H C Verma solutions, MAGNETIC FIELD DUE TO A CURRENT, Chapter-35, Concepts of Physics, Part-II

Magnetic Field Due to a Current

Questions for Short Answer

    1.  An electric current flows in a wire from north to south. What will be the direction of the magnetic field due to this wire at a point east of the wire? West of the wire? Vertically above the wire? Vertically below the wire?       

ANSWER: We stretch the thumb of the right hand along the current i.e. along the south. Now we curl our fingers to pass through the point where the direction of the magnetic field is required. The direction of fingers at that point gives the direction of the magnetic field. 

   According to this rule, we find that at a point east of the wire the direction of the magnetic field is upwards.  

   At a point west of the wire, the direction of the magnetic field is downward.  

      At a point vertically above the wire, the direction of the magnetic field is towards the west. 

      At a point vertically below the wire, the direction of the magnetic field is towards the east. 

    2.  The magnetic field due to a long straight wire has been derived in terms of µₒ, i, and d. Express this in terms of εₒ, c, i, and d.     

ANSWER: The magnetic field due to a long wire at a point is derived as, 

B =µₒi/(2πd) 

But µₒ = 1/εₒc²

Hence B =(1/2πεₒc²)i/d

 

    3.  You are facing a circular wire carrying an electric current. The current is clockwise as seen by you. Is the field at the center coming towards you or going away from you?   

ANSWER: When facing a circular wire carrying an electric current, if the current is clockwise, the field at the center is going away from the viewer. It can be verified by the right-hand thumb rule.     

    4.  In Ampere's law ∮B.dl =µₒi, the current outside the curve is not included on the right-hand side. Does it mean that the magnetic field B calculated by using Ampere's law gives the contribution of only the currents crossing the area bounded by the curve?    

ANSWER: Though the currents outside the curve are not included on the right-hand side in Ampere's law ∮B.dl =µₒi, the magnetic field B calculated by using this law gives the contribution of all the currents crossing or not crossing the area bounded by the curve. In fact, the left-hand side values for all the currents not crossing the area bounded by the curve are zero. 

    5.  The magnetic field inside a tightly wound, long solenoid is B = µₒni. It suggests that the field does not depend on the total length of the solenoid, and hence if we add more loops at the ends of a solenoid the field should not increase. Explain qualitatively why the extra added loops do not have a considerable effect on the field inside the solenoid.    

ANSWER: Let us explain it by a diagram as below. 

Diagram for Q-5

    Figure (a) shows a cross-section of a current-carrying loop. In the upper section of the loop, the current is going into the plane, hence the direction of the magnetic field will be clockwise around it. In the lower section, the current is coming out of it hence the magnetic field will be anticlockwise around it. At the midpoint between these two sections P, the direction of the magnetic field is the same and horizontal and hence add up. 

      Now consider a cross-section of a long solenoid in figure (b). Suppose in all the upper sections of the wire the current is going in, while in the lower sections current is coming out. Let us draw magnetic field lines around each wire that touches the axis of the solenoid. Consider the magnetic field at a point P on this axis. The maximum contribution of magnetic fields at this point is by only four sections of wire that are colored. Other loops of the solenoid at either end do not have any effect on the magnetic field at point P. So adding more loops at the ends does not increase the magnetic field inside the solenoid.        

    6.  A long straight wire carries a current. Is Ampere's law valid for a loop that does not enclose the wire? That encloses the wire but is not circular?   

ANSWER: Yes, Ampere's law is valid for a loop that does not enclose the wire. Since there is no current-carrying wire inside the loop, the right-hand side value will be zero. The left-hand side closed integral ∮B.dl will also be zero. 

   It is even valid when the loop enclosing the wire is not circular.    

    7.  A straight wire carrying an electric current is placed along the axis of a uniformly charged ring. Will there be a magnetic force on the wire if the ring starts rotating about the wire? If yes, in which direction?    

ANSWER: When the ring starts rotating, charges move in a circular motion and are equivalent to a loop, carrying current. Hence the magnetic field due to the ring at wire is along the wire. So the force on the wire, F =ilB.sinδ   

Here δ =0 or π, So F =0.  

    So there will be no magnetic force on the wire.     

    8.  Two wires carrying equal currents i each, are placed perpendicular to each other, just avoiding contact. If one wire is held fixed and the other is free to move under the magnetic forces, what kind of motion will result?   

ANSWER: The magnetic field around the fixed wire will be anticlockwise if the current is coming towards the viewer. So for the other wire, the magnetic field on it will be perpendicular to the plane containing both wires but it will have opposite directions on either side of the fixed wire. See the diagram below.

Blue circle shows magnetic field around fixed wire
Red arrows are magnetic forces on the moving wire

     From the right-hand rule, the magnetic force on the moving wire will be in the plane containing both wires and also perpendicular to it. Its direction will be opposite on either side of the fixed wire. So the free wire will have a circular motion. Just the moment it rotates further after overlapping, the direction of current is opposite to the present in the free wire. Thus the direction of forces on it reverses and it rotates in the opposite direction. So the movement of the free wire will be periodic forward and reverse circular motion. 

   

    9.  Two proton beams going in the same direction repel each other whereas two wires carrying currents in the same direction attract each other. Explain.   

ANSWER: There is a marked difference between the two cases. The proton beams are net moving charges. Though these moving charges produce magnetic fields around it that have an attractive effect the electrostatic repulsion is predominant and the proton beams repel each other. 

    In a current-carrying wire, there is no net charge on the wire. Free or lose electrons move under an electric field created by a battery inside the wire. So there is an absence of electrostatic field outside the wire and only magnetic fields are present due to which two wires carrying current in the same direction attract each other.      

    10.  In order to have a current in a long wire, it should be connected to a battery or some such device. Can we obtain the magnetic field due to a straight, long wire by using Ampere's law without mentioning this other part of the circuit?     

ANSWER: Yes, we can obtain the magnetic field due to a long, straight wire by using Ampere's law without mentioning the battery or other devices. We only need to know the current in the wire. We take amperes loop enclosing this wire only.   

    11.  Quite often, connecting wires carrying currents in opposite directions are twisted together in using electrical appliances. Explain how it avoids unwanted magnetic fields.    

ANSWER: Wires carrying currents in the opposite directions have magnetic fields in counterclockwise directions. So if the wire on the left has magnetic fields in an anticlockwise direction and on the right wire has in a clockwise direction, after half a twist the wires exchange positions, and hence the direction of magnetic fields are also exchanged. So in a long stretch, the effect of the net magnetic field is zero.   

    12.  Two current-carrying wires may attract each other. In absence of other forces, the wires will move towards each other increasing the kinetic energy. Does it contradict the fact that the magnetic force cannot do any work and hence cannot increase the kinetic energy?     

ANSWER: When the magnetic force is acting on a moving charge in a magnetic field, no work is done by the magnetic force because it is always perpendicular to the displacement. 

    For other instances, the magnetic force can either do work if stored potential energy is converted or a supply of electrical energy is done through the appliances, for example, electric motor. 

    In the given case also, when the wires move towards each other, the moving electrons in the wire have a horizontal component of velocity which results in a magnetic force on them that opposes the drift velocity of electrons and reduces the current. To maintain the attraction between wires (and the current), the external sources have to increase the potential difference across each wire. So magnetic forces can not increase the kinetic energy itself, external energy has to be supplied for the purpose.

OBJECTIVE - I

    1.  A vertical wire carries a current in the upward direction. An electron beam sent horizontally towards the wire will be deflected

(a) towards right

(b) towards left

(c) upwards

(d) downwards      

ANSWER: (c).    

Explanation:  The magnetic field due to the wire at the electron beam will be from left to right. For a positive charge going towards the wire, the force on it will be downwards from the right-hand rule. Since the nature of the charge is opposite (electrons) hence the direction of force and the movement will also be opposite. Thus the beam will be deflected upwards. Option (c) is correct.     

    2.  A current-carrying, straight wire is kept along the axis of a circular loop carrying a current. The straight wire

(a) will exert an inward force on the circular loop

(b) will exert an outward force on the circular loop

(c) will not exert any force on the circular loop

(d) will exert a force on the circular loop parallel to itself. 

ANSWER: (c). 

Explanation: The magnetic field due to the current in the loop will be along its axis. And the magnetic field due to the wire will be circular along or opposite to the current in the ring. The angle between the wire and the magnetic field, θ may be zero or 180°. The same is the case for the ring. The force on the wire due to this magnetic field or the force on the loop due to the wire = ilBsinθ =0. So the wire will not exert force on the loop. Option (c) is correct.

    3.  A proton beam is going from north to south and an electron beam is going from south to north. Neglecting the earth's magnetic field, the electron beam will be deflected 

(a) towards the proton beam

(b) away from the proton beam 

(c) upwards

(d) downwards.      

ANSWER: (a).   

Explanation: First, both beams have opposite electrostatic charges that will result in the attraction of beams. Second, both beams may be viewed as current in straight wires in the same direction that will also result in the attraction of both beams. So the electron beam will be deflected towards the proton beam. Option (a) is correct. 

    4.  A circular loop is kept in that vertical plane that contains the north-south direction. It carries a current that is towards the north at the topmost point. Let A be the point on the axis of the circle to the east of it and B a point on this axis to the west of it. The magnetic field due to the loop 

(a) is towards the east at A and towards the west at B 

(b) is towards the west at A and towards the east at B 

(c) is towards the east at both A and B

(d) is towards the west at both A and B.       

ANSWER: (d).  

Explanation: When the loop is viewed from point A the current is clockwise, so the magnetic field is going away from the viewer towards the loop. Similarly when viewed from point B the current in the loop is anticlockwise, so the magnetic field is coming towards the viewer from the loop. Thus the direction of the magnetic field on the axis of the loop is from A to B. This means the magnetic field at both A and B is towards the west. Option (d) is correct.      

    5.  Consider the situation shown in figure (35-Q1). The straight wire is fixed but the loop can move under magnetic force. The loop will 

(a) remain stationary

(b) move towards the wire 

(c) move away from the wire

(d) rotate about the wire.     

Figure for Q-5

ANSWER: (b).   

Explanation: Due to the magnetic field of the straight wire the direction of forces on the opposite sides of the loop will be opposite and in the plane of the loop. The magnitudes of the forces on horizontal sides will be equal, thus canceling each other. For the vertical sides, the magnitude of the force on the nearer side will be greater than the outer side. Since the direction of the force on the nearer side is towards the wire, the net force on the loop will be towards the wire. The loop will move towards the wire. Option (b) is correct.     

