Thursday, September 23, 2021

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

Magnetic Field Due to a Current


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.   


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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


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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- 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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