It is possible that,   

(a) E ॥ B, v ॥ E

(b) E is not parallel to B 

(c) v ॥ B but E is not parallel to B.

(d) E ॥ B but v is not parallel to E.   

     

ANSWER: (a),(b).   

EXPLANATION: If the direction of velocity, electric, and magnetic fields are all parallel, the magnetic force will be zero on the particle. Only the electric force will act parallel to the velocity. It will move along a straight line without deflection. Option (a) is correct.                    If the particle moves in a magnetic field, not in a direction parallel to this field, a magnetic force perpendicular to both the velocity and the magnetic field will act on the particle. If an electric field is present such that it applies an electric force that is equal and opposite to the magnetic force on the charge, then the resultant force on the particle is zero. The charge will move in a straight line undeflected. Option (b) is correct.

     If the direction of velocity is parallel to the magnetic force, no magnetic force is on the charged particle. Now, if the direction of the electric field is not parallel to the magnetic force/velocit,y then there will be an electric force on the particle that is not along the line of velocity, hence it will deflect. Option (c) is not possible.

     In the last case, since the velocity is not parallel to the magnetic field, a magnetic force will act on it that is perpendicular to both B and v. It can move undeflected only if the electric field is along the magnetic force i.e., perpendicular to B, but given that E || B. So in this situation, the condition cannot be fulfilled. So option (d) is not possible.

                        

    7.  If a charged particle goes unaccelerated in a region containing electric and magnetic fields,   

(a) E must be perpendicular to B. 

(b) v must be perpendicular to E.

(c) v must be perpendicular to B. 

(d) E must be equal to vB.          

ANSWER: (a), (b).   

EXPLANATION: A particle can move unaccelerated only when the resultant force acting on it is zero. If there are both electric and magnetic fields present in a region, only in the following situation resultant force be zero.  

   q|E| = q|vxB| -------- (i) 

  So the line of action of E is the same as the line of action of vxB, that is, E ⟂ B and v ⟂ E. Options (a) and (b) are correct.

 From (i), it is clear that there is no necessity for v to be perpendicular to B. Option (c) is incorrect.

  Also from (i), E = vB.sinθ, and since θ is not necessarily 90°, option (d) is not correct.

 

    8.  Two ions have equal masses, but one is singly ionized, and the other is doubly ionized. They are projected from the same place in a uniform magnetic field with the same velocity perpendicular to the field.

(a) Both ions will go along circles of equal radius. 

(b) The circles described by the single ionized charge will have a radius double that of the other circle. 

(c) The circles do not touch each other.

(d) The two circles touch each other.        

ANSWER: (b), (d).   

EXPLANATION: Centripetal force,

qvB =mv²/r 

→r =mv/qB

m, v, and B are constant here, so the radius of the circle for a  2q charged ion will be r/2. Thus, the circle described by the singly charged ion will have a radius double that of the doubly charged ion. Option (b) is correct; option (a) is wrong.  

   Since both ions are projected from the same place, the two circles will touch each other. Option (d) is correct, option (c) is wrong.       

    9.  An electron is moving along the positive X-axis. You want to apply a magnetic field for a short time so that the electron may reverse its direction and move parallel to the negative X-axis. This can be done by applying the magnetic field along,

(a) Y-axis 

(b) Z-axis

(c) Y-axis only

(d) Z-axis only.     

ANSWER: (a), (b).   

EXPLANATION: Since the force exerted due to a magnetic field is perpendicular to the velocity and also perpendicular to the field itself, a magnetic field applied along the Y-axis will move the electron on a circular path in the XZ plane. If applied along the Z-axis, the electron will move on a circular path in the XY plane. In both of these cases, the magnetic field may be withdrawn just after the electron has moved half a circle, resulting in reversal of the movement of the electron along parallel to the negative X-axis. So options (a) and (b) are correct; the other two are wrong. 

    10.  Let E and B denote electric and magnetic fields in a frame S and E' and B' in another frame S' moving with respect to S at a velocity v. Two of the following equations are wrong. Identify them. 

(a) Bᵧ' =Bᵧ +vE_z/c²

(b) Eᵧ' =Eᵧ +vB_z/c²

(c) Bᵧ' =Bᵧ +vE_z

(d) Eᵧ' =Eᵧ +vB_z. 

ANSWER: (b), (c).   

EXPLANATION: Since the magnitude of the electric force =qE and that of the magnetic force =qvBsinθ. In case they are equal,

  E =vBsinθ.

In case that θ =90°, sinθ =1. Then,

  E =vB and B =E/v 

   So the dimensions of E are the same as vB, and the dimensions of B are the same as E/v. The dimensions of the last term of the R.H.S. of the second equation do not match with E. This equation is wrong. Option (b) is correct.

   The same is the case with the third equation. The last term does not dimensionally match with B. This equation is also incorrect. So option (c) is true.

   The first and the last equations are dimensionally correct. Since it is given that two equations are wrong that we have found out as (b) and (c). There is no necessity to check the facts of the remaining two equations.   

 

Exercises

    1.  An alpha particle is projected upward with a speed of 3·0x10⁴ km/s in a region where a magnetic field of magnitude 1·0 T exists in the direction south to north. Find the magnetic force that acts on the α-particle.    

ANSWER: B =1·0 T, v =3·0x10⁴ km/s =3·0x10⁷ m/s. Charge on the alpha particle, q =2x1·6x10⁻¹⁹ C =3·2x10⁻¹⁹ C. The angle between the directions of velocity and magnetic field is θ =90°. 

   The magnitude of the magnetic force acting on the alpha particle,

   F =qvB.sinθ

  =3·2x10⁻¹⁹*3·0x10⁷*1.0 N

  =9.6x10⁻² N.

     From the right-hand rule, we get the direction of this force towards the west. 

 

    2.  An electron is projected horizontally with a kinetic energy of 10 keV. A magnetic field of strength 1.0×10⁻⁷ T exists in the vertically upward direction.

(a) Will the electron deflect towards the right or towards the left of its motion?  

(b) Calculate the sideways deflection of the electron in traveling through 1 m. Make appropriate approximations.         

ANSWER: (a) Since the magnetic force is the vector cross product of velocity and magnetic field, from the right-hand rule, we can find out the direction of the magnetic force on a positive charge. In the given situation, the direction of magnetic force for a positive charge is towards the right, but it is an electron that has a negative charge, so the force on it will be opposite, i.e., towards the left. Hence, the answer is -towards the left. 

    (b) Since the electron moves with a very high speed, let us assume that the direction of force remains constant during this small distance. Let the speed of the electron =v. Kinetic energy =½mv². It is given 10 keV =10*10³*1.6x10⁻¹⁹ J 

→½mv² =1.6x10⁻¹⁵   

→v² =3·2x10⁻¹⁵/9·11x10⁻³¹  

→v² =3·51x10¹⁵  

→v =5.92x10⁷ m/s  

Magnitude of magnetic force on the electron,  

F =qvB

  =1·6x10⁻¹⁹*5·92x10⁷*1·0x10⁻⁷ N 

  =9·472x10⁻¹⁹ N                

Acceleration of the electron perpendicular to the initial velocity,   

a =F/m  

  =9·472x10⁻¹⁹/9·11x10⁻³¹ m/s² 

  =1.04x10¹² m/s² 

The time taken by the electron in traveling through 1 m, 

t =Distance/speed   

  =1/5·92x10⁷ s 

  =1.69x10⁻⁸ s

At this time, the transverse distance traveled, sideways deflection,  

s =ut +½at², here u =0, so  

s =½at² 

  =½*1·04x10¹²*(1.69x10⁻⁸)² m  

  =1·49x10⁻⁴  m 

  ≈1·5x10⁻² cm.                                   

 

    3.  A magnetic field of (4.0x10⁻³ k) T exerts a force of (4.0i +3.0j)x10⁻¹º N on a particle having a charge of 1.0x10⁻⁹ C and going in the X-Y plane. Find the velocity of the particle.     

ANSWER: The magnetic force, 

F = (4·0 i +3·0 j)x10⁻¹º N 

  =Fₓ i + Fᵧ j 

So Fₓ =4·0x10⁻¹º N  

and Fᵧ =3·0x10⁻¹º N  

Since F =qvB for v⟂B, 

v =F/qB, Here 

vᵧ =Fₓ/qB,  (Since velocity will be perpendicular to both force and field)

  =4·0x10⁻¹⁰/(1·0x10⁻⁹*4·0x10⁻³) m/s

 =100 m/s 

and vₓ =Fᵧ/qB 

  =3·0x10⁻¹⁰/(1·0x10⁻⁹*4·0x10⁻³) m/s

  =75 m/s, but from right hand rule, its direction will be in the direction of the negative X-axis. So the velocity of the particle is,  

v =(-75 i +100 j) m/s.      

 

 

    4.  An experimenter's diary reads as follows: "A charged particle is projected in a magnetic field of (7.0i -3.0j)x10⁻³T. The acceleration of the particle is found to be (⃞i +7.0j)x10⁻⁶ m/s²". The number to the left of i in the last expression was not readable. What can this number be?    

ANSWER: Let the number to the left of i in the last expression be equal to x. The acceleration will now be written as,  

α =(x i +7·0 j)x10⁻⁶ m/s²  

Since the magnetic force and hence the resulting acceleration is always perpendicular to the magnetic field, the dot product of the magnetic field B and the acceleration α will be zero. Thus,

α.B =0

{(xi +7·0j)x10⁻⁶}.{(7·0i -3·0j)x10⁻³}=0

→7x -21 =0,{Since i.j =0 and i.i=j.j=1}

→7x =21 

→x =3·0.   