    6.  A charged particle is moved along a magnetic field line. The magnetic force on the particle is

(a) along its velocity

(b) opposite to its velocity

(c) perpendicular to its velocity

(d) zero.     

ANSWER: (d).   

Explanation: The force on a charged particle in a magnetic field is,  

F =q*vxB    

Here the direction of the magnetic field and the velocity is the same, hence the cross product of v and B will be zero. So the magnetic force on the particle will be zero. Option (d) is correct.     

    7.  A moving charge produces

(a) electric field only

(b) magnetic field only

(c) both of them

(d) none of them.   

ANSWER: (c).  

Explanation: A charge always produces an electric field whether static or moving. But a moving charge also produces a magnetic field. Hence a moving charge produces both types of fields. Option (c) is correct.    

    8.  A particle is projected in a plane perpendicular to a uniform magnetic field. The area bounded by the path described by the particle is proportional to

(a) the velocity

(b) the momentum 

(c) the kinetic energy

(d) none of these.   

ANSWER: (c).       

Explanation: Since the plane of projection is perpendicular to the magnetic field, the particle will describe a circular path under the centripetal magnetic force. The radius of this circle is given as, 

r = mv/qB

The area bounded by this path,  

A = πr² =π{mv/qB}² ={πm²/q²B²}v² 

    So for a given mass and charge of a particle, the area bounded by this path is proportional to the square of the velocity. But this option is not given. So let us see the kinetic energy, 

U =½mv²

→mv² =2U

Now, A =(πm/q²B²)*2U =KU

Where K is a constant for a given particle. So the area is proportional to the kinetic energy. Option (c) is correct.        

    9.  Two particles X and Y having equal charge, after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii R₁ and R₂ respectively. The ratio of the mass of X to that of Y is 

(a) (R₁/R₂)^½

(b) R₁/R₂

(c) (R₁/R₂)²

(d) R₁R₂. 

ANSWER: (c).   

Explanation: The radius of the circular path is given as,    

R =mv/qB

So R² =m²v²/q²B² ={2m/q²B²}*½mv² 

or, R² ={2m/q²B²}*qV =2mV/qB² 

or, m =qR²B²/2V  

Where V is the potential difference through which the particle is accelerated. KE = qV =½mv².      

Hence the mass of the first particle,

m₁ =qR₁²B²/2V

And the mass of the second particle,

m₂ =qR₂²B²/2V

So, m₁/m₂ =(R₁/R₂)².

Option (c) is correct.             

    10.  Two parallel wires carry currents of 20 A and 40 A in opposite directions. Another wire carrying a current antiparallel to 20 A is placed midway between the two wires. the magnetic force on it will be

(a) towards 20 A

(b) towards 40 A 

(c) zero

(d) perpendicular to the plane of the currents.     

ANSWER: (b)   

Explanation: It is clear that the current in the middle wire is in the same direction as the 40 A wire and opposite the 20 A wire. So the 20 A wire will repel the third wire and the 40 A wire will attract it under their respective magnetic forces. So the net magnetic force on the third wire in the middle will be towards 40 A wire. Option (b) is correct.    

    11.  Two parallel long wires carry currents i₁ and i₂ with i₁>i₂. When the currents are in the same direction, the magnetic field at a point midway between the wires is 10 µT. If the direction of i₂ is reversed, the field becomes 30 µT. The ratio i₁/i₂ is

(a) 4

(b) 3

(c) 2

(d) 1. 

ANSWER: (c)  

Explanation: Suppose the midpoint is at a distance d from each wire. Since the current-carrying wires are long, the field at the midpoint due to the first wire,  

B₁ =µₒi₁/2πd

and the field due to second wire,

B₂ =µₒi₂/2πd

     When the currents in wires are in the same direction the direction of the magnetic fields at the midpoint will be opposite to each other.

Diagram for Q-11

 So B₁ -B₂ =10 µT   --------- (i)

    When the currents are in opposite directions, the fields due to each wire at the midpoint will be in the same direction. So, 

   B₁ +B₂ =30 µT ------------ (ii)

Solving (i) and (ii) we get,

B₁ =20 µT

B₂ =10 µT.

Also B₁/B₂ = (µₒi₁/2πd)/(µₒi₂/2πd)

or, B₁/B₂ =i₁/i₂

So, i₁/i₂ =B₁/B₂ =20/10 =2.

Option (c) is correct. 

    12.  Consider a long straight wire of cross-sectional area A carrying a current i. Let there be n free electrons per unit volume. An observer places himself on a trolley moving in the direction opposite to the current with a speed v =i/(nAe) and separated from the wire by a distance r. The magnetic field seen by the observer is very nearly 

(a) µₒi/2πr

(b) zero

(c) µₒi/πr

(d) 2µₒi/πr.

ANSWER: (a) 

Explanation: Suppose the drift speed of electrons =v. Charge passing through a cross-section per unit time,

  = nAve 

Hence the current, i =nAev  

→v =i/nAe. 

It means that the observer is moving parallel to electrons in the long wire with the same speed. Since the frame of the observer is not accelerating, the magnetic field due to the current-carrying wire at a point will be seen as if stationary, it will not change. Thus the magnetic field at distance r from the wire 

= µₒi/2πr.  

Option (a) is correct.   

OBJECTIVE - II

    1.  The magnetic field at the origin due to a current element idl placed at a position r is  

(a) (µₒi/4π)(dlxr/r³)

(b) -(µₒi/4π)(rxdl/r³)

(c) (µₒi/4π)(rxdl/r³)

(d) -(µₒi/4π)(dlxr/r³)         

ANSWER: (c), (d).    

Explanation:  When the current element is at origin, the magnetic field due to this at a point having position vector r is  

dB =(µₒi/4π)(dlxr/r³)

  Here the current element is at the position vector r, the magnetic field at the origin will be 

  dB =(µₒi/4π)(rxdl/r³)  

  Option (c) is correct.

Also the cross-product, 

 rxdl = -dlxr

Hence also, dB =-(µₒi/4π)(dlxr/r³) 

So option (d) is also correct.                

    2.  Consider three quantities x=E/B, y =√(1/µₒεₒ) and z =l/CR. Here l is the length of a wire, C is a capacitance and R is resistance. All other symbols have standard meanings.  

(a) x, y have the same dimensions.

(b) y, z have the same dimensions.

(c) z, x have the same dimensions.

(d) None of the three pairs have the same dimensions.    

ANSWER: (a), (b), (c).    

Explanation:  x =E/B  

 qE has a dimension of force 

  qvB has also the dimension of force. 

Hence [qE] = [qvB] 

→[qE]/[qB] =[v] 

→[E/B] = [v]

→[x] = dimensions of velocity. 

Now, y =√(1/µₒεₒ) 

→y =√(εₒc²/εₒ),  Since µₒ=1/εₒc² 

→y =√c² =c =speed of light 

   =dimensions of velocity.

The third one is, z =l/CR 

 CR =(Q/V)R =QR/V =Q/(V/R) 

→CR =Q/I =Q/(Q/t) =t 

So z =l/t =v 

So z has dimensions of velocity.

Thus options (a), (b) and (c) are correct.                

    3.  A long straight wire carries a current along Z-axis. One can find two points in the X-Y plane such that

(a) the magnetic fields are equal  

(b) the directions of the magnetic fields are the same

(c) the magnitudes of the magnetic fields are equal

(d) the field at one point is opposite to that at the other point.        

ANSWER: (b), (c), (d).    

Explanation:  In the given situation the direction of the magnetic field at any point in the X-Y plane will be perpendicular to the line joining the point and Z-axis and in the X-Y plane. So the lines having equal magnitudes of the magnetic field will be concentric circles having centers at Z-axis in the X-Y plane. Option (c) is correct. The direction of the field at any point on such circles will be along the tangent at that point. So no two points in the X-Y plane will have the same magnitude and direction. Option (a) is not correct. 

  At any point on a radial line from the origin in the X-Y plane, the direction of the magnetic field will be the same. Option (b) is correct. 

   At the diametrically opposite points on the equal magnetic field circles, the fields are opposite. Option (d) is correct.        

    4.  A long straight wire of radius  R carries a current distributed uniformly over its cross-section. The magnitude of the magnetic field is 

(a) maximum at the axis of the wire

(b) minimum at the axis of the wire

(c) maximum at the surface of the wire

(d) minimum at the surface of the wire.      

ANSWER: (b), (c).    

Explanation:  The magnitude of the magnetic field, 

B =µₒi/2πd

So maximum magnitude of the magnetic field will be at the surface of the wire. Option (c) is correct. 

   At the axis, the direction of the magnetic fields due to the current around it will be equal in every direction, resulting in the net-zero magnitude of the field. Hence the option (b) is correct. 

     

    5.  A hollow tube is carrying an electric current along its length distributed uniformly over its surface. the magnetic field 

(a) increases linearly from the axis to the surface

(b) is constant inside the tube

(c) is zero at the axis

(d) is zero just outside the tube?     

ANSWER: (b), (c).    

Explanation:  Like the gravitational field inside a shell, the magnetic field inside a hollow tube is constant. And it will be zero at the axis. Hence the options (b) and (c) are correct. 

  

    6.  In a coaxial straight cable the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero  

(a) outside the cable

(b) inside the inner conductor

(c) inside the outer conductor

(d) in between the two conductors.        

ANSWER: (a).    

Explanation:  Consider a circle of radius r outside the cable as an Ampereian loop. Then from Ampere's law, 

∮B.dl =µₒi

→B∮dl =µₒ(i-i)

→B*2πr =0

→B =0.  

So outside the cable, the magnetic field is zero. Option (a) is correct. 

  Adjusting the value of r according to the three other options, we can see that the sum of the currents enclosed by the circle is not zero. Hence other three options are not correct.        

    7.  A steady electric current is flowing through a cylindrical conductor.

(a) The electric field at the axis of the conductor is zero.

(b) The magnetic field at the axis of the conductor is zero.

(c) The electric field in the vicinity of the conductor is zero. 

(d) The magnetic field in the vicinity of the conductor is zero.        

ANSWER: (b), (c).    

Explanation:  There is an electric field inside the conductor created by a battery or other source due to which the free electrons flow as current. Option (a) is not correct.  

   The magnetic field in the vicinity of the conductor =µₒi/2πd, not zero. Option (d) is incorrect. 

   The magnetic field at the axis due to all current elements will have equal magnitudes and direction all around it in a cross-section, thus canceling each other. The net magnetic field is zero. Option (b) is correct.  

