Thursday, January 27, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, OBJECTIVE-I, Concepts of Physics, Part-II

Electromagnetic Induction


OBJECTIVE-I


    1.  A rod of length l rotates with a small but uniform angular velocity ⍵ about its perpendicular bisector. A uniform magnetic field exists parallel to the axis of rotation. The potential difference between the center of the rod and an end is

(a) zero

(b) ⅛⍵Bl²

(c) ½⍵Bl²

(d) B⍵l². 


ANSWER: (b)     


EXPLANATION: Small emf developed across a small length dr is 

dƐ =vBdr = ωrBdr

→Ɛ = ∫ωrBdr

     =ωB∫rdr

     =ωB[r²/2] 

{between limits r =0 to r =l/2.}

    =½ωB*l²/4

    =⅛ωBl². 

Hence option (b) is correct. 





    2.  A rod of length l rotates with uniform angular velocity ω about its perpendicular bisector. A uniform magnetic field B exists parallel to the axis of rotation. The potential difference between the two ends of the rod is 

(a) zero

(b) ½Blω² 

(c) Blω²

(d) 2Blω².       


ANSWER: (a)     


EXPLANATION: Each end will be at ⅛ωBl² higher or lower potential than the center depending upon the directions of the magnetic field and the rotation. So both the ends are at equal potential and hence the potential difference between the ends is zero. Option (a) is correct. 

     




    3.  Consider the situation shown in Figure (38-Q2). If the switch is closed and after some time it is opened again, the closed loop will show 

(a) an anticlockwise current-pulse 

(b) a clockwise current pulse   

(c) an anticlockwise current pulse and then a clockwise current pulse. 

(d) a clockwise current pulse and then an anticlockwise current pulse.     
Figure for Q-3


ANSWER: (d)     


EXPLANATION: When the switch is closed, the current in the left loop increases from zero to a constant value in a very very short time. Hence the magnetic field around the wire changes for a very short time and this changing magnetic field induces a current-pulse in the closed-loop. According to Lenz's law direction of this current pulse will be to oppose the flux through it. Since the direction of flux through the loop due to the nearest wire-current is 'coming out of paper', a clockwise current pulse in the closed loop will produce flux 'going into the paper' through it. 

    When the switch is opened again, the magnetic flux density through the ring begins to decrease. Hence the induced current pulse in the closed-loop opposes the change and tries to maintain the flux density. In this effort, the induced current in the loop produces a flux in the same direction. Since the direction of the flux before opening the switch was 'coming out of the paper', the same will be the direction of the flux due to the induced current pulse in the closed loop. Such flux can be produced by an anticlockwise current pulse in the loop. Hence option (d) is correct. 


     




    4.  Solve the previous question if the closed loop is completely enclosed in the circuit containing the switch.    


ANSWER: (c)     


EXPLANATION: In this case, the direction of increasing clockwise current in the outer loop will produce increasing magnetic flux inside the loop 'going into the plane of the loop'. To oppose it, the direction of the induced current pulse in the closed loop will be anticlockwise so that the flux inside it is coming out of the plane. 

         When the switch is opened again, the magnetic flux through the closed loop begins to decrease. The induced current in the loop opposes this decrease in flux, i.e. it tries to maintain the flux by producing the flux in the same direction. So there will be a clockwise induced current pulse in the closed-loop. Option (c) is correct. 


          




    5.  A bar magnet is released from rest along the axis of a very long, vertical copper tube. After some time the magnet 

(a) will stop in the tube 

(b) will move with almost constant speed

(c) will move with an acceleration g

(d) will oscillate.        


ANSWER: (b)     


EXPLANATION:  After the release from rest, the magnet will fall with acceleration g. This will result in the changing flux linked with the copper tube. It will result in a current in the tube that will resist this change in flux. So a magnetic force will act on the bar magnet opposite to the direction of movement. This opposing force will increase till a velocity is reached when this force is equal to the weight of the magnet. At this stage the net force on the magnet being zero, there will be no acceleration, and the magnet will fall with almost a constant speed. Option (b) is correct. 


   




    6.  Figure (38-Q3) shows a horizontal solenoid connected to a battery and a switch. A copper ring is placed on a frictionless track, the axis of the ring being along the axis of the solenoid. As the switch is closed, the ring will 

(a) remain stationary 

(b) move towards the solenoid 

(c) move away from the solenoid 

(d) move towards the solenoid or away from it depending on which terminal (positive or negative) of the battery is connected to the left end of the solenoid.       
Figure for Q-6


ANSWER: (c)     


EXPLANATION: As the switch is closed, the magnetic flux through the solenoid and hence through the loop will change. The induced current in the loop will be such that it will oppose the increase in the flux through it, and it can be achieved by having the direction of the current opposite to that in the solenoid. This will produce the magnetic flux opposite to that produced by the solenoid so that the net flux is reduced. Since the currents in the solenoid and the loop are opposite, they will repel each other and the loop will move away from the solenoid. Option (c) is correct. 