 

    5.  A 10 g bullet having a charge of 4.00 µC is fired at a speed of 270 m/s in a horizontal direction. A vertical magnetic field of 500 µT exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100 m. Make appropriate approximations.    

ANSWER: Time to travel 100 m will be, 

t =100/270 s =0·37 s. We have to calculate the distance traveled by the bullet perpendicular to the velocity. Since the speed is very high, we assume that the magnetic force on the bullet remains perpendicular to the initial direction of the velocity during this time. The magnetic force,  

F =qvB, since theta is 90°. 

Acceleration of the bullet along F,

α =F/m =qvB/m   

The initial velocity along F is zero, hence the distance traveled in time t i.e., the required deflection is,

s =ut +½αt² =½αt² 

 =qvBt²/2m  

 =4x10⁻⁶*270*500x10⁻⁶*0·37²/(2*0.01)

 =3·7x10⁻⁶ m.            

 

    6.  When a proton is released from rest in a room, it starts with an initial acceleration α₀ towards the west. When it is projected towards the north with a speed v₀, it moves with an initial acceleration 3α₀ toward the west. Find the electric field and the maximum possible magnetic field in the room.     

ANSWER: When the proton is released from rest, there will be no magnetic force acting on it. Since the acceleration initially is towards the west, there is an electric field present towards the west. The electric force on the proton, 

F =qE =eE

→α₀ =F/m =eE/m,  

(where E is the electric field present, m is the mass of the proton, and e is the charge on it.)

→E = mα₀/e towards the west. 

   The maximum magnetic field will exert a maximum magnetic force, for which the angle between velocity and the magnetic field should be 90°. The value of magnetic force, F' =qvB =ev₀B. Since in the second case, the acceleration is still towards the west, it means the magnetic force is also acting towards the west, and the net acceleration produced by it,

 =3α₀ -α₀ =2α₀. 

So, F' =m*2α₀ 

→ev₀B =m*2α₀ 

→B =2mα₀/ev₀ 

 From the right-hand rule, its direction is downward.            

 

    7.  Consider a 10 cm long portion of a straight wire carrying a current of 10 A, placed in a magnetic field of 0.1 T making an angle of 53° with the wire. What magnetic force does the wire experience?    

ANSWER: Length, l =10 cm =0.1 m, 

Current, i =10 A, Magnetic field, B =0.1 T,  Angle between the wire and the magnetic field, θ =53°.  

  Hence, the force on the wire, 

F = ilBsinθ 

  =10*0.1*0.1*sin53°

  =0.08 N

  Its direction will be perpendicular to both the direction of the magnetic field and the wire.     

       

    8.  A current of 2A enters at the corner of a square frame abcd of side 20 cm and leaves the opposite corner b. A magnetic field B =0.1 T exists in the space in a direction perpendicular to the plane of the frame, as shown in Figure (34-E1). Find the magnitude and direction of the magnetic forces on the four sides of the frame.    

The figure for Q-8

ANSWER: It is clear from the figure that the current will be divided into equal parts and enter into branches dcb and dab, hence the current in each of these two parts will be 1 A.

   The magnitude of the magnetic force on each side =ilBsinθ 

 =1*0.20*0.1*sin90°

 =0.02 N

The current in dc and ab is towards the right, and the direction of the magnetic field is up on the plane of the paper; hence, the force on these two will be downward (from the right-hand rule)

   Similarly, the direction of the magnetic force on da and cb will be towards the left.  

  

  

    9.  A magnetic field of strength 1·0 T is produced by a strong electromagnet in a cylindrical region of radius 4·0 cm as shown in figure (34-E2). A wire carrying a current of 2·0 A is placed perpendicular to and intersecting the axis of the cylindrical region. Find the magnitude of the force acting on the wire.     

The figure for Q-9

ANSWER: The length of the wire in the magnetic field =2*4 cm =8 cm = 0.08 m

 B =1 T

 i =2.0 A, the angle between the wire and the magnetic field is 90°, hence the force on the wire,    

 F =ilB 

   =2.0*0.08*1 N

  =0.16 N

Perpendicular and towards the inside of the plane of the paper.      

  

    10.  A wire of length l carries current i along the X-axis. A magnetic field exists which is given as B =B₀(i +j +k) T. Find the magnitude of the magnetic force acting on the wire.     

ANSWER:  The magnetic field has equal magnitude along each axis, equal to B₀. The component alongthe  X-axis will not have an effect on the wire because the current is also in the same direction. The other two components make an angle of 90° with the wire, hence the force on the wire due to each of these components

F =ilB₀

The direction of force due to the field component along the Y-axis is along the Z-axis, and due to the field component alongthe  Z-axis is towards the negative Y-axis. So the two components of the force are perpendicular to each other. The magnitude of the resultant force 

=√{(ilB₀)²+(ilB₀)²}

=√2B₀il  

    11.  A current of 5·0 A exists in the circuit shown in the figure (34-E3). The wire PQ has a length of 50 cm and the magnetic field in which it is immersed has a magnitude of 0·20 T. Find the magnetic force acting on the wire PQ.   

The figure for Q-11

ANSWER: Given, current i =5·0 A, the length of the wire, l =50 cm =0·50 m, the strength of the magnetic field, B =0·20 T. 

   Since the directions of the current and the magnetic field are at right angles, the magnetic force on the wire PQ, 

   F =ilB  

     =5·0*0·50*0·20 N

     =0·50 N, towards the battery in the plane of paper/circuit (From the right-hand rule).         

    12.  A circular loop of radius a, carrying a current i, is placed in a two-dimensional magnetic field. The center of the loop coincides with the center of the field (figure 34-E4). The strength of the magnetic field at the periphery of the loop is B. Find the magnetic force on the wire.    

The figure for Q-12

ANSWER: The magnetic field B is perpendicular to the wire and hence current, everywhere. So the magnetic force on the wire,   

F = ilB 

   = i*2πa*B 

   =2πaiB  

 Its direction will be perpendicular to the plane of the loop and going into it.             

 

 

    13.  A hypothetical magnetic field existing in a region is given by B =Bₒeᵣ, where eᵣ denotes the unit vector along the radial direction. A circular loop of radius a, carrying a current i, is placed with its plane parallel to the X-Y plane and the center (0, 0, d). Find the magnitude of the magnetic force acting on the loop.    

ANSWER: Let us first draw a diagram to understand the situation as below:-

The diagram for Q-13

C is the center of the loop, and A is a point on the loop in the YZ plane (for example). The direction of the unit vector eᵣ is radial along OA. ∠COA =θ (say). We resolve the magnetic field B in two components along perpendicular directions, Bz perpendicular to the plane of the loop and By along the radius of the loop. The direction of the magnetic force on a very small length dl of the loop due to Bz will be towards the center of the loop, and hence the net force on the whole loop due to the components of the magnetic field along the Z-axis will be zero. The direction of the magnetic force on dl due to the component By will be parallel to the Z-axis. The total force on the loop due to By is

F =ilxBy  

F =|F| =ilBy 

   =i*(2πa)*Bₒ.sinθ  

   =2πaiBₒ*{a/√(a²+d²)}

   =2πa²iBₒ/√(a²+d²) 

along either of the Z-axis directions, depending upon whether the current in the loop is clockwise or anticlockwise.      

 

    14.  A rectangular wire loop of width 'a' is suspended from the insulated pan of a spring balance as shown in Figure (34-E5). A current 'i' exists in the anticlockwise direction in the loop. A magnetic field B exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.     

The figure for Q-14

ANSWER: The magnetic forces on the two vertical lengths of the rectangular loop will be horizontal and equal and opposite, so they will not contribute to the tension of the spring. The direction of the magnetic force on the lower width of the loop will be upwards and equal to =iaB. 

  When the current in the loop is reversed, the directions of the horizontal forces on the two lengths of the loop are also reversed, but they are still equal and opposite, canceling each other. Thus, they still have no effect on the tension of the spring. The direction of the magnetic force on the lower width is also reversed, and it is downwards with the same magnitude. So now force on it =-iaB. 

   Change in the spring tension 

=iaB-(-iaB) = 2iaB.        

 

    15.  A current loop of arbitrary shape lies in a uniform magnetic field B. Show that the net magnetic force acting on the loop is zero.   

ANSWER: Let us consider a very small length of the arbitrary loop at a point. The magnetic force dF =idlxB. Here, dl and B are vectors. The total force on the loop,

F =∫dF =∫idlxB = i*∫dlxB 

Since dl is a vector, its integration or summation between two points will be a vector joining those two points. Since dl is being integrated over a loop here, the two points here are the same. So the distance between these two points is zero, thus the magnitude of ∫dl =0. So the total magnetic force on the loop,

F = i*∫dlxB =0. 

 

  

 

 

    16.  Prove that the force on a current-carrying wire, joining two fixed points a and b in a uniform magnetic field, is independent of the shape of the wire.     

ANSWER: Consider an infinitesimally small section of the wire dL. In a uniform magnetic field B, the small force on the wire dL is,  

dF = i*dLxB 

The diagram for Q-16

The total force on the wire 

F = i*∫dLxB   ------------- (i)

dL is a vector, hence ∫dL is the sum of all vectors having magnitude dL. All these small vectors are arranged touching tip and toe between points a and b. Hence, their vector summation will be a vector having direction ab and magnitude of straight length ab.  Thus ∫dL = L. The total force is now from (i),

F = i*∫dLxB = i*LxB. 