   Since a current-carrying conductor is electrically neutral, the electric field in its vicinity is zero from the Gauss law. Option (c) is correct.    

EXERCISES

    1.  Using the formula F = qvxB and B =µₒi/2πr, show that the SI units of the magnetic field B and the permeability constant µₒ may be written a N/A-m and N/A² respectively.    

ANSWER: F =qvBsinθ 

Sinθ is a ratio, hence no unit. Now,

B = F/qv

   =N/C*(m/s)

   =N-s/(A-s)m

   =N/A-m

 Coulomb, C has been replaced by ampere-second A-s. 

B =μₒi/2πr

μₒ =2πrB/i

    = m(N/A-m)/A

   = N/A²

2π is unitless.   

    2.  A current of 10 A is established in a long wire along the positive Z-axis. Find the magnetic field B at the point (1 m, 0, 0).    

ANSWER: If the current is coming towards the viewer, the magnetic field around the wire is anticlockwise. So the direction of the magnetic field at the given point will be parallel to the positive Y-axis.   

Diagram for Q-2

   The magnitude of the magnetic field,

B =μₒi/2πr  

   = (4π*10⁻⁷)*10/(2π*1) T  

   =2x10⁻⁶ T 

   =2 µT, along the positive Y-axis.   

 

    3.  A copper wire of diameter 1.6 mm carries a current of 20 A. Find the maximum magnitude of the magnetic field B due to the current.    

ANSWER: B =µₒi/2πr 

Since B is inversely proportional to r, so B will increase as we go near the wire. It will be maximum at the surface because further reducing r we go inside the wire where the field begins to decrease. 

For maximum B, 

r =1.6/2 mm =0.8 mm =8x10⁻⁴ m 

So, B =(4π*10⁻⁷)*20/(2π*8x10⁻⁴) T 

    =5x10⁻³ T  

    =5 mT.  

 

    4.  A transmission wire carries a current of 100 A. What would be the magnetic field at a point on the road if the wire is 8 m above the road?    

ANSWER: Here, i =100 A, r =8 m. Hence,

The magnetic field at the road, 

B =µₒi/2πr  

  =(4π*10⁻⁷)*100/(2π*8) T

  = (200/8)*10⁻⁷ T 

  =25x10⁻⁷ T

  =2.5 µT.    

 

    5.  A long, straight wire carrying a current of 1.0 A is placed horizontally in a uniform magnetic field B =1.0x10⁻⁵ T pointing vertically upward (figure 35-E1). Find the magnitude of the resultant magnetic field at the points P and Q, both situated at a distance of 2.0 cm from the wire in the same horizontal plane.    

The figure for Q-5

ANSWER: i =1.0 A, r =2.0 cm =0.02 m

Hence the magnitude of the magnetic field, 

B' =µₒi/2πr 

  =(4π*10⁻⁷)*1.0/(2π*0.02) T 

  =1.0x10⁻⁵ T. 

   If the current is coming towards the viewer, the magnetic field around the wire is anticlockwise. So the direction of B' at point P is vertically upward while it is vertically downward at point Q. 

  So resultant magnetic field at P =B+B'

 =1.0x10⁻⁵ +1.0x10⁻⁵ T

 =2.0x10⁻⁵ T 

 =20x10⁻⁶ T

 =20 µT. 

The resultant magnetic field at point Q, 

=B -B' 

=1.0x10⁻⁵ -1.0x10⁻⁵ T

=zero.        

 

    6.  A long, straight wire of radius r carries a current i and is placed horizontally in a uniform magnetic field B pointing vertically upward. The current is uniformly distributed over its cross-section. (a) At what points will the resultant magnetic field have maximum magnitude? (b) What will be the minimum magnitude of the resultant magnetic field?     

ANSWER: (a) The maximum magnitude of the magnetic field due to the current in the wire will be near its surface and it will be, B' =µₒi/2πr where r is the radius of the wire. The resultant magnetic field will be maximum at points where the directions of the existing uniform magnetic field and the magnetic field due to the current are the same. If we see along the current, such points will be at the leftmost points of the wire and the resultant magnetic field will be B+B' 

 =B +µₒi/2πr. 

(b) Minimum magnitude of the resultant magnetic field will be at points where the directions of B and B' are opposite i.e.

=|B-B'| 

For the condition that the resultant field is zero, B =B'   

→B =µₒi/2πd

→d =µₒi/2πB

If the radius of the wire is equal to d, r =d, then the resultant magnetic field will be zero at the rightmost points on the surface of the wire while looking along the current. If r<d then the resultant is zero (minimum numerical value) at distance d from the wire. 

  If r>d, then the points at distance d will be inside the wire but we have to get points outside the wire. At the surface, the resultant magnetic field will be 

=B-µₒi/2πr and it will be the minimum magnitude because as we go farther from the wire the subtracting quantity µₒi/2πr will get lesser and the resultant magnitude B-µₒi/2πd will increase.           

   

    7.  A long, straight wire carrying a current of 30 A is placed in an external, uniform magnetic field of 4.0x10⁻⁴ T parallel to the current. Find the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire. 

ANSWER: External magnetic field B =4.0x10⁻⁴ T. Current in the wire, i =30 A, distance of the point d =2.0 cm =0.02 m.

  The magnetic field due to the current, B' =µₒi/2πd 

  =(4πx10⁻⁷)*30/(2π*0.02)  

  =3.0x10⁻⁴ T, perpendicular to the current or B. 

Hence the resultant magnitude of the magnetic field =√(B²+B'²)

  =√{(4.0x10⁻⁴)²+(3.0x10⁻⁴)²} 

  =√(5x10⁻⁴)² 

  =5.0x10⁻⁴ T.      

 

    8.  A long vertical wire carrying a current of 10 A in the upward direction is placed in a region where a horizontal magnetic field of magnitude 2.0x10⁻³ T exists from south to north. Find the point where the resultant magnetic field is zero.   

ANSWER: External magnetic field, B=2.0x10⁻³ T, from south to north. To the west of the wire, the direction of the magnetic field due to the current B' will be from north to south. At the point to the west of wire where the magnitude of B' =B, the resultant magnetic field will be zero. Let such point is at distance d from the wire, then  

µₒi/2πd =B 

→d =µₒi/2πB 

  =(4πx10⁻⁷)*10/(2π*2x10⁻³) 

  =1.0x10⁻³ m 

  =1.0 mm, west to the wire.      

 

    9.  Figure (35-E2) shows two parallel wires separated by a distance of 4.0 cm and carrying equal currents of 10 A along opposite directions. Find the magnitude of the magnetic field B at the points A₁, A₂, A₃, A₄.    

The figure for Q-9

ANSWER: Point A₁ 

Distance of the point from the left wire, d =2 cm =0.02 m 

Magnetic field due to left wire,

B =µ₀i/2πd 

  =(4πx10⁻⁷)*10/(2π*0.02) T

  =1.0x10⁻⁴ T, upwards 

Distance of the point from the right wire, d =6 cm =0.06 m 

Magnetic field due to the right wire, 

B' =(4πx10⁻⁷)*10/(2π*0.06)  

 =3.33x10⁻⁵ T, downwards.

Hence the net magnetic field =B-B' 

 =1.0x10⁻⁴ -3.33x10⁻⁵ T 

 =6.67x10⁻⁵ T or 0.67x10⁻⁴ T.  

Point A₂

Here the directions of magnetic fields due to both wires are the same hence they would add up. Distance from the left wire, d =1 cm =0.01 m and distance from the right wire, d' =3 cm =0.03 m. So the net magnetic field at this point.

=µₒi/2πd +µₒi/2πd'

=(µₒi/2π)(1/d +1/d') 

=(4πx10⁻⁷*10/2π)(1/0.01+1/0.03)

=20x10⁻⁷(133.33) T

=2.67x10⁻⁴ T  

Point A₃

This point is the midpoint on the line joining both wires and the directions and the magnitudes of the magnetic fields due to each wire is the same hence they add up. Net magnetic field 

=2µₒi/2πd 

=(4πx10⁻⁷*10)/(π*0.02) T 

=2.0x10⁻⁴ T. 

Point A₄

This point is equidistant from each wire hence the magnitude of the magnetic field due to each wire is the same, say B. The direction of the field due to each wire will be perpendicular to the line joining the wire to the point as shown below:- 

diagram for Q-9

  Clearly the angle between both fields =90°. Net magnetic field, 

B' =√2B 

  =√2µₒi/2πd  

d =√2*2 cm =0.02√2 m 

So, B' =√2µₒi/(2π*0.02√2)

  =µₒi/(0.04π)

 =(4πx10⁻⁷*10)/(0.04π) T 

 =1.0x10⁻⁴ T.                    

  

    10.  Two parallel wires carry equal currents of 10 A along the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point that is 2.0 cm away from each of these wires.    

ANSWER: Suppose A and D two parallel wires separated by 2.0 cm carry equal current i =10 A coming towards the viewer. The magnetic field around each wire will be anticlockwise. Since the given point is equidistant from the wires, the magnitude of the magnetic field due to each wire will be equal, say B. The direction of each field will be perpendicular to the line joining the point and the wire in a plane perpendicular to the wires containing the line joining. See the diagram below:- 

Diagram for Q-10

 The given point and the wires make an equilateral triangle in a plane that is perpendicular to the wires, hence each angle is =60°. Angle DCF =90°, so angle ACF =30°. Angle ACE =90°, so angle ECF =60°. Since both fields B are equal in magnitude, their resultant CG will bisect the angle ECF, so angle FCG =30°. Now the angle ACG =30°+30° =60°. Since alternate angles ACG and CAD are equal, the direction of resultant CG is parallel to the line joining the wires AD. The magnitude of the resultant, 

B' =√{B²+B²+2B²cos60°} 

 =√{2B²(1+0.5)}

 =√{3B²}

 =B√3

 =(µₒi/2πd)*√3

 =(4πx10⁻⁷*10√3/(2π*0.02) T

 =1.7x10⁻⁴ T.    

    11.  Two long, straight wires, each carrying a current of 5 A, are placed along the X - and Y - axes respectively. The currents point along the positive directions of the axes. Find the magnetic fields at the points (a) (1 m, 1 m), (b) (-1 m, 1 m), (c) (-1 m, -1 m) and (d) (1 m, -1 m).   

ANSWER: Given, i = 5 A. If the current is coming towards the viewer, the magnetic field direction around the wire is anticlockwise.  

