Sunday, February 20, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, EXERCISES, Q11 to Q20, Concepts of Physics, Part-II

Electromagnetic Induction


EXERCISES, Q11 to Q2O


    11.  A conducting loop of face area A and resistance R is placed perpendicular to a magnetic field B. The loop is withdrawn completely from the field. Find the charge that flows through any cross-section of the wire in the process. Note that it is independent of the shape of the loop as well as the way it is withdrawn.       


ANSWER: Initial magnetic flux through the coil, φ = BA. 

After the withdrawal the magnetic flux =0. 

Change in flux, Δφ =BA -0 =BA. 

Let the time of withdrawal = Δt. 

emf induced Ɛ =Δφ/Δt

     →Ɛ = BA/Δt

Hence the current in the loop,

 i = Ɛ/R

→i = BA/(R*Δt)

→i*Δt = BA/R

→The flow of charge through any cross-section of wire = BA/R.

            




 

    12.  A long solenoid of radius 2 cm has 100 turns/cm and carries a current of 5 A. A coil of radius 1 cm has 100 turns and a total resistance of 20 Ω is placed inside the solenoid coaxially. The coil is connected to a galvanometer. If the current in the solenoid is reversed in direction, find the charge flown through the galvanometer.        


ANSWER: Magnetic field inside the solenoid, B =µₒni.

  The magnetic flux through the inner coil,

 φ =B*(πr²)*n =πn'r²B

where n' = the total number of turns in the inner coil.

When the current in the solenoid is reversed, the magnetic field becomes -B.

Hence the magnetic flux through the coil now is 

φ' =-πn'r²B

Hence the change in the flux,

  Δφ =φ -φ'

    =2πn'r²B.

Suppose this change occurs in a time Δt, then the emf developed in the coil,

E =Δφ/Δt

For a resistance R of the coil, the current in the coil,

i' =E/R.

  So the charge flown through the galvanometer 

  Q =i'*Δt

     =EΔt/R

     =Δφ/R

     =2πn'r²B/R

     =2πn'r²*µₒni/R

Given values, n =100, r =1 cm =0.01 m, r' =2 cm =0.02 m, n' =100 turns/cm =10⁴ turns/m, i =5 A, R =20 Ω.

Hence Q=2π*100*0.0001*4π*10⁻⁷*10⁴*5/20 C

 =2π²*10⁻⁵ C

 =1.97x10⁻⁴ C

 ≈2x10⁻⁴ C. 





 

    13.  Figure (38-E3) shows a metallic square frame of an edge 'a' in a vertical plane. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the figure. Two boys pull the opposite corners of the square to deform it into a rhombus. They start pulling the corners at t = 0 and displace the corners at a uniform speed u. (a) Find the induced emf in the frame at the instant when the angles at these corners reduce to 60°. (b) Find the induced current in the frame at this instant if the total resistance of the frame is R. (c) Find the total charge that flows through a side of the frame by the time the square is deformed into a straight line.         
Figure for Q-13


ANSWER: (a) The projection of an edge of the square perpendicular to the velocity

  =a.sinθ

(Where θ is the angle between an edge and the direction of velocity.) 

This length will be the length effective for the generation of emf.  

The emf generated in one edge 

 =B*u*(a.sinθ) 

 =auBsinθ

emf generated in four edges 

Ɛ = 4auBsinθ

(Given that the angle at the corners =60°, hence θ = 30°.)

  =4auB*sin30° 

  =2auB. 


(b) Total resistance of the frame = R. 

Hence the induced current in the frame at this moment, 

i = Ɛ/R =2auB/R.  


(c) Initial flux =Ba² 

Final flux =B*0 =0 since the frame is now a straight line and its area will be zero. Change in flux in a time interval say Δt,

Δφ =Ba²-0 =Ba²

Emf induced E' =Δφ/Δt 

Induced current i =E'/R 

→i =Δφ/RΔt

→i*Δt =Δφ/R

Hence the charge flew through one side,

Q =i*Δt

   =Δφ/R

   =Ba²/R.

 

 

 



 

    14.  The north pole of a magnet is brought down along the axis of a horizontal circular coil (figure 38-E4). As a result, the flux through the coil changes from 0.35 weber to 0.85 webers in an interval of half a second. Find the average emf induced during this period. Is the induced current clockwise or anticlockwise as you look into the coil from the side of the magnet?        
Figure for Q-14


ANSWER: Δφ =0.85 -0.35 weber

→Δφ =0.50 weber

Δt = 0.50 s

Hence the induced emf

E =Δφ/Δt

   =0.50/0.50 V

   =1.0 V.

From Lenz's law, the induced current will oppose the cause that has induced it. The cause here is that the downward magnetic flux through the loop is increasing. So the induced current will produce a magnetic flux that is upward. Such flux will be produced by a current in the loop that is anticlockwise when viewed from the side of the magnet.

 


 



 

    15.  A wire-loop confined in a plane is rotated in its own plane with some angular velocity. A uniform magnetic field exists in the region. Find the emf induced in the loop.       


ANSWER: When the loop is rotated in its own plane with some angular velocity, there is no change in the magnetic flux through the loop. Since there is no change in the magnetic flux through the loop, there will be no emf induced. 

    




 

    16.  Figure (38-E5) shows a square loop of side 5 cm being moved towards right at a constant speed of 1 cm/s. The front edge enters the 20 cm wide magnetic field at t = 0. Find the emf induced in the loop (a) at t = 2 s. (b) t = 10 s, (c) t =22 s and (d) t =30 s.       
The figure for Q -16


ANSWER: (a) At t = 0, the magnetic flux through the loop =0.

At t = 2 sec, the area of loop inside the field =2 cm*5 cm =10 cm² =0.001 m².