So the magnetic force is always as if on a straight length of wire ab, whatever be the shape of the wire. 

   

 

    17.  A semicircular wire of radius 5·0 cm carries a current of 5·0 A. A magnetic field B of magnitude 0·50 T exists along the perpendicular to the plane of the wire. Find the magnitude of the magnetic force acting on the wire.    

ANSWER: The magnetic force on the semicircular wire will be equivalent to the force on a wire of the diameter of the loop. Hence, the magnitude of the magnetic force, 

F =iLB =i*(2r)*B 

  =5·0*(2*5·0/100)*0·5 N 

  =0·25 N.        

 

    18.  A wire carrying a current i, is kept in the X-Y plane along the curve y =A.sin(2πx/λ). A magnetic field B exists in the z-direction. Find the magnitude of the magnetic force on the portion of the wire between x =0 and x =λ.    

ANSWER: The magnetic force on a current-carrying curved wire length between points a and b is the same as the magnetic force on a straight wire length ab. For the curve length in the given problem, the straight length between two points x =0 and x = λ is equal to λ. Hence, the magnitude of the magnetic force on the given portion of the current-carrying wire,

F = i*λ*B {Since the direction of B is perpendicular to λ.} 

   =iλB.

 

 

    19.  A rigid wire consists of a semicircular portion of radius R and two straight sections (figure 34-E6). The wire is partially immersed in a perpendicular magnetic field B as shown in the figure. Find the magnetic force on the wire if it carries a current i.   

The figure for Q-19

ANSWER: Whatever be the shape of the current-carrying wire in a magnetic field, the magnitude of the magnetic force is always equal to the force on a straight length of the wire between endpoints. In the given figure, the straight length between endpoints, L =2R. The direction of the magnetic field is perpendicular to the plane of the wire, hence the magnitude of the magnetic force,

F =iLB

  =i*2R*B

  =2iRB.  

From the right-hand rule, the direction of this force is upward in the plane of the wire.    

 

    20.  A straight, horizontal wire of mass 10 mg and length 1·0 m carries a current of 2·0 A. What minimum magnetic field B should be applied in the region so that the magnetic force on the wire may balance its weight?   

ANSWER: The minimum required magnetic field B should be perpendicular to the length of the wire. In this case, the magnitude of the magnetic force on the wire will be, 

F =iLB.  

F should be equal to the weight of the wire to balance it =10 mg  

=1x10⁻⁵ kg force

=1x10⁻⁵*9·8 N

=9·8x10⁻⁵ N. 

i =2·0 A, L =1·0 m,  

Hence B =F/iL 

  =9·8x10⁻⁵/(2·0*1·0) T 

  =4·9x10⁻⁵ T.     

    21.  Figure (34-E7) shows a rod PQ of length 20·0 cm and mass 200g suspended through a fixed point O by two threads of length 20·0 cm each. A magnetic field of strength 0·500 T exists in the vicinity of the wire PQ as shown in the figure. The wires connecting PQ with the battery are loose and exert no force on PQ. (a) Find the tension in the threads when the switch S is open. (b) A current of 2·0 A is established when switch S is closed. Find the tension in the threads now.     

The figure for Q-21

ANSWER: (a) When the switch S is open, there is no current in the rod PQ. With no current, the threads have to balance the weight of the rod PQ.  

   Weight, W =mg

      = 0.20*9.8 N = 1.96 N  

Let the tension in each of the threads =T  

Balancing the forces in the vertical direction, 

2T .sin 60° =W

→  T = W/√3 

      =1.96/√3 N

      =1.13 N

(b) When the switch S is closed, the current in the rod PQ, i =2.0 A, 

Force due to the magnetic field, 

F = ilB,  

(because the current and the magnetic fields are perpendicular to each other).

→F  =2.0*0.20*0.500 N

  =0.20 N

From the right-hand rule, this force is downward. Hence, the total force downward =W+F

      = 1.96+0.20

      = 2.16 N

Hence, the tension in each thread 

 = 2.16/√3 =1.25 N   

    22.  Two metal strips, each of length l, are clamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m lies on them perpendicularly as shown in Figure (34-E8). A vertically upward magnetic field of strength B exists in the space. The metal strips are smooth but the coefficient of friction between the wire and the floor is µ. A current i is established when the switch S is closed at the instant t =0. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach?    

The figure for Q-22

ANSWER: The force on the wire will be towards the right and equal to,   F =ibB.  

The acceleration of the wire, 

a =ibB/m

Velocity, v, at the end of the strips, 

v² = u² +2as 

   = 0 +2ibBl/m

→ v =√(2iblB/m)

When the wire moves on the floor, the force of friction on it, F' =µmg

Retardation, a' =F'/m=µg 

Suppose it goes upto a distance =d, then from v² = u² +2as 

Here v =0, s =d, a = -a', and    

u = v  = √(2iblB/m) 

Hence, d  = u²/2a'

 →  d = 2iblB/2mµg 

        = ilbB/µmg 

 

 

    23.  A metal wire PQ of mass 10 g lies at rest on two horizontal metal rails separated by 4·90 cm (figure 34-E9). A vertically downward magnetic field of magnitude 0·800 T exists in the space. The resistance of the circuit is slowly decreased, and it is found that whenthe  resistance goes below 20·0 Ω, the wire PQ starts sliding on the rails. Find the coefficient of friction.    

The figure for Q-23

ANSWER: m = 10 g =0.010 kg 

Force of friction on the wire when trying to move F' =µmg =µ*0.010*9.8 =0.098µ 

  When the wire just starts to move, the magnetic force on the wire is equal to the force of friction. 

But the magnetic force,F =ilB

 F = (6/20)*(4.90/100)*0.800 N  

    =0.0118 N  

So, 0.098µ =0.0118

  → µ =0.12       

        

 

    24.  A straight wire of length l can slide on two parallel plastic rails kept in a horizontal plane with a separation d. The coefficient of friction between the wire and the rails is µ. If the wire carries current 'i', what minimum magnetic field should exist in the space in order to slide the wire on the rails?    

ANSWER: For the magnetic field, B, to be minimum, its direction should be perpendicular to the horizontal plane. 

   The magnitude of the magnetic field, 

F =ilB 

If the mass of the wire =m, its weight =mg. 

When the wire tends to slide, the force of friction, F' =µmg

For the wire to slide on the rails, F =F'

→ilB =µmg

→B = µmg/il     

 

    25.  Figure (34-E10) shows a circular wire loop of radius a, carrying a current i, placed in a perpendicular magnetic field B. (a) Consider a small part dl of the wire. Find the force on this part of the wire exerted by the magnetic field. (b) Find the force of compression in the wire.    

The figure for Q-25

ANSWER: (a) Since the wire is placed in a perpendicular magnetic field, the force on a small part dl of the wire carrying a current i will be equal to   

dF =i*dl*B =idlB 

From the figure, the magnetic field is going inside the plane of the loop, so from the right-hand rule, its direction will be towards the center.     

(b) Let the force of compression in the loop = T.

  Considering the section that divides the loop into two equal halves. On the two ends of a half, the force of compression = T at each end.  Suppose the part dl is situated at an angle θ from the horizontal axis as shown in the figure below.

Diagram for Q-25

     The dl length subtends an angle dθ at the center. Equating the forces along T,    

2T = ∫dF*cosθ

→2T = ∫idlB*cosθ   

→2T =iB∫cosθ*adθ  =iaB∫cosθdθ  

→2T =iaB[ sinθ]

{The limits of integration are from -π/2 to π/2}   

→2T =iaB [ sin π/2 -sin(-π/2)] =2iaB

→T = iaB   

 

    26.  Suppose the radius of the cross-section of the wire used in the previous problem is r. Find the increase in the radius of the loop if the magnetic field is switched off. The Young's modulus of the material of the wire is Y.    

ANSWER: The length of the wire =circumference of the loop =2πa,  

Area of the cross-section of the wire, A =πr²

  When compression is induced in the wire, the stress in the wire = T/πr²

Suppose the decrease in the length =dl, 

From the theory of elasticity,

Stress/strain = Y

(T/πr²)/(dl/2πa) = Y

dl =2aT/r²Y

Now the length of the wire

 =2πr -2aT/r²Y

The radius of the loop now       

=r -aT/πr²Y

The radius of the loop decreases by

aT/πr²Y. 

   So the radius will increase by the same length when compression vanishes due to the absence of the magnetic field. Putting the value of T in the expression, the increase in the radius of the loop    

=a*iaB/πr²Y

=ia²B/πr²Y      

  

 

  

    27.  The magnetic field existing in a region is given by 

B = Bₒ{1+x/l}k.  

A square loop of edge l and carrying a current i, is placed with its edges parallel to the X-Y axes. Find the magnitude of the net magnetic force experienced by the loop.   

ANSWER: Let us first draw a diagram. 

Diagram for Q-27

   Each side of the square loop is equal to l. Assume the current in the loop as shown.  

Force on the side AB,  

F₁ =ilB

  =ilB₀{1 +l/l} 

  =2ilB₀    (towards right)

Force on the side CO 

F₃ =ilB₀{1 +0/l} =ilB₀ (towards left)

Force on the side BC

F₂ =ilB₀{1 +(l/2)/l} =3ilB₀/2, upward.