Figure for Q-11

 From this rule, the direction of the magnetic field at A due to the wire along the X-axis is perpendicular to the paper and coming out. The magnitude can be found by putting i =5 A and d =1 m in B =µ₀i/2πd. The magnetic field at A due to the wire along Y-axis is perpendicular to the paper and going in with the same magnitude. So both the fields are equal in magnitude but opposite in direction, thus canceling out. The resultant magnetic field is zero at A.  

   At point B, the distance of the point from the wires and the current in the wires is the same and hence magnitudes are equal. From the rule, the directions of the magnetic fields at B are the same due to each wire and hence they will add up. So, the net magnetic field at point B is,

=2*µ₀i/2πd 

=µ₀i/πd 

=(4πx10⁻⁷)*5/π*1 T

=2.0x10⁻⁶ T  

=2.0 µT, perpendicularly coming out of the XY plane at point B.         

   At point C, the magnetic fields due to each wire are equal in magnitude but opposite in direction, hence the resultant magnetic field at point C (-1 m, -1 m) is zero. 

   At point D, the magnitude of magnetic fields due to each wire is equal and also in the same direction (going into the plane). The resultant magnetic field at D will be, 

=2µₒi/2πd =µₒi/πd 

=(4πx10⁻⁷)*5/π*1 T 

=2.0 µT, perpendicularly going into the XY plane at point D.     

  

    12.  Four long, straight wires, each carrying a current of 5.0 A, are placed in a plane as shown in figure (35-E3). The points of intersection form a square of side 5.0 cm. (a) Find the magnetic field at the center P of the square. (b) Q₁, Q₂, Q₃ and Q₄ are points situated on the diagonals of the square and at a distance from P that is equal to the length of the diagonal of the square. find the magnetic field at these points.   

Figure for Q-12

ANSWER: (a) Distance of center P from each wire is the same and the magnitude of current in each wire is also the same, hence the magnitude of the magnetic field due to each wire is the same. As we see the magnetic fields due to each set of the parallel wire at P is equal but opposite in direction, thus the resultant field at center P is zero. 

     (b) The distance of these points from the nearer wires is equal to r =2.5 cm and from the farther wires is r' =7.5 cm (From geometry). 

   Point Q₁: From the rule, the direction of magnetic fields due to each wire is the same here hence all will add up. The resultant field, 

B =2*µₒi/2πr +2*µₒi/2πr'

 =(µₒi/π)*(1/r +1/r') 

 =(4πx10⁻⁷*5/π)(1/2.5 +1/7.5)*100 T

 =(2x10⁻⁶)(40+13.3) T

 ≈1.1x10⁻⁴ T, coming out of the plane of the figure perpendicularly. 

     Point Q₂: The magnetic field at this point by each set of parallel wires is equal in magnitude but opposite in direction. Hence the net magnetic field is zero. 

   Point Q₃: The direction of magnetic fields at this point due to a set of parallel wires is the same and magnitude equal. Hence they will add up. The location of the point is similar to the point Q₁ hence the same resultant magnitude 1.1x10⁻⁴ T but the direction is perpendicularly going in the plane of the figure.    

   Point Q₄: Its position is similar to point Q₂. The direction of the magnetic field due to the horizontal parallel set of wires is going into the plane of the figure while it is just opposite due to the vertical parallel set of wires. The magnitudes being equal. Hence the net magnetic field is zero. 

  

    13.  Figure (35-E4) shows along wire bent at the middle to form a right angle. Show that the magnitudes of the magnetic fields at the points P, Q, R and S are equal and find this magnitude.  

Figure for Q-13

ANSWER: The magnitude of the magnetic field at the point P due to the vertical wire is zero because it lies inside and on the axis of the wire. Due to the semi-infinite horizontal wire the magnitude of the magnetic field =µₒi/4πd. 

   Hence the net magnitude of the magnetic field at point P =µₒi/4πd. 

     The same is the case with point R that lies on the axis of the vertical wire and the field is only due to the horizontal wire. So the net magnitude of the magnetic field at point R is only due to the horizontal semi-infinite wire =µₒi/4πd. 

   Now point Q and S lie on the axis of the horizontal wire hence the magnetic fields due to this wire at these points are zero. The magnetic fields at these points are only due to the semi-infinite vertical wire. Distance of the points from the vertical wire =d. Hence net magnetic field at each of these points =µₒi/4πd.           

 

    14.  Consider a straight piece of length x of a wire carrying a current i. Let P be a point on the perpendicular bisector of the piece, situated at a distance d from its middle point. Show that for d>>x, the magnetic field at P varies as 1/d² whereas for d<<x, it varies as 1/d.   

ANSWER: The magnitude of the magnetic field at a point on the perpendicular bisector of a current-carrying wire of length x is, 

  B =µₒix/{2πd√(x²+4d²)}

When d>>x, then x² is negligible in compaision to d², hence for the expression x²+4d² we can write only 4d². so now,   

B =µₒix/{2πd√(4d²)} 

  =µₒix/{2πd*2d} 

  =µₒix/{4πd²} 

→B ∝ 1/d². {except d, rest are constant at the given time}

For d<<x, we can ignore 4d² in comparision to x². Hence the magnetic field now,

  B =µₒix/{2πd(√x²)}

    =µₒix/(2πdx)

    =µₒi/(2πd)

Hence, B ∝ 1/d.

 

 

    15.  Consider a 10 cm long piece of a wire which carries a current of 10 A. Find the magnitude of the magnetic field due to the piece at a point that makes an equilateral triangle with ends of the piece.  

ANSWER: Clearly the point lies on the perpendicular bisector of the length of the wire. Here the length of the wire 'a' =10 cm =0.10 m, current in the wire i =10 A and the distance of the point from the wire d =a*sin60° =√3a/2. See diagram below.     

Diagram for Q-15

    Now the magnetic field at the given point, 

B =µₒia/{2πd√(a²+4d²)} 

 =µₒia/{2πd√(a²+4*3a²/4)

 =µₒia/{2πd√(4a²)}

 =µₒia/(4πad)

 =µₒi/(4πd)

 =4π*10⁻⁷*10/(4π*√3*0.10/2) T

 =2x10⁻⁵/√3 T 

 =1.15x10⁻⁵ T

 =11.5x10⁻⁶ T

 =11.5 µT.    

 

    16.  A long, straight wire carries a current i. Let B₁ be the magnetic field at a point P at a distance d from the wire. Consider a section of length l of this such that the point P lies on a perpendicular bisector of the section. Let B₂ be the magnetic field at this point due to this section only. Find the value of d/l so that B₂ differs from B₁ by 1%.   

ANSWER: B₁ =µₒi/2πd. The magnetic field due to length l of the wire at distance d on its perpendicular bisector is,  

B₂ =µₒil/{2πd√(l²+4d²)} 

For the given condition,        

B₂ =0.99*B₁

→B₂/B₁ =0.99

→l/√(l²+4d²) =0.99 

→l²/(l²+4d²) =0.99²

→(l²+4d²)/l² =1/0.99²

→1 +4(d/l)² =1/0.99²

→(d/l)² =¼{1/0.99² -1} =¼(0.02)

→d/l =√{0.005}

     =0.07 

 

 

 

 

 

    17.  Figure (35-E5) shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. Find the magnetic field B at the center of the loop assuming uniform wires. 

Figure for Q-17

ANSWER: Distance of the center of the loop from each side of the square is, 

d =a/2.  

      Potential difference between points A and C will be the same for each wire ABC and ADC. Let current in ABC =i₁ and in ADC =i₂. Hence, 

  i₁*r =i₂*2r 

→i₁ =2i₂. 

But i =i₁ +i₂ =3i₂ 

→i₂ =i/3,

and i₁ =2i/3 

   The magnetic field due to AB and BC at the center will be the same in magnitude and direction. So total magnetic field at center due to AB and BC, 

B'=2*µₒi₁a/{2πd√(a²+4d²)} 

(Since the center is on the perpendicular bisector of each side) 

=µₒ(2i/3)a/{π(a/2)√(a²+a²)}

=4µₒi/{3πa√2} 

=2√2µₒi/(3πa). Going into the paper.

    Similarly, the magnetic fields due to the sides AD and DC at the center will be equal in magnitude and direction. So the magnitude of the magnetic field due to AD and DC at the center is,

B"=2*µₒi₂a/{2πd√(a²+4d²)}

=µₒ(i/3)a/{π(a/2)√(a²+a²)}

=2µₒi/(3πa√2)  

=√2µₒi/(3πa). Coming out of the paper. 

  So, the net magnetic field due to the loop will be =B' -B" 

=2√2µₒi/(3πa) -√2µₒi/(3πa)  

=√2µₒi/(3πa), going into the paper.   

              

 

    18.  Figure (35-E5) shows a square loop of edge 'a' made of a uniform wire. A current i enters the loop at point A and leaves it at point C. Find the magnetic field at point P which is on the perpendicular bisector of AB at a distance a/4 from it.  

Figure for Q-18

ANSWER: It is clear from the figure that the current on each side of the loop is i/2. Given point, P is at the middle (equidistant) of two parallel current-carrying equal wire segments AD and BC. Since equal currents flow in these two parallel wires hence the magnetic fields due to each at P is equal in magnitude but opposite in directions. Thus net magnetic field at P due to AD and BC is zero. 

   Due to wire AB, the magnetic field at P is,  

B =µₒ(i/2)a/{2π(a/4)√(a²+4a²/16)} 

 =µₒi/{πa√5/2}

=2µₒi/(√5πa), coming out of the paper.

   The magnetic field due to the side DC at point P is, 

B' =µₒ(i/2)a/{2π(3a/4)√(a²+4*9a²/16)}

=µₒi/{3πa√13/2}

=2µₒi/(3√13πa), going into the paper.

  Hence the net magnetic field at point P,

=B -B'

=2µₒi/(√5πa) -2µₒi/(3√13πa)

=(2µₒi/πa)*{1/√5 -1/(3√13)},

Coming out of the paper.       

 

    19.  Consider the situation described in the previous problem. Suppose the current i enters the loop at point A and leaves it at point B. Find the magnetic field at the center of the loop.  

ANSWER: Let the resistance of each side of the wire loop =r. Then the resistance of wire ADCB =3r and that of AB =r. Let current in ADCB =i₁ and in AB =i₂. Equating the potential difference between A and B, we get  

   i₁*3r =i₂*r 

→i₁ =i₂/3.

But i =i₁ +i₂ =i₂/3 +i₂

→i =4i₂/3

→i₂ =¾i 

and i₁ =¼i.