         




    7.  Consider the following statements 

(A) An emf can be induced by moving a conductor in a magnetic field. 

(B) An emf can be induced by changing the magnetic field. 


(a) Both A and B are true

(b) A is true but B is false 

(c) B is true but A is false 

(d) Both A and B are false.         


ANSWER: (a)     


EXPLANATION: When a conductor is moved in a magnetic field, a magnetic force is acted upon the electrons due to the component of the speed perpendicular to the magnetic field. This results in an induced emf in the conductor. Statement A is correct. 

   Due to a change in the magnetic field, there is a flux change inside a loop placed there. This induces an emf in the loop. Statement B is also correct. 

      Option (a) is correct.  


       




    8.  Consider the situation shown in the figure (38-Q4). The wire AB is slid on the fixed rails with a constant velocity. If the wire AB is replaced by a semicircular wire, the magnitude of the induced current will

(a) increase

(b) remain the same 
(c) decrease
(d) increase or decrease depending on whether the semicircle bulges towards the resistance or away from it.  
Figure for Q-8


ANSWER: (b)     


EXPLANATION: The induced emf does not depend on the shape of the wire because all the electrons in the wire AB or semicircular wire move in the same direction. Hence option (b) is correct.        




    9.  Figure (38-Q5a) shows a conducting loop being pulled out of a magnetic field with a speed v. Which of the four plots shown in figure (38-Q5b) may represent the power delivered by the pulling agent as a function of the speed v?    
Figure for Q-9

ANSWER: (b)     


EXPLANATION:  In this loop, the emf will be induced in the wire length 'l' that is perpendicular to the velocity. This emf is 

  ξ =Blv. 

If the resistance of the circuit is R, then the current in the loop, 

  i =ξ/R =Blv/R 

Power P =ξi

→P =Blv*Blv/R =(B²l²/R)*v² 

So, P ∝ v² 

This is an equation of a parabola symmetrical to the P-axis and passing through the origin. Since we are interested in positive v, plot b is appropriate. Hence option (b) is correct.


          




    10.  Two circular loops of equal radii are placed coaxially at some separation. The first is cut and a battery is inserted in between to drive a current in it. The current changes slightly because of the variation in resistance with temperature. during this period, the two loops  

(a) attract each other 

(b) repel each other 

(c) do not exert any force on each other 

(d) attract or repel each other depending on the sense of the current.          


ANSWER: (a)     


EXPLANATION: When there is a constant current in the loop, a constant magnetic flux is generated by it and the same constant flux passes through the other loop. When due to the temperature variation the resistance increases, the current decreases. So the flux generated by the current begins to decrease. The induced current in the other loop will be such that it will oppose the change of the flux. In order to keep the flux the same, the induced current in the loop will be in the same sense. Since the current in both the loops is in the same sense, they will attract each other. Option (a) is correct. 


 




    11.  A small conducting circular loop is placed inside a long solenoid carrying a current. The plane of the loop contains the axis of the solenoid. If the current in the solenoid is varied, the current induced in the loop is 

(a) clockwise 

(b) anticlockwise 

(c) zero 

(d) clockwise or anticlockwise depending on whether the resistance is increased or decreased.         


ANSWER: (c)     


EXPLANATION: When the current in the solenoid is changed, the magnetic flux along the axis of the solenoid changes. Since the plane of the small conducting loop is parallel to this flux, there is no flux change perpendicular to the plane of the loop. Hence no current is induced in the loop. Option (c) is correct.        




    12.  A conducting square loop of side l and resistance R moves in its plane with a uniform velocity v perpendicular to one of its sides. A uniform and constant magnetic field B exists along the perpendicular to the plane of the loop as shown in figure (38-Q6). The current induced in the loop is 

(a) Blv/R clockwise 

(b) Blv/R anticlockwise

(c) 2Blv/R anticlockwise 

(d) zero.        
Figure for Q-12


ANSWER: (d)     


EXPLANATION: There will not be any emf induced in the sides parallel to the direction of the velocity. The other two sides will have equal emf induced but there will be no current in the loop because the ends connected of these two sides are equipotential. Hence option (d) is correct.      


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Links to the Chapters




CHAPTER- 38- Electromagnetic Induction


CHAPTER- 34- Magnetic Field

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

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

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

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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