(Since B varies linearly along the X-axis, here we have taken an average magnetic force at x =l/2)

Similarly, the force on the side OA

F₄ =ilB₀{1 +(l/2)/l} =3ilB₀/2, downward

 The forces along the Y-axis F₂ = F₄, are equal and opposite, so no net force along the Y-axis.

  Net force along X-axis =F₁ -F₃   

  =2ilB₀ -ilB₀  

  =ilB₀.         

 

    28.  A conducting wire of length l, lying normal to a magnetic field B, moves with a velocity v as shown in the figure (34-E11). (a) Find the average magnetic force on a free electron of the wire. (b) Due to the magnetic force, electrons concentrate at one end, resulting in an electric field inside the wire. The redistribution stops when the electric force on the free electrons balances the magnetic force. Find the electric field developed inside the wire when the redistribution stops. (c) What potential difference is developed between the ends of the wire?      

The figure for Q-28

ANSWER: (a) The magnetic force on a particle moving with a velocity v and having a charge q in a magnetic field of strength B is    

  F =qvB, when the directions of v and B are perpendicular to each other.  

   The free electrons in the wire will have an average velocity v in the direction of movement of the wire. Charge on an electron =e. Hence, the average magnetic force on a free electron in the wire = evB.     

  

(b) Suppose the electric field developed inside the wire = E.  

Electric force on an electron =eE.  

When the electric and magnetic fields balance each other inside the wire, we have,   

eE =evB 

→E =vB.  

(c) Let the potential difference developed between the ends of the wire = V. 

Since E =dV/dr =V/l, (for a uniform electric field)  

→V =lE = lvB.             

 

    29.  A current is passed through a silver strip of width d and area of cross-section A. The number of free electrons per unit volume is n. (a) Find the drift velocity v of the electrons. (b) If a magnetic field B exists in the region as shown in figure (34-E12), what is the average magnetic force on the free electrons? (c) Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor, which opposes the magnetic force on the electrons. Find the magnitude of the electric field that will stop further accumulation of electrons. (d) What will be the potential difference developed across the width of the conductor due to the electron accumulation? The appearance of the transverse emf, when a current-carrying wire is placed in a magnetic field, is called the Hall effect.  

The figure for Q-29

ANSWER: (a) Assume the length of the strip = l, Volume of the strip =Al. The total number of free electrons in the strip =Aln. 

  The charge of these free electrons 

 q =Alne. 

The current in the strip, i =q/t 

→i =Alne/t =Ane*(l/t)  

→i =Anev, (v is the drift velocity of the electron) 

→v =i/(Ane). 

(b) Average magnetic force on free electrons, F = evB 

→F =e*(i/Ane)*B

  =iB/An.            

From the right-hand rule, the direction of this force will be upward in the figure. 

(c) Let this balancing electric field inside the strip = E.  

The electric force on an electron =eE. 

For the given condition,

eE =F 

→eE =iB/An

→E =iB/Ane.  

(d) Let the potential difference developed across the width of the strip = V.   

Since E =dV/dr, here, E =V/d,  

→iB/Ane =V/d 

→V = iBd/Ane.         

  

    30.  A particle having a charge of 2·0x10⁻⁸ C and a mass of 2·0x10⁻¹⁰ g is projected with a speed of 2·0x10³ m/s in a region having a uniform magnetic field of 0·10 T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.    

ANSWER: For the particle given that,

Charge, q =2·0x10⁻⁸ C and mass, 

m = 2·0x10⁻¹⁰ g =2·0x10⁻¹³ kg, The perpendicular Magnetic field, B =0·10 T, velocity of the particle, v =2·0x10³ m/s.  

  Hence, the magnetic force on the particle F =qvB    

    Since this force will always be perpendicular to the direction of velocity, the particle will move in a circle with F as the centripetal force.  

Thus, F = mv²/r, where r is the radius of this circle. 

→r =mv²/F

   =mv²/qvB    

   =mv/qB

   =2x10⁻¹³*2x10³/(2x10⁻⁸*0·10) m 

  =2·0x10⁻¹ m 

  =0·20 m 

  =20 cm.  

   Let the time period be =T. In this time, the particle travels a distance equal to the circumference of the circle, d =2πr. Hence 

T =d/v =2πr/v 

   =2πmv/vqB 

   =2πm/qB

   =2*3.14*2x10⁻¹³/(2x10⁻⁸*0·10) 

   =6.3x10⁻⁴ s               

    31.  A proton describes a circle of radius 1 cm in a magnetic field of strength 0·10 T. What would be the radius of the circle described by an α-particle moving with the same speed in the same magnetic field?      

ANSWER: The magnetic force on a moving charged particle is perpendicular to its motion, so it provides the centripetal force to move the particle in a circle. If m =mass, v =speed, r =radius of the circle, B =strength of the magnetic field, and q =charge on the particle, then,

 qvB = mv²/r   

→r =mv/qB

It shows that for constant v and B, the radius of the circle is directly proportional to m and inversely proportional to q. An alpha particle has a mass four times that of a proton and a charge twice that of a proton. Hence, the radius of the circle described by an alpha particle,

  R =4mv/2qB =2*mv/qB =2r

Given r =1 cm, hence

R =2 cm.         

 

    32.  An electron having a kinetic energy of 100 eV circulates in a path of radius 10 cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.      

ANSWER: Kinetic energy, K =½mv²,   

→v² =2K/m

Since, B =mv/qr  

→B² =m²*2K/mq²r² =2mK/q²r²

  =2mK/e²r²

Putting values,

B²=2*9.11X10⁻³¹*100*1.60x10⁻¹⁹/(1.60x10⁻¹⁹)²*(0.1)²

 =11.39X10⁻⁸

→B =3.4X10⁻⁴ T    

   Time period of the electron,   

T =2πr/v, so

Frequency = 1/T =v/2πr

 =√(2K/m)/2πr   

 =√(K/2π²mr²)

=√{100*1.6X10⁻¹⁹/2*π²*9.11x10⁻³¹*(0.1)²}

= √{89X10¹²}

= 9.4x10⁶            

    33.  Protons having kinetic energy K emerge from an accelerator as a narrow beam. The beam is bent by a perpendicular magnetic field so that it just misses a plane target kept at a distance l in front of the accelerator. Find the magnetic field.      

ANSWER: The proton will move in a circle, and its radius is =l.   

We have seen in problem 32,  

B² =2mK/q²r²  

Here, q =e, r =l, and let the mass of a proton =mₒ, then 

B² =2mₒK/e²l²     

→B =√(2mₒK)el            

    34.  A charged particle is accelerated through a potential difference of 12 kV and acquires a speed of 1·0x10⁶ m/s. It is then injected perpendicularly into a magnetic field of strength 0·2 T. Find the radius of the circle described by it.       

ANSWER: Let the charge on the particle =q. In moving it through a potential difference of V, work done on it =qV. This work done is equal to the kinetic energy gained =½mv². So,  

qV =½mv²

→q =½mv²/V

We have the radius of the circle

r =mv/qB

 =mv*(2V/mv²)/B

 =2V/vB

Put values,

r =2*12000/(1x10⁶*0.2)

  = 12x10⁻² m

 = 12 cm.    

   

    35.  Doubly ionized helium ions are projected with a speed of 10 km/s in a direction perpendicular to a uniform magnetic field of magnitude 1·0 T. Find (a) the force acting on an ion. (b) the radius of the circle in which it circulates and (c) the time taken by an ion to complete the circle.      

ANSWER: (a) Given, q =2e, 

v =1.0X10⁴ m/s, B =1.0 T.

The force acting on an ion,

F =qvB

 =2*1.6X10⁻¹⁹*1.0X10⁴*1.0

 =3.2x10⁻¹⁵ N

(b) The radius of the circle in which it circulates,

r =mv/qB

The mass of a helium ion will be about four times the mass of a proton, so

r =4*1.67x10⁻²⁷*1.0x10⁴/(2*1.6x10⁻¹⁹)*1.0 m

  =2.1X10⁻⁴ m.  

(c) Time taken by an ion to complete the circle  

T =distance/speed

 =2πr/v

 =2*3.14*2.1x10⁻⁴/1.0x10⁴ s

 =1.31x10⁻⁷ s.                           

    36.  A proton is projected with a velocity of 3x10⁶ m/s perpendicular to a uniform magnetic field of 0·6 T. Find the acceleration of the proton.      

ANSWER: Given, v =3x10⁶ m/s, B =0.6 T. So the force on the proton perpendicular to its motion, 

 F = evB

   =1.6x10⁻¹⁹*3x10⁶*0.6 N

   =2.88x10⁻¹³ N

Acceleration of the proton in the direction of force,

a =F/m

 =2.88x10⁻¹³/1.67x10⁻²⁷ m/s²

 =1.72x10¹⁴ m/s²           

    37.  (a) An electron moves along a circle of radius 1 m in a perpendicular magnetic field of strength 0·50 T. What would be its speed? Is it reasonable?

    (b) If a proton moves along a circle of the same radius in the same magnetic field, what would be its speed?       

ANSWER: (a) Given, r = 1 m,

 B =0.50 T 

We have the radius of the circle

r =mv/qB  

→v =rqB/m

  =1*1.6x10⁻¹⁹*0.50/9.11x10⁻³¹

  =8.8x10¹⁰ m/s.