   The center point of the loop is equidistant from each side, d =a/2 and also at the perpendicular bisectors. The magnitudes and the directions of the magnetic fields due to each of the side AD, DC and CB is at the center is the same. Hence total magnetic field at the center due to these three sides, 

B =3*µₒ(¼i)a/{2π(a/2)√(a²+4a²/4)} 

 =3µₒi/(4√2πa), going into the paper.

  The magnetic field at the center due to the wire AB is,

B' =µₒ(¾i)a/{2π(a/2)√(a²+4a²/4)}

  =3µₒi/(4√2πa), coming out of the paper.

  Hence the net magnetic field at the center =B -B'

  =3µₒi/(4√2πa) -3µₒi/(4√2πa)

  =Zero.   

 

    20.  The wire ABC shown in figure (35-E7) forms an equilateral triangle. Find the magnetic field B at the center O of the triangle assuming the wires to be uniform.  

Figure for Q-20

ANSWER: The current coming at A will divide equally into the two sections of the loop. From symmetry the magnetic field due to sides AB and AC at O will be equal and opposite. Hence the total magnetic field due to AB and AC at O is zero.     

   Similarly, the length of each base half is equal, the direction of current in them is equal but opposite. From symmetry the magnetic field due to each half at O is equal but from the rule the direction of them is opposite. Thus the total magnetic field due to each half of the base wire is zero at O. 

    So the resultant magnetic field at center O of the triangle is zero.        

 

    21.  A wire of length l is bent in the form of an equilateral triangle and carries an electric current i. (a) Find the magnetic field B at the center. (b) If the wire is bent in the form of a square, what would be the value of B at the center?      

ANSWER: (a) The direction of the magnetic field at the center of the equilateral triangle will be perpendicular to the plane of the triangle due to each of the sides. Hence the magnitude of the net magnetic field at the center will be

B =3*µₒia/{2πd√(a²+4d²)} 

Diagram for Q-21

    Here, a =l/3,

  d= OD = ⅓AD =⅓(l/3)*sin60 =√3l/18

So,

B=3µₒil/{3*2π√3l√(l²/9 +12l²/18²)/18}

 =µₒi/{2√3πl√48/18²} 

 =µₒi*18²/2*4(√3)²πl 

 =µₒi*18²/24πl 

 =13.5µₒi/πl  

 =27µₒi/2πl.  

   (b) The direction of the magnetic field at the center of the square wire will still be the same i.e. perpendicular to the plane of the square. The magnitude of the magnetic field will be 

B =4*µₒia/{2πd√(a²+4d²)} 

here, a =l/4, d =½*l/4 =l/8, so,

B =(4*µₒil/4)/{2π(l/8)√(l²/16+4l²/64)}

 =µₒi/{πl√2/16} 

 =16µₒi/√2πl 

 =8√2µₒi/πl.               

 

 

    22.  A long wire carrying a current i is bent to form a plane angle α. Find the magnetic field B at a point on the bisector of this angle situated at a distance x from the vertex.    

ANSWER: The distance of given point P from each side of the angle, 

d =xSin(α/2) 

Diagram for Q-22

 The direction and magnitude of the magnetic field at P will be the same due to each of the wires. The direction will be perpendicular to the plane of the angle. Hence the magnitude of the magnetic field at P will be double that due to one side wire. 

B =2*µₒi(cosθ₁ -cosθ₂)/4πd

Here θ₁ =α/2, θ₂ =π, so,

B=2µₒi.{cos(α/2)+1}/4πxSin(α/2)

=µₒi{2cos²(α/4)}/2πx{2Sin(α/4)cos(α/4)}

=µₒi{cos(α/4)/sin(α/4)}/2πx

=µₒiCot(α/4)/2πx. 

 

    23.  Find the magnetic field B at the center of a rectangular loop of length l and width b, carrying a current i.    

ANSWER: The direction of the magnetic field due to each side of the rectangle will be the same at the point i.e. perpendicular to the plane of the rectangle. 

Diagram for Q-23

    The magnitude of the magnetic field due to each width will be the same and also due to each length will also be the same. 

    Hence the net magnetic field at the center, 

B =2µₒil/{2πd√(l²+4d²)}+2µₒib/{2πd'√(b²+4d'²)}

=µₒil/{π½b√(l²+4b²/4)}+µₒib/{π½l√(b²+4l²/4)}

={2µₒi/π√(l²+b²)}(l/b +b/l) 

=2µₒi(l²+b²)/{πlb√(l²+b²)} 

=2µₒi√(l²+b²)/πlb.          

 

 

 

    24.  A regular polygon of n sides is formed by bending a wire of total length 2πr which carries a current i. (a) Find the magnetic field B at the center of the polygon. (b) By letting n→∞, deduce the expression for the magnetic field at the center of a circular current.     

ANSWER: (a) Length of each side of the polygon a =2πr/n. The angle is subtended at the center by each side, ß =2π/n. Half of this angle, ß/2 =π/n. If the distance of the center from each side = d. 

    Then tan(ß/2) =(a/2)/d =a/2d 

→d =a/2tan(ß/2) =a/2tan(π/n) 

The direction of the magnetic field at the center due to each side is the same i.e. perpendicular to the plane of the polygon. Hence the net magnitude of the magnetic field at the center, 

B=n*µₒia/{2πd√(a²+4d²)} 

=nµₒi(2πr/n)/{2π(a/2tan(π/n))√(a²+a²/tan²(π/n)} 

=µₒir{2tan²(π/n)/{a²√(1+tan²(π/n)}

=2µₒir.tan²(π/n)/{a²sec(π/n)} 

=2µₒirSin(π/n)tan(π/n)/a² 

=2µₒirSin(π/n)tan(π/n)/(2πr/n)² 

=µₒin²Sin(π/n).tan(π/n)/2π²r. 

(b) For a circle, n→∞. So π/n →0. The expression can be written as,  

B =µₒi{sin(π/n)/(π/n)}²/2rcos(π/n) 

when π/n →0, cos(π/n) ≈1 and also the expression 

sin(π/n)/(π/n) → 1. Thus,

B =µₒi/2r.              

 

    25.  Each of the batteries shown in figure (35-E8) has an emf equal to 5 V. Show that the magnetic field B at the point P is zero for any set of values of the resistances.     

Figure for Q-25

ANSWER: Let us assume the currents and resistances as shown in the diagram below. 

Diagram for Q-25

    Applying Kirchoff's loop law in the left loop anticlockwise. 

i₁r₁+5 +i₂r₂-5 = 0

→i₁r₁ +i₂r₂ =0

Similarly, i₂r₂ +i₃r₃ =0

But i₃+i₁ =i₂

So the first one is now,

i₁r₁ +r₂i₃+r₂i₁ =0

→(r₁+r₂)i₁ +r₂i₃ =0 -------- (i)

The second one becomes,

r₂i₃+r₂i₁+i₃r₃ =0

→r₂i₁+(r₂+r₃)i₃ =0

→{r₂²/(r₂+r₃)}i₁ +r₂i₃ =0 ------(ii)

Subtracting (ii) from (i), we get

{(r₁+r₂) -r₂²/(r₂+r₃)}i₁ =0

→i₁ =0

Putting it in (i), we get. i₃ =0.

Now i₂ =i₃+i₁ =0.

So whatever be the resistances in the circuit, there will be no current. Hence the magnetic field at any point (and hence also point P) will be zero.

   

    26.  A straight long wire carries a current of 20 A. Another wire carrying equal current is placed parallel to it. If the force acting on a length of 10 cm of the second wire is 2.0x10⁻⁵ N, what is the separation between them?    

ANSWER: Given, i₁ =i₂ =20 A. The force per unit length of the wire is given as =µₒi₁i₂/2πd, where d is the separation of the wire. Hence the force on 10 cm of the second wire,

0.10*µₒ*(20)²/2πd =2x10⁻⁵

→d =(4πx10⁻⁷/2π)*400x10⁴/2 m

 =40 m.     

 

    27.  Three coplanar parallel wires, each carrying a current of 10 A along the same direction, are placed with a separation of 5.0 cm between the consecutive ones. Find the magnitude of the magnetic force per unit length acting on the wires.     

ANSWER: Parallel wires carrying currents in the same direction attract each other. Since the separations and the currents are the same between the middle and side wires, each side wire will attract the middle one towards them with equal force. So the forces per unit length acting on the middle wire is equal and opposite. Thus net force is zero. 

   The force per unit length on a side wire is due to another side wire and the middle wire. Their directions are also the same i.e. towards the middle wire. Hence the net magnetic force per unit length on a side wire 

=µₒi²/2πd +µₒi²/{2π(2d)} 

=1.5µₒi²/2πd 

=1.5*(2x10⁻⁷)*10²/0.05 N 

=6.0x10⁻⁴ N.        

  

    28.  Two parallel wires separated by a distance of 10 cm carry currents of 10 A and 40 A along the same direction. Where should a third current be placed so that it experiences no magnetic force?   

ANSWER: Since the parallel wires carry current in the same direction the third current should be put in between them so that the force is equal and opposite. Let the third current i is placed at x cm distance from the 10 A wire and 10-x cm from the 40 A wire. Force per cm length on it due to 10 A wire, 

=µₒ*10i/2πx N 

Due to 40 A wire, 

=µₒ*40i/2π(10-x) N 

Equating these two we get, 

1/x =4/(10-x) 

→10-x =4x

→5x =10

→x =2 cm.  

So the third current should be placed 2 cm from the 10 A current and 10-2 =8 cm from the 40 A current.             

 

    29.  Figure (35-E9) shows a part of an electric circuit. The wires AB, CD, and EF are long and have identical resistances. The separation between neighboring wires is 1.0 cm. The wire AE and BF have negligible resistances and the ammeter reads 30 A. Calculate the magnetic force per unit length of AB and CD.      

Figure for Q-29

ANSWER: Since three parallel wires have equal resistances, the total current of 30 A will be equally divided and each wire will carry 10 A. Wires AE and BF are perpendicular hence will not affect the three wires. 

  Magnetic force per unit length of AB due to middle wire, 

=µₒi²/2πd and due to bottom wire, 

=µₒi²/4πd. Both forces downwards. Hence net force per unit length of AB

=µₒi²/2πd +µₒi²/4πd  

=3µₒi²/4πd 

=3*(1x10⁻⁷)*10²/(0.01) N/m 

=3.0x10⁻³ N/m. Downward.  

   The force per unit length on the middle wire CD due to AB and EF will be equal and opposite, hence net foce zero.           

