Monday, August 2, 2021

H C Verma solutions, MAGNETIC FIELD, Chapter-34, Exercises, Q1 -Q10, Concepts of Physics, Part-II

Magnetic Field


Exercises, Q1 to Q10


    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, 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 1.0x10⁻⁷ 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 wih a very high speed, let us assume that the direction of force remain 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

In this time the transverse distance traveled i.e. 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 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

  It's 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 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 inside 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 along X-axis will not have an effect on the wire because the current is also in the same direction. Other two components make an angle of 90° with the wire, hence the force on the wire due to each of these

F =ilB₀

The direction of force due to the field component along Y-axis is along Z-axis and due to the field component along Z-axis is towards 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 

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