Wednesday, November 10, 2021

H C Verma solutions, MAGNETIC FIELD DUE TO A CURRENT, Chapter-35, EXERCISES, Q31_Q40, Concepts of Physics, Part-II

Magnetic Field Due to a Current


EXERCISES, Q31 to Q40


    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. 

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Links to the Chapters



CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




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


CHAPTER- 8 - Work and Energy

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CHAPTER- 7 - Circular Motion

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For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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