  

  

    30.  A long straight wire is fixed horizontally and carries a current of 50.0 A. A second wire having linear mass density 1.0x10⁻⁴ kg/m is placed parallel to and directly above this wire at a separation of 5.0 mm. What current should this second wire carry such that the magnetic repulsion can balance its weight?    

ANSWER: To have a magnetic repulsion the current in the second wire should be in opposite direction. Let the magnitude of the current be i. d =5 mm =0.005 m. The magnetic force of repulsion per m =1x10⁻⁴*9.8 N =9.8x10⁻⁴ N.  

So, µₒ*50i/(2π*0.005) =9.8x10⁻⁴ 

→2x10⁻⁷*i =9.8x10⁻⁸ 

→i =4.9x10⁻¹ A 

→i =0.49 A.        

 

    31.  A square loop PQRS carrying a current of 6.0 A is placed near a long wire carrying 10 A as shown in figure (35-E10). (a) Show that the magnetic force acting on the part PQ is equal and opposite to that on the part RS. (b) Find the magnetic force on the square loop.   

The figure for Q-31

ANSWER: (a) The direction of the magnetic field B on the sides PQ and RS will be the same. The magnetic force on a very small length dl of the wire on PQ and RS, dF =idlxB. The directions of vector dl on these two wires will be opposite. Hence the directions of dF on these two sides will be opposite. The direction of B is perpendicular to the loop and going in. Hence the direction of dF on RS is downward in the figure and on PQ it is upward. Let us calculate the magnitude of the force on each of PQ and RS. The magnetic field at a point due to long wire, 

      B =µₒi/2πr, where r is the distance of the point from the wire. 

    So, dF =i.dl.B =i'*dr*µₒi/2πr  

     =(µₒi'i/2π)*dr/r

The total force on each of the wire PQ and RS,

F =∫dF  =(µₒi'i/2π)∫dr/r 

Limit of integration, r =1 cm =0.01 m to r =3 cm =0.03 m.

→ F =(µₒi'i/2π)[ln r]  

  =2x10⁻⁷*6*10*[ln 0.03 -ln 0.01] 

  =12x10⁻⁶*ln 3 

  =13.2x10⁻⁶ N =13.2 µN. 

This is the magnitude of the magnetic force on each of PQ and RS but opposite in direction.  

(b) On the square loop, the forces on PQ and RS will cancel out each other. Since the current in PS is opposite to the current in the long wire, they will repel each other. Thus the force on PS of the loop is towards the right. 

  The force per unit length of the wire in RS,  

dF/dl =µₒi'i/2πd 

Force on RS,  

F₁ =(µₒi'i/2πd)*RS 

 =(2x10⁻⁷*6*10/0.01)*0.02 N 

 =12x10⁻⁴*0.02 N 

 =2.4x10⁻⁵ N, towards the right. 

Force on the side QR, 

  F₂ =(2x10⁻⁷*6*10/0.03)*0.02 N 

  =8x10⁻⁶ N

 =0.80x10⁻⁵ N, towrds left (due to parallel currents). 

Hence net cforce on the loop.

=F₁ -F₂ 

=2.4x10⁻⁵ -0.8x10⁻⁵ N 

=1.6x10⁻⁵ N, towards the right.                      

    32.  A circular loop of one turn carries a current of 5.00 A. If the magnetic field B at the center is 0.200 mT, find the radius of the loop.        

ANSWER: Current, i =5.0 A, number of turns, n =1, magnetic field inside B =0.200 mT =2.0x10⁻⁴ T.  

  If the radius of the loop is 'a', then 

B =µₒi/2a

2.0x10⁻⁴ = 4πx10⁻⁷*5.0/2a  

→a =½πx10⁻² m 

     =1.57x10⁻² m 

     =1.57 cm.          

 

    33.  A current-carrying circular coil of 100 turns and radius 5.0 cm produces a magnetic field of 6.0x10⁻⁵ T at its center. Find the value of the current.       

ANSWER: Magnetic field at the center due to n circular loop, B=µₒni/2a 

Hence, i =2Ba/µₒn.

Putting the values from the problem,

i =2*6x10⁻⁵*0.05/{(4πx10⁻⁷)*100}  

 =0.6/4π A 

 =0.048 A   

 =48 mA.    

 

    34.  An electron makes 3x10⁵ revolutions per second in a circle of radius 0.5 angstrom. Find the magnetic field B at the center of the circle.      

ANSWER: For an electron, 

Charge q =1.602x10⁻¹⁹ C, 

n = number of revolution per second.

   At any section in its path, nq coulomb of charge passes per second that is equal to a current i =nq =3x10⁵*1.602x10⁻¹⁹ A

→i =4.806x10⁻¹⁴ A   

 

The radius of the circle,

  a =0.5 angstrom   

   =5x10⁻¹¹ m 

The magnetic field at the center, 

B =µₒi/2a 

 =4πx10⁻⁷*4.806x10⁻¹⁴/2*5x10⁻¹¹ T

 =60.4x10⁻¹¹ T  

 =6.0x10⁻¹⁰ T.

 

    35.  A conducting circular loop of radius 'a' is connected to two long, straight wires. The straight wires carry current 'i' as shown in figure (35-E11). Find the magnetic field B at the center of the loop.    

The figure for Q-35

ANSWER: Since the upper and lower semicircular wires of the loop are similar their resistances will also be the same. Hence the current of the long wire will be equally divided. Thus the current in each of the semicircular wires will be =i/2.  

   The magnetic field due to each of them will be equal in magnitude due to similarity. But from the direction rule, we find that their directions are opposite. Hence the net magnetic field at the center of the loop will be zero.       

 

    36.  Two circular coils of radii 5.0 cm and 10 cm carry equal currents of 2.0 A. The coils have 50 and 100 turns respectively and are placed in such a way that their planes, as well as the centers, coincide. Find the magnitude of the magnetic field B at the common center of the coils if the currents in the coils are (a) in the same sense (b) in the opposite sense.      

ANSWER: (a) The magnetic field due to the smaller coil at the center, 

B =µₒni/2a 

 =4πx10⁻⁷*50*2/(2*0.05) T 

 =4πx10⁻⁴ T

 

The magnetic field due to the larger coil at the center,

B' =4πx10⁻⁷*100*2/(2*0.10) T  

  =4πx10⁻⁴ T. 

We see that B =B'.

  When the current in each coil is in the same sense, the direction of the magnetic field is the same and hence they will add up. So the net magnetic field =B+B'

 =8πx10⁻⁴ T. 

(b) When the current in the coils is in the opposite sense, the directions of the magnetic fields at the center will be opposite. Hence net magnetic field at the center =B-B' =0 (zero).      

 

    37.  If the outer coil of the previous problem is rotated through 90° about a diameter, what would be the magnitude of the magnetic field B at the center?       

ANSWER: In this case, the two magnetic fields B and B' will be at 90° in a plane that is perpendicular to the planes of the coils and passes through their centers. The magnitude of the resultant magnetic field at the center will be  

  =√(B² +B'²) 

  =√(2B²)

  =B√2

  =4πx10⁻⁴*√2 T 

  =17.8x10⁻⁴ T

  ≈1.8 mT.         

 

    38.  A circular loop of radius 20 cm carries a current of 10 A. An electron crosses the plane of the loop with a speed of 2.0x10⁶ m/s. The direction of motion makes an angle of 30° with the axis of the circle and passes through its center. Find the magnitude of the magnetic force on the electron at the instant it crosses the plane.      

ANSWER: The magnetic field at the center of the loop,  

B =µₒi/2a 

  =4πx10⁻⁷*10/(2*0.20) T

 =πx10⁻⁵ T, along the axis. 

Speed of the electron, v =2x10⁶ m/s. 

The angle between speed and magnetic field θ =30°. 

The magnitude of the magnetic force on the electron,

F =qvBsinθ

  =evBsinθ  

=1.602x10⁻¹⁹*2x10⁶*πx10⁻⁵*sin30° =1.6πx10⁻¹⁸ N

=16πx10⁻¹⁹ N.              

 

    39.  A circular loop of radius R carries a current I. Another circular loop of radius r(<<R) carries current i and is placed at the center of the larger loop. The planes of the two circles are at a right angle to each other. Find the torque acting on the smaller loop.      

ANSWER: The magnetic field at the center due to the loop with radius R, 

 B =µₒI/2R.  

   The smaller current-carrying loop is placed in this magnetic field B. Area of the smaller loop, A =πr². The angle θ between the area vector of the smaller loop and the magnetic field B is given as 90°. Therefore the net torque on the smaller loop 

  Γ =iABsinθ

  =iπr²(µₒI/2R)sin90°

  =µₒπiIr²/2R. 

 

    40.  A circular loop of radius r carrying a current i is held at the center of another circular loop of radius R(>>r) carrying a current I. The plane of the smaller loop makes an angle of 30° with that of the larger loop. If the smaller loop is held fixed in this position by applying a single force at a point on its periphery, what would be the minimum magnitude of this force?        

ANSWER: The situation is just like the problem (39) with a difference that the angle θ =30°. 

  Hence the torque on the smaller loop,

Γ =µₒπiIr²/4R

Let the required minimum force be F, perpendicular to the plane of the smaller loop. To balance the torque the moment of this force about the common diameter = Fd, where d is the perpendicular distance from the point on the periphery to the common diameter. Keeping Fd constant, minimum F will be when d is maximum. The maximum value of d =r. 

Diagram for Q-40

Hence in this case 

Fd =Fr

Equating this moment with the torque. we get

Fr =µₒπiIr²/4R

→F =µₒπiIr/4R. 

 

    41.  Find the magnetic field B due to a semicircular wire of radius 10.0 cm carrying a current of 5.0 A at its center of curvature.      

ANSWER: If the magnetic field due to a very small length dl at the center is dB, then  

dB =(µₒi/4π)dlxr/r³  

Since dl and 'r' are at right angles, the magnitude  

dB =(µₒi/4π)dl/r² 

Now the total magnetic field at the center 

B =∫dB

 =(µₒi/4πr²)∫dl

 =(µₒi/4πr²)*πr

 =µₒi/4r. 

 =4πx10⁻⁷*5/(4*0.10) T

 =15.7x10⁻⁶ T

 =1.6x10⁻⁵ T.

        

 

    42.  A piece of wire carrying a current of 6.00 A is bent in the form of a circular arc of radius 10.0 cm, and it subtends an angle of 120° at the center. Find the magnetic field B due to this piece of wire at the center.      