    It does not sound reasonable because the speed of light is 3x10⁸ m/s, and according to the theory of relativity, no matter should have a speed more than the speed of light.               

  

(b) The charge on a proton is the same as the electron but opposite in nature, and the mass is 1840 times more than the electron. So the mass of the proton,

m' =1840m  

The speed of a proton in the same condition,

v' =rqB/1840m 

   =v/1840

   =8.8x10¹⁰/1840 m/s   

   =4.8x10⁷ m/s.        

 

    38.  A particle of mass m and positive charge q, moving with a uniform velocity v, enters a magnetic field B as shown in figure (34-E13). (a) Find the radius of the circular arc it describes in the magnetic field. (b) Find the angle subtended by the arc at the center. (c) How long does the particle stay inside the magnetic field? (d) Solve the three parts of the above problem if the charge q on the particle is negative.      

Figure for Q-38

ANSWER: (a) The particle enters the magnetic field perpendicularly, hence the magnetic force at right angles to its movement on it.    

F =qvB,

This force acts as a centripetal force for the circular motion of the particle. If r is the radius of the circular arc, then,   

mv²/r =qvB

→r = mv/qB  

(b)  The direction of the velocity at the entrance will be tangent to the arc.

Diagram for Q-38b

 From geometry, the angle between the radius and the chord of the arc will also be equal to theta. In the isosceles triangle OAB, the angle subtended by the arc at the center  =π -2θ  

(c) Distance traveled by the particle inside the magnetic field = length of the arc, S =(π -2θ)*2πr/2π 

   =(π -2θ)r

The time taken by the particle in traveling this distance = S/v

  =(π-2θ)r/(qBr/m)

 =(π -2θ)m/qB

{ Since v =qBr/m}

(d) (i) If the charge on the particle is negative, the radius of the circular arc will still be the same, i.e.  

     

 r =mv/qB, only the movement will now be clockwise, opposite. 

(ii) As we see in the diagram, the angle between the chord and the radius is also θ. Hence, the angle subtended by the smaller arc (that is outside the magnetic field and imaginary) at the center =π -2θ

Diagram for Q-38c

   So the angle subtended at the center by the larger arc (the actual path traveled by the particle) is, 

 =2π -(π- 2θ) 

 =π +2θ

(iii) The distance traveled by the particle =the length of the arc inside the magnetic field, which is equal to

 {(π +2θ)/2π}*2πr

={π+2θ}r

Time taken to travel this distance

={π+2θ}r/v

=(π+2θ)m/qB

{Since r =mv/qB so r/v =m/qB}

 

                 

 

    39.  A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find the angle of deviation (figure 34-E14) of the particle as it comes out of the magnetic field if the width d of the region is very slightly smaller than

(a) mv/qB  (b) mv/2qB  (c) 2mv/qB. 

Figure for Q-39

ANSWER: (a) As we have seen, the radius of the arc inscribed by the particle inside the magnetic field is mv/qB. Since the width of the magnetic field, d is slightly less than mv/qB, the particle will inscribe only a quarter of a circle. In this case, the direction of the final speed will be perpendicular to the direction at entry.

Diagram for Q-39 (a)

   Hence, the angle of deviation when the particle comes out of the magnetic field =π/2. 

(b) When d =mv/2qB =r/2.   

Let the angle subtended by the arc traversed by the particle = θ.

  From geometry as in the figure below,

Diagram for Q-39 (b)

Sinθ =(r/2)/r =1/2 =Sin π/6

θ =π/6

The angle of deviation here is the angle between the tangents at the entry and exit of the particle. It will be equal to the angle subtended by the chord at the center = θ =π/6.

(c) Since the width d of the magnetic field is more than the radius of the described circle, the particle will not exit from the right face, instead it will describe a semicircle and come out from the entry face. Since the velocities at entry and exit are just opposite, the angle deviation will be equal to π.

     

 

    40.  A narrow beam of singly charged carbon ions, moving at a constant velocity of 6·0x10⁴ m/s, is sent perpendicularly in a rectangular region having a uniform magnetic field B = 0·5 T (figure 34-E15). It is found that two beams emerge from the field in the backward direction, the separation from the incident beam being 3.0 cm and 3.5 cm. Identify the isotopes present in the ion beam. Take the mass of an ion = A(1·6x10⁻²⁷) kg, where A is the mass number.      

The figure for Q-40

ANSWER: (a)  Given, B =0.50 T,

 v =6.0X10⁴ m/s,

 q =e =1.6X10⁻¹⁹ C,

for the first beam, r =3/2 cm =1.5 cm =0.015 m

Hence mass of each atom in the first emerging beam,

m = rqB/v 

   = reB/v

   =0.015*1.6X10⁻¹⁹*0.5/6x10⁴ kg

   =2.0x10⁻²⁶ kg

Dividing it by the mass of a proton,

m/mₒ =2.0x10⁻²⁶/1.67X10⁻²⁷ = 1.2*10

   =12

    So one atom of this carbon beam is twelve times heavier than a proton. A carbon atom has 6 protons, thus the other 6 particles are neutrons. It is a ¹²C isotope beam.   

(b) For the other emerging beam,

 r =3.5/2 cm =1.75 cm.   

m/mₒ =12*(1.75/1.5) = 14

   So this isotope has 6 protons and 8 neutrons in its nucleus; it is a ¹⁴C isotope beam.    

    41.  Fe⁺ ions are accelerated through a potential difference of 500 V and are injected normally into a homogeneous magnetic field B of strength 20·0 mT. Find the radius of the circular paths followed by the isotopes with mass numbers 57 and 58. Take the mass of an ion = A(1·6x10⁻²⁷) kg where A is the mass number.      

ANSWER: Let the mass of an ion of Fe =m, since it is singly charged, charge on it q =e. If the velocity of the ion after traveling a potential difference V =v, then the kinetic energy of the ion,

½mv² =qV =eV

→v² =2eV/m

→v =√(2eV/m)  

    Since the ions are injected normally into the uniform magnetic field, they will describe circular paths inside it. The radius of the circle will be given as,   

r =mv/qB =mv/eB   

→r =(m/eB)*√(2eV/m)  

  =√(2mV/eB²) 

Putting the values,  

r=√(2A*1.6x10⁻²⁷*500/1.6x10⁻¹⁹*0.02²)

=√(A*25x10⁻³)  

=0.5√(A/10) m

=50√(A/10) cm 

For the isotope 57, 

r =50√(57/10) cm =119 cm 

For the isotope 58,  

r =50√(58/10) cm =120 cm.  

 

   

    42.  A narrow beam of singly charged potassium ions of kinetic energy 32 keV is injected into a region of width 1.00 cm having a magnetic field strength of 0.500 T as shown in the figure (34-E16). The ions are collected at a screen 95.5 cm away from the field region. If the beam contains isotopes of atomic weights 39 and 41, find the separation between the points where these isotopes strike the screen. Take the mass of a potassium ion = A(1.6x10⁻²⁷) kg where A is the mass number.        

The figure for Q-42

ANSWER: The angle of deviation of the ions will be the same as the angle subtended by the circular arc (path) inside the magnetic region at the center of the circle. Let the angle of deviation =δ. 

Diagram for Q-42

From the figure, sin δ =d/r 

DH ≈d/2 =1/2 cm =0.50 cm  

DF =DH+HF =0.50 +95.50 cm 

  =96 cm.  

      After deflection, the ion strikes at point E on the screen. F is the point on the screen where the ion would have stricken without deflection. From triangle EFD, 

EF =DF*tan δ 

  = DF*sin δ/cos δ

  =DF*(d/r)/√{1-(d/r)²}

  =DF*d/√(r²-d²) ---------- (i)

We know DF = 96 cm, d =1·0 cm, q =e. We need to find r. We know,

Kinetic energy U =½mv² 

→v =√(2U/m)  

Also, r =mv/qB =mv/eB

  =(m/eB)√(2U/m)   

  =√(2Um/e²B²) 

=√{2*32x10³x1.6x10⁻¹⁹xA*1.6x10⁻²⁷/(1.6x10⁻¹⁹)²*0·5²}

=0.16√(A/10) 

For isotope of A.W. 39, 

r =0.16√(39/10) m =0.316 m 

  =31.6 cm  

For the isotope of A.W. 41,

r =0.16√(41/10) m =0.324 m  

  =32·4 cm.  

     So from (i), the distance EF for the heavier isotope 

=96*1/√(32.4²-1²) cm 

=2.96 cm 

  For the lighter isotope, the distance EF,

=96*1/√(31·6²-1²) 

=3.04 cm 

Hence the separation between the points where these isotopes strike the screen, 

=3·04 -2·96 cm 

=0·08 cm  

=0·80 mm                       

  

    43.  Figure (34-E17) shows a convex lens of focal length 12 cm lying in a uniform magnetic field B of magnitude 1.2 T parallel to its principal axis. A particle having a charge of 2.0x10⁻³ C and mass 2.0x10⁻⁵ kg is projected perpendicular to the plane of the diagram with a speed of 4.8 m/s. The particle moves along a circle with its center on the principal axis at a distance of 18 cm from the lens. Show that the image of the particle goes along a circle and find the radius of that circle.    

The figure for Q-43

ANSWER: The radius of the circle in which the particle moves, 

r =mv/qB  

 =2x10⁻⁵*4·8/(2x10⁻³*1·2) 

 =4/100 m = 4 cm  

 Given that the center of this circle lies on the principal axis. At any instant, let us draw the image of the radius joining the particle and the center (assuming it as an object).  

Diagram for Q-43

   Since the focal length of the convex lens =12 cm and the object is at 18 cm, the object is placed between focus and the center of curvature. The image will be inverted, beyond the center of curvature on the other side and it will be a real image. Since the particle distance from its center Q remains the same in the plane of the circle, the image distance P'Q' will also remain the same from the principal axis. Only the image will be on the other side of the principal axis. So the image will also move in a circle with its center Q'. 

  To find the radius of that circle we need to know the distance of the image radius OQ' =v from the lens. u = -18 cm, f =12 cm. From the lens formula, 

1/v -1/u =1/f   

→1/v -1/(-18) =1/12 

→1/v =1/12 -1/18 =1/36 

→v =36 cm.   

From the triangle PQO and P'Q'O,  

P'Q' =OQ'*PQ/OQ  

   =36*4/18 cm 

   =8 cm.      

          

 

    44.  Electrons emitted with negligible speed from an electron gun are accelerated through a potential difference V along the X-axis. 

The figure for Q-44

    These electrons emerge from a narrow hole into a uniform magnetic field B directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles as shown in the figure (34-E18). Show that these paraxial electrons are refocused on the X-axis at a distance 

√{(8π²mV)/(eB²)}.        

ANSWER: Since the direction of the magnetic field is along the X-axis, the velocity of electrons moving along the X-axis will be unaffected. We resolve the velocity of other electrons emerging with a divergent angle say δ from the X-axis into two perpendicular components - one along the X-axis v.cosδ and another perpendicular to the X-axis v.sinδ. The electrons with a velocity component v.sinδ perpendicular to the magnetic field will move along a circular path with a speed v.sinδ. So at any instant, the electron will have two speeds, one along the X-axis =v.cosδ another v.sinδ along a circular path. Ignoring the v.cosδ, let us first find the time taken by the electron to touch the X-axis after emerging from the gun and following a path of a circular arc. This arc will be a semicircle.  

   The length of this semicircle = 2πr. Time taken by the electron to travel this distance,  

t =2πr/v.sinδ,

  But r =mv/eB, 

and K.E. of an electron, eV =½mv² 

→v² =2eV/m, 

  so r =(m/eB)√(2eV/m) 

→r =√(2mV/eB²) 

Now time t =2π√(2mV/eB²)/v.sinδ

=2π√(2mV/eB²)/{sinδ*√(2eV/m)}

=(2π/sinδ)*√{2mV*m/(2eV*eB²)}

=2πm/eB.sinδ 

    In this time the distance traveled by the electron along the X-axis

d =v.cosδ*t

=v.cosδ*2πm/eB.sinδ

=√(2eV/m)*2πm/eBtanδ

=√(8π²mV/eB²)*δ/tan 

Since δ is a slightly divergent angle that means it is very very small. 

   Hence tanδ ≈δ, →δ/tanδ =1. Now,  

d =√{(8π²mV)/(eB²)}. 

     We note that the expression for d is free from δ, hence all the paraxial electrons will be refocussed at distance d from the hole.     

 

  

    45.  Two particles each having a mass m are placed at a separation d  in a uniform magnetic field B as shown in figure (34-E19). They have opposite charges of equal magnitude q. At time t =0, the particles are projected towards each other each with a speed v. Suppose the Coulomb force between the charges is switched off. (a) Find the maximum value vₘ of the projection speed so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if v =vₘ/2? (c) At what instant will a collision occur between the particles if v =2vₘ? (d) Suppose v = 2vₘ and the collision between the particles is completely inelastic. Describe the motion after the collision.       

The figure for Q-45

ANSWER: (a) Both the particles will deflect upward in the figure and move in circular paths. Just before the collision the radius of each circular path will be =d/2. See diagram below:-

Diagram for Q-45(a)

         We know r =mv/qB. Hence, 

d/2 =mvₘ/qB

→vₘ =qBd/2m.

(b) When v =vₘ/2 =qBd/4m, The radius of the circular paths will be,

r =mv/qB =(m/qB)*(qBd/4m)

→r =d/4.

    The minimum separation will be when both the particles have moved a quarter of the circular path and it will be equal to,

d -d/4 -d/4

=d -d/2

=d/2.

   The maximum separation will be when both the particles have moved three quarters of the circular path. The maximum sparation will be equal to,

d +d/4 +d/4

=d +d/2

=3d/2.

 

(c) When v =2vₘ =2*qBd/2m 

→v =qBd/m,

radius of the circular path, r =mv/qB 

→r =(m/qB)*(qBd/m) 

→r =d. 

The diagram is shown for
one particle only with charge q,
 other will have a similar motion

  The collision will occur when both the particles have moved a horizontal distance of d/2. In the figure, GP =d/2. Suppose at this instant the arc traveled by the particles make an angle δ at the center of the circular path. From geometry in the figure,  

sin δ =GP/PE =(d/2)/d =1/2 =sin π/6

So δ =π/6. 

Length of this arc =rδ =d*π/6 =πd/6.

Time spent in covering this arc length, 

t =distance/speed

  =(πd/6)/(qBd/m)  

  =πm/6qB.

(d) Since the motion is completely inelastic, both the particles will stick togather after collision. Just bfore this instant, each particle will have a horizonat component of velocity =v.cosδ but opposite in direction. Suppose after collision, the joint horizontal velocity =v'. Applying the conservation of momentum principle in horizontal direction,

mv.cosδ +m(-v.cosδ) =2m*v' 

→2mv' =0 

→v' =0. 

So after sticking togethe the joint particle has no velocity in the horizontal direction. Now suppose that the joint velocity in upward direcion (in the plane of figure) =v". Just before the collision, each particle has an upward velocity,

 = v.sin δ 

From the conservation of momentum in upward direction, 

2mv" =mv.sin δ +mv.sin δ

→v" =v.sin δ 

  =2vₘ*sin π/6

  =2vₘ*(1/2)

  =vₘ

      So after collision the joint mass will move upward in the plane of figure from the point of collision P in a straight line with a velocity vₘ.  

   

       

    46.  A uniform magnetic field of magnitude 0.20 T exists in space from east to west. With what speed should a particle of mass 0.010 g and having a charge 1.0x10⁻⁵ C be projected from south to north so that it moves with a uniform velocity?     

ANSWER: From the right hand rule, the magnetic force on the particle will be in the vertically upward direction. The particle will move with a uniform velocity only if the weight of the particle balances the magnetic force on it so that the resultant force on the particle is zero. 

   Let the required speed of the particle be v. Then,

qvB =mg

→v =mg/qB

  =(1.0x10⁻⁵)*9.8/(1.0x10⁻⁵)*0.2 m/s

  =49 m/s. 

 

 

    47.  A particle moves in a circle of diameter 1.0 cm under the action of a magnetic field of 0.40 T. An electric field of 200 V/m makes the path straight. Find the charge/mass ratio of the particle.     

ANSWER: The magnetic force on the particle F =qvB.  

  Since the radius of the circular path under the magnetic field is,

r =mv/qB  →v =qBr/m =qBd/2m,

{r =d/2} 

 So F =qB*qBd/2m

    =q²B²d/2m

   Since the given magnetic field makes the path straight, so the coulumb force is equal in magnitude but opposite in direction on the particle. The coulumb force,

F' =qE  

Here F =F'

→q²B²d/2m =qE

→q/m =2E/B²d

 =2*200/(0.40²*0.01)

 =2.5x10⁵ C/kg.  

It is the charge/mass ratio of the particle.

 

 

    48.  A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of 2.0x10⁵ m/s. The velocity is perpendicular to both fields. When the electric field is switched off, the proton moves along a circle of radius 4.0 cm. Find the magnitude of the electric and the magnetic fields. Take the mass of the proton =1.6x10⁻²⁷ kg.     

ANSWER: Given, v =2.0x10⁵ m/s,

 q =e =1.6x10⁻¹⁹ C, under magnetic field only, r =4.0 cm =0.04 m. Then, 

E =? and B =? 

  r =mv/qB =mv/eB   -------- (i)

→B =mv/er

  =1.6x10⁻²⁷*2x10⁵/(1.6x10⁻¹⁹*0.04) T

  =0·05 T.  

 

Magnetic force F =evB 

  Under the both fields in action, the path of the proton is a straight line. So both forces are equal and opposite. 

  Coulumb force, F' =eE

So, F' =F  

→eE =evB

→E =vB

  =2·0x10⁵*0.05 V/m 

  =1·0x10⁴ V/m.               

 

 

    49.  A particle having a charge of 5.0 µC and mass of 5.0x10⁻¹² kg is projected with a speed of 1.0 km/s in a magnetic field of magnitude 5.0 mT. The angle between the magnetic field and the velocity is sin⁻¹(0.90). Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.     

ANSWER: The angle between the magnetic field and the velocity,

 δ =sin⁻¹(0·90)

→sin δ =0·90.

  Since the velocity is not perpendicular to the magnetic field, it will have uniform velocity along the magnetic field 

=v.cos δ

     Due to the perpendicular component of the velocity v.sin δ, it will have an circular motion. But oweing to the two types of motion the circular path will not be closed and the actual path will be a helix.

     For the circular motion under magnetic field, 

 r =mv.sinδ/qB

→d =2mv/qB

=2*5·0x10⁻¹²*10³*0·90/(5·0x10⁻⁶*5x10⁻³) m

=0·36 m

=36 cm =Diameter of the helix.