ANSWER: In this case the ∫dl 

=(120°/360°)*2πr

=⅔πr

Hence B=(µₒi/4πr²)∫dl

→B =(µₒi/4πr²)*⅔πr

  =µₒi/6r

  =4πx10⁻⁷*6/(6*0.10) T

 =4πx10⁻⁶ T

 =12.6x10⁻⁶ T

 =1.26x10⁻⁵ T.

 

 

    43.  A circular loop of radius r carries a current i. How should a long, straight wire carrying a current 4i be placed in the plane of the circle so that the magnetic field at the center becomes zero?       

ANSWER: The direction of the magnetic field at the center of the loop will be coming towards the viewer if the current is anticlockwise. The direction of the magnetic field around a straight wire is anticlockwise if the current is coming towards the viewer. To keep the direction of the magnetic field due to the wire in the plane of the loop opposite at the center of the loop, the direction of the current in the wire and the nearest portion of the loop should be opposite. 

Diagram for Q-43

To make the net magnetic field zero at the center, both magnitudes should be equal. Hence, 

µₒi/2r = µₒ(4i)/2πd, where d is the distance of the long wire from the center of the loop. 

→d =4r/π.     

 

    44.  A circular coil of 200 turns has a radius of 10 cm and carries a current of 2.0 A. (a) Find the magnitude of the magnetic field B at the center of the coil. (b) At what distance from the center along the axis of the coil will the field B drop to half its value at the center? (∛4 =1·5874...).    

ANSWER: (a) n =200, radius a =10 cm =0.10 m, i =2.0 A. The magnitude of the magnetic field at the center of the loop 

 B =µₒni/2a 

 =(4πx10⁻⁷*200*2)/(2*0.10) T 

 =2.51x10⁻³ T  

 =2.51 mT 

(b) The magnetic field at a point distance d from the center along the axis of the loop is, 

B' =µₒnia²/{2(a²+d²)^3/2} 

In this case, B' =B/2  

→µₒni/4a =µₒnia²/{2(a²+d²)^3/2} 

→(a²+d²)^3/2 = 2a³ 

→(a²+d²)³ =4a⁶ 

→a²+d² =a²∛4 

→d² =(1.5874 -1)a²

→d² =0.5874a² 

→d =0.766a =0.766*10 cm  

      =7.66 cm.                

 

    45.  A circular loop of radius 4.0 cm is placed in a horizontal plane and carries an electric current of 5.0 A in the clockwise direction as seen from above. Find the magnetic field (a) at a point 3.0 cm above the center of the loop (b) at a point 3.0 cm below the center of the loop.     

ANSWER: Since the current in the horizontal loop is in the clockwise direction when seen from above, the direction of the magnetic field in each case will be downward. Also, the points above and below the center of the loop are equidistant from the center, the magnitude of the magnetic field will also be the same. This magnitude, 

B =µₒia²/{2(a²+d²)¹∙⁵} 

 =(4πx10⁻⁷*5.0*0.04²)/{2(0.04²+0.03²)¹∙⁵} T 

 =3.2πx10⁻⁹/{2(0.05²)¹∙⁵} T

 =1.6πx10⁻⁹/0.05³ T 

 =4.0x10⁻⁵ T. Downwards in both the cases.    

 

 

    46.  A charge of 3.14x10⁻⁶ C is distributed uniformly over a circular ring of radius 20.0 cm. The ring rotates about its axis with an angular velocity of 60 rad/s. Find the ratio of the electric field to the magnetic field at a point on the axis at a distance of 5.00 cm from the center.     

ANSWER: In one revolution of the ring, 2π radians are covered. So per second, 60/2π revolutions of the ring will take place. Hence the equivalent current in the ring 

i =Charge passing through a section per second 

=60*3.14x10⁻⁶/2π A

=3x10⁻⁵ A  

The radius of the ring, a =0.20 m  

The electric field at this point, 

E =(1/4πεₒ)*Qd/(a²+d²)¹·⁵  

The magnetic field at this point 

B =µₒia²/{2(a²+d²)¹·⁵}

 =(1/4πεₒc²)4πia²/{2(a²+d²)¹·⁵} 

{Substituting for µₒ} 

Hence, 

E/B=2Qdc²/4πia² 

   =Qdc²/2πia² 

=3.14x10⁻⁶*0.05*(3x10⁸)²/(2π*3x10⁻⁵*0.20²) m/s

=1.88x10¹⁵ m/s.   

  The unit of this ratio,

E/B = (N/C)/(N/A-m) 

  = A-m/C 

  = (C/s)*m/C

  =m/s

 

    47.  A thin, long, hollow cylindrical tube of radius r carries current 'i' along its length. Find the magnitude of the magnetic field at a distance r/2 from the surface (a) inside the tube (b) outside the tube.     

ANSWER: (a) Consider an amperes circular loop of radius r/2 with its center on the axis of the tube. 

Diagram for Q-47

The circulation, 

∮B.dl =µₒi

→B*2πr/2 = 0, since current inside this loop is zero.

→B =0.

So the magnitude of the magnetic field at r/2 distance inside the tube is zero.

(b) Taking the ampere circular loop through the point outside the tube at r/2 distance from the surface and the center on the axis of the tube, the circulation,

∮B.dl =µₒi

→B*2π(3r/2) =µₒi

→B =µₒi/(3πr).      

 

    48.  A long cylindrical tube of inner and outer radii a and b carries a current 'i' distributed uniformly over its cross-section. Find the magnitude of the magnetic field at a point (a) just inside the tube (b) just outside the tube.      

ANSWER: (a) Take an ampere circular loop with its center on the axis of the cylindrical tube. The circulation ∮B.dl  along this loop is,

∮B.dl =µₒi

→B*2πa =µₒi

{Here enclosed current, i = 0}

→2πBa =0

→B =0.

(b) Similarly taking an ampere circular loop just outside the tube with its center on the axis of the tube. The circulation ∮B.dl along it is,

∮B.dl =µₒi

→B*2πb =µₒi, {Enclosed current =i}

→B =µₒi/2πb.

 

 

    49.  A long cylindrical wire of radius b carries a current 'i' distributed uniformly over its cross-section. Find the magnitude of the magnetic field at a point inside the wire at a distance 'a' from the axis.     

ANSWER: When we take an ampere circular loop through the given point with its center on the axis, the portion of the current through this loop,   

i' =i*(πa²/(πb²) =ia²/b²  

The circulation ∮B.dl over this loop is,

∮B.dl = µₒi'

→B*2πa =µₒia²/b²

→B =µₒia/2πb².      

 

    50.  A solid wire of radius 10 cm carries a current of 5.0 A distributed uniformly over its cross-section. Find the magnetic field B at a point at a distance (a) 2 cm  (b) 10 cm and (c) 20 cm away from the axis. Sketch a graph of B versus x for 0<x<20 cm.     

ANSWER: Taking the ampere circular loop through a point with the center of this loop on the axis of the wire. Due to symmetry, the magnitude of B at each point of this loop will be the same. The current enclosed by this loop when the point is inside the wire,

i' =i*(πx²)/(πr²)

 =ix²/r²

 The circulation ∮B.dl over this loop

∮B.dl =µₒi'

→B*2πx =µₒix²/r²

→B =µₒix/2πr²   ------- (i)

(a) For x =2 cm =0.02 m

  r =10 cm =0.10 m, i =5.0 A

Hence from (i),

B =(4πx10⁻⁷*5*0.02)/(2π*0.10²) T

 =2.0x10⁻⁶ T

 =2.0 µT

(b) For x = r =10 cm =0.10 m

B =µₒi/2πr

  =(4πx10⁻⁷*5)/(2π*0.10) T

 =10x10⁻⁶ T

 =10.0 µT

(c) When the point is outside the wire, i.e. x>r, the current enclosed inside the ampere loop =i. The circulation ∮B.dl over this loop gives,

∮B.dl =µₒi

→B*2πx =µₒi

→B =µₒi/2πx ------- (ii)  

For x =20 cm =0.20 m, 

B =(4πx10⁻⁷*5)/(2π*0.20) T

 =5x10⁻⁶ T

 =5.0 µT.  

Graph of B vs x 

From (i) we see that when the point is inside the wire B=µₒix/2πr² =K*x

Hence B varies directly with x, the graph is linear. 

From (ii) when the point is outside the wire B =µₒi/2πx =K'/x. 

So B is inversely proportional to x. The graph is a non-linear one with B =0 at x =∞. See below    

Graph for Q-50

 

    51.  Sometimes we show an idealized magnetic field that is uniform in a given region and falls to zero abruptly. One such field is represented in figure (35-E12). Using Ampere's law over the path PQRS, show that such a field is not possible.     

Figure for Q-51

ANSWER: From Ampere's law, the closed integral of B.dl over the loop PQRS should be µₒ times the current through the loop. ∮B.dl over PQRS is the same as ∮B.dl over the length PS. (Because the direction of field and length over PR and RS are perpendicular and the field over QR is zero.) Hence,

 ∮B.dl =B*PS =µₒ*i =µₒ*0 =0

  Because 'i' through the loop is zero.

It gives B = 0 which is not correct. So the magnetic field can not fall to zero abruptly.

 

    52.  Two large metal sheets carry surface currents as shown in figure (35-E13). The current through a strip of width dl is Kdl where K is a constant. Find the magnetic field at a point P, Q, and R.  

Figure for Q-52

ANSWER: From symmetry, the magnitudes of the magnetic fields at P and R will be the same (say B'), and their directions parallel to the plates because the plates are large. The magnetic field at Q due to each plate will be towards the right. So both the magnitudes will be added due to the same direction. Let net magnitude =B. The direction of the magnetic field everywhere will be parallel to plates because the plates are large.

Diagram for Q-52

 We assume an Ampereian loop EFGH through P and R as shown in the figure. The magnetic field is perpendicular to EH and FG. If EF =L, then from ampere's law  

∮B.dl =µₒi
→2B'L =µₒ(KL-KL) =0, Since the currents in the sheets are equal and opposite.

→B' =0.

So the magnetic fields at P and R are zero.

Diagram for Q-52

Take another Ampereian loop ABCD about the upper plate through P and Q as shown in the figure. Here, ∮B.dl =µₒi 

→B'L+BL =µₒ*KL

→BL =µₒKL

→B =µₒK. (towards the right)

 

 

    53.  Consider the situation of the previous problem. A particle having charge q and mass m is projected from the point Q in a direction going into the plane of the diagram. It is found to describe a circle of radius r between the two plates. Find the speed of the charged particle.  