To find out the pitch of the helix we first need to know the time period of one revolution. Distance traveled in one revolution =πd =0·36π m.

   Speed =v.sinδ =1000*0·9 m/s

    =900 m/s. 

   Time needed to cover one revolution

t =Distance/speed

=0·36π/900 s

=1·25x10⁻³ s

 The pitch of the helix will be the distanced moved along the magnetic field in this time t. Uniform component of the velocity along the magnetic field =v.cos δ =v√(1-sin²δ)

=1000*√{1 -(0·90)²} m/s

=436 m/s

Hence pitch =436*1·25x10⁻³ m

   =0·545 m

   ≈55 cm.

  

 

    50.  A proton projected in a magnetic field of 0.020 T travels along a helical path of radius 5.0 cm and pitch 20 cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton =1.6x10⁻²⁷ kg.      

ANSWER: Let the component of velocity along the magnetic field =v. Given that pitch of the helical path =20 cm =0·20 m. 

   Hence time taken to complete one revolution, t =0·20/v.  

   Radius of the helical path =5 cm 

=0·05 m 

  The distance covered in one revolution =2πr 

=2π*0·05 m =0·10π m. 

 So the speed of the particle in the circular motion, 

  v' =distance/time 

→v' =0·10π/(0·20/v) 

    =πv/2 m/s. 

Given B =0·020 T

For the circular motion of the proton under magnetic field, 

r =mv'/eB =πmv/2eB 

→v =2eBr/πm

=2*1.6x10⁻¹⁹*0·02*0·05/(π*1.6x10⁻²⁷)

=6·4x10⁴ m/s. 

As assumed in the beginning it is the velocity component along the magnetic field. 

  The velocity component perpendicular to the magnetic field, v' =πv/2 

→v' =π*6·4x10⁴/2

  =1·0x10⁵ m/s.    

    51.  A particle having mass m and charge q is released from the origin in a region in which electric field and magnetic field are given by  

B =-Bₒj and E =Eₒk

Find the speed of the particle as a function of its z-coordinate. 

ANSWER: Along the Z-axis, only the electric force is acting on the particle =qEₒ 

So the acceleration along the Z-axis,  

a =qEₒ/m 

So from the v² = u² +2as, the speed v of the particle after a distance z, 

v² =0² +2(qEₒ/m)z 

→v =√(2qEₒz/m).             

 

 

    52.  An electron is emitted with negligible speed from the negative plate of a parallel plate capacitor charged to a potential difference of V. The separation between the plates is d and a magnetic field B exists in the space as shown in figure (34-E20). Show that the electron will fail to strike the upper plate if

d > {2mₑV/eBₒ²}^½.   

The figure for Q-52

ANSWER: Electric field between the plates, E =V/d. 

  Coulomb Force on an electron =eE

  =eV/d

Acceleration due to this force = eV/mₑd

Speed after a distance x, from v² = u²+2as; 

v² =0 +2eVx/mₑd 

→v =√(2eVx/mₑd) 

Radius of the circle due to this speed, 

=mₑv/eB 

=(mₑ/eB)*√(2qVx/mₑd) 

=√(2mₑVx/edB²) 

The maximum distance it can travel towards the upper plate is the radius of this circle. For the x =d, 

Radius =√(2mₑVd/edB²) 

  =√(2mₑV/eB²) 

So if d >√(2mₑV/eB²), the electron will fail to strike the upper plate.         

    53.  A rectangular coil of 100 turns has a length of 5 cm and a width of 4 cm. It is placed with its plane parallel to a uniform magnetic field and a current of 2 A is sent through the coil. Find the magnitude of the magnetic field B, if the torque acting on the coil is 0·2 N-m.   

ANSWER: Area of the coil A =5*4 cm²

→A =20 cm² =0.002 m²,  

i =2 A, Number of turns n =100.

   The directions of the area vector and the magnetic field are mutually perpendicular. Hence, 

The torque =n*i*A*B 

  =100*2*0.002*B 

  =0.4B

But given this torque =0.2 N-m

So, 0.4B =0.2 

→B =0·5 T.  

  

    54.  A 50 turn circular coil of radius 2·0 cm carrying a current of 5·0 A is rotated in a magnetic field of strength 0·20 T. (a) What is the maximum torque that acts on the coil? (b) In a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?    

ANSWER: Number of turn, n =50, B=0·20 T, i =5·0 A, Radius r =2.0 cm =0.02 m, Hence torque Γ =niAB*sinδ, where δ is the angle between the area vector and the magnetic field.

(a) This torque will be maximum if the value of sinδ is maximum. Maximum value of sinδ will be 1 for δ=90°.

 So maximum torque on the coil, 

Γₘₐₓ=niAB

=50*5*{π*(0.02)²}0·2 N-m

≈0·063 N-m

=6·3x10⁻² N-m.     

(b) Now given that for a certain angle, 

Γ =½Γₘₐₓ 

so, niABsinδ =Γ 

→50*5*{π(0.02)²}*0.2*sinδ =½*0.063

→0.04π sinδ =0.063

→sinδ =0.5

→δ =30° 

It is the angle between area vector (perpendicular to the plane of the coil) and the direction of magnetic field. Hence the angle betwen the plane of coil and the magnetic field =90°-30° =60°.   

 

 

    55.  A rectangular loop of sides 20 cm and 10 cm carries a current of 5·0 A. A uniform magnetic field of magnitude 0·20 T exists parallel to the larger side of the loop. (a) What is the force acting on the loop? (b) What is the torque acting on the loop?   

ANSWER: (a) Since the magnetic field is parallel to the larger sides, there will be no force on the larger sides. The shorter sides will be perpendicular to the magnetic field hence the force on them will be perpendicular to the plane of the loop but opposite in direction because the direction of current is opposite in two shorter sides. 

  Since the direction of forces on the loop are equal in magnitude but opposite in direction, net force on the loop is zero.

(b) The forces acting on the loop are equal in magnitude and opposite in direction but their line of action is not the same. So these two forces make a couple and apply a torque on the loop. The direction of magnetic field and the area vector are perpendicular to each other here. Hence the torque acting on the loop =iAxB  

Its magnitude =iAB

   =5*(0.20*0.10)*0.2 N-m 

  =0·02 N-m.        

Its direction will be perpendicular to both A and B i.e. parallel to the shorter side. 

 

 

    56.  A circular coil of radius 2.0 cm has 500 turns in it and carries a current of 1.0 A. Its axis makes an angle of 30° with the uniform magnetic field of magnitude 0·40 T that exists in the space. Find the torque acting on the coil.   

ANSWER: Radius of the coil r =2·0 cm =0·02 m. 

Number of turns n =500,   

Current i =1·0 A,  

B =0·40 T 

δ =30° 

The torque acting on the coil,

=niAB.sinδ 

=500*1.0*π(0.02)²*0.40*sin30° 

≈0·13 N-m.           

 

 

    57.  A circular loop carrying a current 'i' has a wire of total length L. A uniform magnetic field B exists parallel to the plane of the loop. (a) Find the torque on the loop. (b) If the same length of wire is used to form a square loop, what would be the torque? Which is larger?    

ANSWER: (a) If the radius of the circular loop = r, then 

2πr =L

→r =L/2π

Area of the loop A =πr²

→A =π(L/2π)² =L²/4π

Hence the torque on the loop,

Γ =iAB

  =i(L²/4π)B 

  =iL²B/(4π).  

(b) For a squar loop of same wire length, the length of ane side =L/4. 

So area of the square =L²/16  

So torque now is 

Γ' = iAB

   = i(L²/16)B 

   =iL²B/16.    

 Let us find the ratio Γ/Γ'.

Γ/Γ' =16/4π =1.27

So the torque in the first case is nearly 1·27 times larger.     

  

    58.  A square coil of edge l having n turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists in a direction parallel to an edge. The total mass of the coil is M. What should be the minimum value of B for which the coil will start tipping over?   

ANSWER: The torque on the coil due to magnetic field Γ =niAB =nil²B. 

  The weight of the coil will act at the center of the coil and the torque due to the weight will be the restoring one. 

  This torque Γ' =Mg*l/2 =Mgl/2. 

Diagram for Q-58

  At the point of tipping,

Γ =Γ' 

nil²B =Mgl/2 

→B =Mg/2nil.        

  

  

    59.  Consider a nonconducting ring of radius r and mass m which has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speed ⍵. (a) Find the equivalent electric current in the ring. (b) Find the magnetic moment µ of the ring. (c) Show that µ =(q/2m)l where l is the angular momentum of the ring about its axis of rotation.   

ANSWER: (a) The charge on the ring passes at a point in one time period T. So the equivalent electric current in the ring i =q/T.

→i =q/(2π/⍵) =q⍵/2π.

(b) Magnetic moment of the ring,

µ =i*A

  =(q⍵/2π)*πr²

  =½q⍵r².

(c) Angular momentum of the ring about its axis of rotation,

l =I⍵, {Where moment of inertia,I=mr²}

 =mr²⍵ 

→⍵ =l/mr²

  Putting this in the expression for magnetic moment, 

µ =½q⍵r²

  =½q*(l/mr²)*r²

  =(q/2m)l. 