ANSWER: Near point Q, a uniform magnetic field is there B =µₒK. This field and the direction of the speed are perpendiculars hence a magnetic force F will act on the charged particle.  

F =qvB 

This will provide the centripetal force to describe a circle. F =mv²/r. So, 

qvB =mv²/r 

→v =qBr/m 

→v =q(µₒK)r/m =µₒKqr/m.           

 

    54.  The magnetic field B inside a long solenoid, carrying a current of 5.00 A, 3.14x10⁻² T. Find the number of turns per unit length of the solenoid.   

ANSWER: B =3.14x10⁻² T, i =5.0 A. Let n = number of turns per unit length. Then for a long solenoid,  

B =µₒni

→n =B/µₒi 

  =3.14x10⁻²/(4πx10⁻⁷*5)

 =5000 turns/m.        

 

    55.  A long solenoid is fabricated by closely winding a wire of radius 0.5 mm over a cylindrical non-magnetic frame so that the successive turns nearly touch each other. What would be the magnetic field B at the center of the solenoid if it carries a current of 5 A?  

ANSWER: The radius of the wire =0.5 mm, hence the thickness of one turn in the solenoid =1.0 mm. The number of turns in 1-meter length (1000 mm) of the solenoid, n =1000.  

   Given current i =5 A. 

The magnetic field at the center of the solenoid, 

B =µₒni 

 =4πx10⁻⁷*1000*5 T 

 =2πx10⁻³ T           

 

    56.  A copper wire having a resistance of 0.01 ohm in each meter is used to wind a 400-turn solenoid of radius 1.0 cm and length 20 cm. Find the emf of a battery which when connected across the solenoid will cause a magnetic field of 1.0x10⁻² T near the center of the solenoid.  

ANSWER: Given B =1.0x10⁻² T. The number of turns =400 in 20 cm length. Hence the number of turns per meter, 

n =400/0.20 =2000. 

We have B =µₒni, 

→i =B/µₒn 

  =1.0x10⁻²/(4πx10⁻⁷*2000) A

 =100/8π A. 

The total length of the wire in the solenoid, 

=400*2π*1.0 cm 

=800π cm =8π m 

The resistance of the wire in solenoid 

R =8π*0.01 Ω

  =0.08π Ω  

Let the emf of the battery =V. 

Then V =iR

   =(100/8π)*0.08π volts 

  = 1 volt.                 

 

    57.  A tightly wound solenoid of radius 'a' and length 'l' has n turns per unit length. It carries an electric current i. Consider a length dx of the solenoid at a distance x from one end. This contains ndx turns and may be approximated as a circular current indx. (a) Write the magnetic field at the center of the solenoid due to this circular current. Integrate this expression under proper limits to find the magnetic field at the center of the solenoid. (b) Verify that if l>>a, the field tends to B =µₒni and if a>>l, the field tends to B =µₒnil/2a. Interpret these results. 

ANSWER: (a) The magnetic field at a point on the axis of a loop having current 'i' and at a distance d from its center is 

B =µₒia²/2(a²+d²)ⁿ, where n =3/2.

For a loop at distance x from one end, the center of the solenoid is at a distance d = l/2-x from the loop. 

For the current 'indx' in the loop,   

dB =µₒa²indx/2{a²+(l/2-x)²}ⁿ 

Integrating between the limits, x =0 to x =l, we have 

B =∫dB =(µₒa²in/2)∫dx/{a²+(l/2-x)²}ⁿ

Let l/2-x =y, then dx =-dy and the integration becomes,

B =-(µₒa²in/2)∫dy/{a²+y²}ⁿ

 =-½(µₒa²in)[{y/a²√(a²+y²)}] 

{The above integration has been derived by substituting y=a.tanδ where n =3/2} 

The limits for y is :- if x =0, y =l/2 and if x =l, y =-l/2. Putting the limits ,  

B=-½µₒin[(-l/2)/√(a²+l²/4)-(l/2)/√(a²+l²/4)]

=½µₒinl/√{(4a²+l²)/4} --(1)

=µₒinl/{l√{(1+(2a/l)²}

=µₒin/√{1+(2a/l)²}.         

(b) When l>>a, (2a/l)²≈0 

In such case,

B =µₒin.  

This is the case of a long solenoid. B is independent of l and a.

When a>>l, 4a²+l² ≈4a² 

From (1),

B =½µₒinl/√{(4a²+l²)/4}

 =µₒinl/√{(4a²+l²)}

=µₒinl/√(4a²)

 =µₒinl/2a.

   This is the case of a short solenoid. Here B depends on both l and a.        

 

    58.  A tightly wound, long solenoid carries a current of 2.00 A. An electron is found to execute a uniform circular motion inside the solenoid with a frequency of 1.00x10⁸ rev/s. Find the number of turns per meter in the solenoid.    

ANSWER: Current i = 2.0 A.  

If the magnetic field inside the solenoid be B, then equating the magnetic force on the electron and the centripetal force,

evB =mv²/r

v =eBr/m.   

Here v is the speed of the electron. Time period of the revolution, 

T =2πr/v 

Hence frequency f =1/T =v/2πr  

f =eBr/{m*2πr}

B =2πmf/e

But for a long solenoid,

B =μₒni

Hence, 

μₒni =2πmf/e

n = 2πf/μₒ(e/m)i

 =2π*10⁸/{4π*10–⁷*1.76x10¹¹*2}

 =1.42x10³

 =1420 turns per m.

              

 

    59.  A tightly wound, long solenoid has n turns per unit length a radius r, and carries a current i. A particle having charge q and mass m is projected from a point on the axis in a direction perpendicular to the axis. What can be the maximum speed for which the particle does not strike the solenoid? 

ANSWER: Since the solenoid is long, the magnetic field inside it 

B = μₒni

Force on a particle having a charge q in this magnetic field,

F =qvB, where v is the speed of the particle. This force is perpendicular to the speed. So it will act as the centripetal force to make the particle describe a circular path. Let the radius of the path =R.

→mv²/R =qvB 

→mv =qRμₒni

v =μₒqRni/m

Maximum speed will be when the radius R is maximum. In order that the particle does not strike the solenoid, R =r/2. It is because of the particle being projected from a point on the axis. The largest circle without striking the solenoid will be between the axis and the side of the solenoid. The diameter of this circle =r, hence the radius, R =r/2. So the maximum speed of the particle 

= μₒq(r/2)ni/m 

= μₒqrni/2m.

           

 

    60.  A tightly wound, long solenoid is kept with its axis parallel to a large metal sheet carrying a surface current. The surface current through a width dl of the sheet is Kdl and the number of turns per unit length of the solenoid is n. The magnetic field near the center of the solenoid is found to be zero. (a) Find the current in the solenoid. (b) If the solenoid is rotated to make its axis perpendicular to the metal sheet, what would be the magnitude of the magnetic field near its center?  

ANSWER: First, let us find out the magnetic field at a point P, a distance d from the large metal sheet. 

Consider the Ampereian loop in the diagram below. P and Q are at a distance d from the plate. Hence the magnitude at these points will be the same =B. 

Diagram for Q-60

From Amperes law,

∮B.dl over the loop =2BL

So, 2BL =µₒi

→2BL =µₒ*KL

→B =µₒK/2. {It is independent of d}  

It can also be derived from the diagram below, 

Diagram for Q-60

The magnetic field at P due to the current Kdl in the sheet at distance l from the root of the perpendicular from P, 

dB =µₒKdl/{2π√(d²+l²)} 

{Since B =µₒi/2πr}

Its component parallel to the plate will be only effective because the perpendicular component will be neutralized by the current equidistant from the point Q. So we have to integrate dBsinδ between the limits δ =0 to π. 

 So the magnetic field at P, 

B =∫dBsinδ 

 =∫µₒKdl.sinδ/{2π√(d²+l²)}

 =(µₒK/2π)∫dl*d/(d²+l²)

 =(µₒK/2πd)∫{d²/(d²+l²)}dl

 =(µₒK/2πd)∫sin²δ.dl ----- (i)

 Since d/l =tanδ, →l =d.cotδ 

→dl =-d.cosec²δdδ

Putting in (i), 

B=(µₒK/2πd)∫sin²δ(-d.cosec²δ.dδ)

 =-(µₒK/2π)∫dδ

 =-(µₒK/2π)[δ], putting limis of δ

 =-(µₒK/2π)[0 -π]

 =µₒK/2. 

(a) The magnetic field at the center of the solenoid due to current 'i' in the solenoid =µₒni. Since the net magnetic field at the center is found to be zero, the magnetic fields due to the sheet and the solenoid must be equal and opposite.

Thus, µₒni =µₒK/2

→i =K/2n.

(b) When the solenoid is kept perpendicular to the metal sheet, two equal magnetic fields, each µₒK/2, are directed perpendicular to each other. Hence the resultant magnetic field,

=√{(µₒK/2)²+(µₒK/2)²}

=√{2(µₒK/2)²}

=√2*µₒK/2

=µₒK/√2.

            

 

    61.  A capacitor of capacitance 100 µF is connected to a battery of 20 volts for a long time and then disconnected from it.  It is now connected across a long solenoid having 4000 turns per meter. It is found that the potential difference across the capacitor drops to 90% of its maximum value in 2.0 seconds. Estimate the average magnetic field produced at the center of the solenoid during this period.   

ANSWER: Q =CV. In 2.0 seconds the remaining potential difference =0.9V, so the charge on the capacitor is now Q' =0.9CV =0.9Q  

Thus in this period of 2.0 s, 0.1Q of charge flows through the solenoid. So average current,  

i =0.1Q/2 =0.05Q =0.05CV 

The average magnetic field at the center of the long solenoid, 

B =µₒni

 =µₒn*0.05CV 

 =0.05*4πx10⁻⁷*4000*100x10⁻⁶*20 T 

 =1.6πx10⁻⁷ T 

 =16πx10⁻⁸ T.  

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CHAPTER 47- The Special Theory of Relativity

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CHAPTER- 39- Alternating Current

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CHAPTER- 33- Thermal and Chemical Effects of Electric Current

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CHAPTER- 32- Electric Current in Conductors

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CHAPTER- 25-CALORIMETRY

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CHAPTER- 24-Kinetic Theory of Gases

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CHAPTER- 18 - Geometrical Optics

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CHAPTER- 10 - Rotational Mechanics

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EXERCISES - Q-16 TO Q-30

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

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CHAPTER- 5 - Newton's Laws of Motion

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CHAPTER- 4 - The Forces

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CHAPTER- 3 - Kinematics - Rest and Motion

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