 

 

    60.  Consider a nonconducting plate of radius r and mass m which has a charge q distributed uniformly over it. The plate is rotated about its axis with an angular speed ⍵. Show that the magnetic moment µ and the angular momentum l of the plate are related as µ =(q/2m)l.    

ANSWER: Time period of rotation, 

T =2π/⍵

   Equivalent current in a ring on the plate dx width at a distance x from the center, 

 di =charge/time period of rotation

  ={(q/πr²)*2πxdx}/(2π/⍵)

  =(q⍵/πr²)*xdx

Magnetic moment of this ring in the circular plate =current*area 

dµ = (q⍵/πr²)xdx*πx²

  =(q⍵/r²)*x³dx  

  Magnetic moment of the plate,

µ =(q⍵/r²)∫x³dx

  =(q⍵/r²)[x⁴/4]

 Putting the limit of integration x =0 to x =r, we get

µ =(q⍵/r²)*r⁴/4

  =¼q⍵r²  

      Angular momentum of the plate,

l =I⍵

 =(½mr²)⍵

→⍵ =2l/mr²

Eliminating ⍵ from the expression of µ,

µ =¼qr²*(2l/mr²)

  =(q/2m)l.

      

    61.  Consider a solid sphere of radius r and mass m which has a charge q distributed uniformly over its volume. The sphere is rotated about a diameter with an angular speed ⍵. Show that the magnetic moment µ and the angular momentum l of the sphere are related as µ =(q/2m)l.  

ANSWER: Charge density in th sphere,

=q/(4πr³/3) =3q/4πr³. 

  Let us consider a very small surface inside the sphere at distance x from the center having an area =xdθ.dx

(See the diagram below)

Diagram for Q-61

When the sphere is rotated about the diameter XY, this small area rotates in a ring having radius x.sinθ and its center on the diameter XY with an angular speed ⍵.

Equivalent current in this ring,

di =(3q/4πr³)*2πx.sinθ*xdθdx/(2π/⍵)

 =(3q⍵/4πr³)x²sinθ.dθ.dx       

Magnetic moment of this ring, 

dµ =di*π(xsinθ)² 

 =(3q⍵/4πr³)x²sinθ*πx²sin²θdθdx

 =(3q⍵/4r³)x⁴sin³θdθdx

We double integrate it between limits x =0 to r and θ =0 to π.

µ =(3q⍵/4r³)∬x⁴sin³θdθdx

 =(3q⍵/4r³)(r⁵/5)∫sin³θdθ

 =(3q⍵r²/20)[¼(-3cosθ+⅓cos3θ)]

=(3q⍵r²/80)[3-⅓+3-⅓]

=(q⍵r²/5)

The angular momemtum of the sphere 

l =I⍵

→l = 2mr²⍵/5

→⍵ =5l/(2mr²)

Replacing ⍵ in the expression of µ,

µ =(qr²/5)*(5l/2mr²)

→µ =(q/2m)l. 

Hence proved.

Note: Integration of sin³xdx

Sin3x = 3sinx -4sin³x

→sin³x =¼(3sinx -sin3x)

→∫sin³xdx =¼∫(3sinx -sin3x)

  =¼[-3cosx +⅓cos3x]+c 

 

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CHAPTER-3 - Kinematics - Rest and Motion

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CHAPTER-7 - Circular Motion

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CHAPTER-8 - Work and Energy

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Exercises - Q 1 to Q 10

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CHAPTER- 9 - Center of Mass, Linear Momentum, Collision

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EXERCISES Q-1 TO Q-10

EXERCISES Q-11 TO Q-20

EXERCISES Q-21 TO Q-30

EXERCISES Q-31 TO Q-42

EXERCISES Q-43 TO Q-54

EXERCISES Q-55 TO Q-64

CHAPTER- 10 - Rotational Mechanics

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES Q-01 TO Q-15

EXERCISES Q-16 TO Q-30

EXERCISES Q-31 TO Q-45

EXERCISES Q-46 TO Q-60

EXERCISES Q-61 TO Q-75

EXERCISES Q-76 TO Q-86


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CHAPTER- 11 - Gravitation


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EXERCISES Q-01 TO Q-10

EXERCISES Q-11 TO Q-20

EXERCISES Q-21 TO Q-30

EXERCISES Q-31 TO Q-39 (With Extra 40th problem)

CHAPTER- 12 - Simple Harmonic Motion 


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OBJECTIVE - I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-58 with EXTRA QUESTIONS Q-59 and Q-60 


CHAPTER- 13 - Fluid Mechanics 


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OBJECTIVE-I

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EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-35



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CHAPTER- 14 - Some Mechanical Properties of Matter 


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EXERCISES - Q -11 TO Q -20

EXERCISES - Q -21 TO Q -32

CHAPTER- 15 - Wave Motion and Waves on a String

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OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-57

CHAPTER- 16 - Sound Waves

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OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-60

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EXERCISES - Q-71 TO Q-80

EXERCISES - Q-81 TO Q-89

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EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-41

CHAPTER- 18 - Geometrical Optics

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OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

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EXERCISES - Q-61 TO Q-70

EXERCISES - Q-71 TO Q-79

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EXERCISES - Q-1 TO Q-12

EXERCISES - Q-13 TO Q-24


CHAPTER- 20 - Dispersion and Spectra

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EXERCISES


CHAPTER- 21 - Speed of Light

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Part-II

Solutions - "Concepts of Physics" Part-II, by H C Verma

CHAPTER- 23 - Heat and Temperature

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OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-34


CHAPTER- 24 - Kinetic Theory of Gases

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OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 to Q10

EXERCISES - Q-11 to Q-20

EXERCISES - Q-21 to Q-30


EXERCISES - Q-31 to Q-40

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EXERCISES - Q-1 to Q-10

EXERCISES - Q11 to Q-18 




CHAPTER- 26 - Laws of Thermodynamics

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CHAPTER- 27 - Specific Heat Capacities of Gases

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EXERCISES - Q-1 to Q-10 

EXERCISES - Q-11 to Q-20

EXERCISES - Q-21 to Q-30

EXERCISES - Q-31 to Q-35

CHAPTER- 28 - Heat Transfer


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EXERCISES - Q-1 to Q-10

EXERCISES - Q-11 to Q-20

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CHAPTER- 29 - Electric Field and Potential

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EXERCISES - Q-1 to Q-10

EXERCISES - Q-11 to Q-20

EXERCISES - Q-21 to Q-30

EXERCISES - Q-31 to Q-40

EXERCISES - Q-41 to Q-50

EXERCISES - Q-51 tO Q-60

EXERCISES - Q-61 to Q-70

EXERCISES - Q-71 to Q-75

CHAPTER- 30 - Gauss's Law

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CHAPTER- 31 - Capacitors

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OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 to Q10

EXERCISES - Q11 to Q20

EXERCISES - Q21 to Q30

EXERCISES - Q31 to Q40

EXERCISES - Q41 to Q50

EXERCISES - Q51 to Q60

EXERCISES - Q61 to Q68

CHAPTER- 32 - Electric Current in Conductors

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OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q30

EXERCISES - Q31 - Q40

EXERCISES - Q41 - Q50

EXERCISES - Q51 - Q60

EXERCISES - Q61 - Q70

EXERCISES - Q71 - Q80

EXERCISES - Q81 - Q84

CHAPTER- 33 - Thermal and Chemical Effects of Electric Current

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OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q12

EXERCISES - Q13 - Q24

CHAPTER- 34 - Magnetic Field

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OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q30

EXERCISES - Q31 - Q40

EXERCISES -Q41 - Q50

EXERCISES -Q51 - Q61

CHAPTER- 35 - Magnetic Field Due to a Current

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OBJECTIVE - I

OBJECTIVE - II

EXERCISES -Q1 - Q10

EXERCISES -Q11 -Q20

EXERCISES -Q21 -Q30

EXERCISES -Q31 -Q40

EXERCISES -Q41 -Q50

EXERCISES -Q51 -Q61

CHAPTER- 36 - Permanent Magnet

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE -II

EXERCISES -Q1 -Q10

EXERCISES -Q11 -Q20

EXERCISES -Q21 -Q25

CHAPTER- 37 - Magnetic Properties of Matter

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES

CHAPTER- 38 - Electromagnetic Induction

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 -Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q30

EXERCISES - Q31 - Q40

EXERCISES - Q41 - Q50

EXERCISES - Q51 - Q60

EXERCISES - Q61 - Q70

EXERCISES - Q71 - Q80

EXERCISES - Q81 - Q90

EXERCISES - Q91 - Q98

 

CHAPTER- 39 - Alternating Current

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q19

CHAPTER- 40 - Electromagnetic Waves

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q9

CHAPTER- 41 - Electric Current Through Gases

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q23

CHAPTER- 42 - Photoelectric Effect and Wave-Particle Duality

Questions for Short Answer

OBJECTIVE -I

OBJECTIVE -II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q35

CHAPTER- 43 - BOHR'S MODEL AND PHYSICS OF THE ATOM

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q30

EXERCISES - Q31 -Q46

CHAPTER- 44 - X-rays

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q27

CHAPTER- 45 - Semiconductors and Semiconductor Devices

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE -II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q35

CHAPTER- 46 - The Nucleus

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES - Q11 - Q20

EXERCISES - Q21 - Q30

EXERCISES - Q31 - Q40

EXERCISES - Q41 - Q53

CHAPTER 47 - The Special Theory of Relativity

Questions for Short Answer 

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 - Q10

EXERCISES -Q11 - Q28

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