Flux through the loop at this time =B*A 

=0.6*0.001 weber 

=6x10⁻⁴ weber.

Change in flux in 2 s =6x10⁻⁴ weber

So emf induced =6x10⁻⁴/2 V 

       =3x10⁻⁴ V. 


(b) From t =5 s to t =10 s, the loop is fully inside the magnetic field and there is no change in magnetic flux through the loop at t =10 s. So no emf induced i.e. induced emf =zero. 


(c) At t =20 s, the loop is fully inside the magnetic field. Magnetic flux through the loop =BA 

  =0.6*0.0025 weber 

  =0.0015 weber. 

At t =22 s, the area of the loop inside the magnetic field =5 cm*3 cm 

   =15 cm² 

   =0.0015 m² 

The magnetic flux through the loop at t =22 s, 

=0.6*0.0015 weber 

=0.0009 weber.

The change in magnetic flux through the loop in 2 s =0.0015 -0.0009 weber 

   =0.0006 weber

The emf induced in the loop at t = 22 s, 

  =0.0006/2 V

  =0.0003 V

  =3x10⁻⁴ V. 


(d) At t = 25 s, the loop is totally out of the magnetic field. Hence at t = 30 s, there is no change in the magnetic flux through the loop. So the induced emf in the loop =zero.                    




 

    17.  Find the total heat produced in the loop of the previous problem during the interval 0 to 30 s if the resistance of the loop is 4.5 mΩ.       


ANSWER: The heat will be produced only when there is an induced current in the loop. The induced current will be from t =0 to t = 5 s and then t =20 s to t =25 s. These are at the entry and exit times. At both of these times, the heat produced will be the same.  

   Between t = 0 to t = 5 s, change in magnetic flux, 

Δφ = 0.6*25/10000 weber

     = 0.0015 weber

emf induced, E =Δφ/Δt

→E =0.0015/5 V 

      =0.0003 V

R = 4.5 mΩ

   =0.0045 Ω

Hence the induced current,

  i =E/R

    =0.0003/0.0045 A

   =1/15 A.

The heat produced =i²Rt

    = (1/15)²*0.0045*5 J

    = 1x10⁻⁴ J

The same will be the heat produced during exit, i.e. t = 20 s to t =25 s.

   Hence total heat produced between  t =0 to t =30 s will be

   =2*1x10⁻⁴ J

   =2x10⁻⁴ J.

 



 

    18.  A uniform magnetic field B exists in the cylindrical region of radius 10 cm as shown in figure (38-E6). A uniform wire of length 80 cm and resistance 4.0 Ω is bent into a square frame and is placed with one side along the diameter of the cylindrical region. If the magnetic field increases at a constant rate of 0.010 T/s, find the current induced in the frame.     
Figure 38-E6


ANSWER: Area of the magnetic fields inside the loop, A =πr²/2  

→A =π(10 cm)²/2 =50π cm²

     =0.005π m². 

Magnetic flux through the loop, φ =BA

dφ/dt =A.dB/dt 

(Since A is constant, and given dB/dt = 0.010 T/s) 

Hence emf induced, E =dφ/dt 

→E =A.dB/dt 

   =0.005π*0.010 V 

   =5πx10⁻⁵ V 

R =4 Ω, hence the induced current, 

i =E/R 

  =5πx10⁻⁵/4 A 

  =3.9x10⁻⁵ A.                





 

    19.  The magnetic field in the cylindrical region shown in figure (38-E7) increases at a constant rate of 20.0 mT/s. Each side of the square loop abcd and defa has a length of 1.00 cm and a resistance of 4.00 Ω. Find the current (magnitude and sense) in the wire ad if (a) the switch S₁ is closed but S₂ is open. (b) S₁ is open but S₂ is closed.  (c) both S₁ and S₂ are open and (d) both S₁ and S₂ are closed.  
Figure for Q-19


ANSWER: Area of each square loop, 

A =(1 cm)² =1 cm² =1x10⁻⁴ m².

Magnetic flux through each square,

φ =BA 

Emf induced in a loop, E =dφ/dt 

→E =A.dB/dt

     =1x10⁻⁴*20x10⁻³ V

     =2x10⁻⁶ V.  


(a) When S₁ is closed and S₂ is open.

The current will be induced only in the left loop.

  The total resistance of one loop, R =4*4 Ω 

→ R = 16 Ω 

Current induced, i =E/R

→ i =2x10⁻⁶/16 A

     = 1.25x10⁻⁷ A.

The magnetic field going into the plane of paper is increasing, hence the induced current in the loop will oppose it. So it will produce a magnetic field that is coming out of the plane of the paper. It will happen when the induced current in the left loop is anticlockwise, So the current in the wire ad will be from a to d.


(b) When S₁ is open but S₂ is closed.

In this case, the current will be induced in the right loop only. Since both the loops are similar, the same magnitude of current will flow in this loop.

 So, i = 1.25x10⁻⁷ A.

   Also in this right loop, the induced current will be anticlockwise. So the current in the wire ad will be from d to a.


(c) Both S₁ and S₂ are open.

In this case, emf will be induced in each loop but there will be no current as there is no circuit complete. Hence the current in the wire ad is zero.  


(d) Both S₁ and S₂ are closed 

     In both of the loops, equal emf will be induced which will further induce an equal anticlockwise current in them. Since the wire ad is common to both loops, the direction of currents will be equal and opposite in it. This will result in zero current in the wire ad.

 


 

 

 

 

    20.  Figure (38-E8) shows a circular coil of N turns and radius a, connected to a battery of emf Ɛ through a rheostat. The rheostat has a total length L and resistance R. The resistance of the coil is r. A small circular loop of radius a' and resistance r' is placed coaxially with the coil. The center of the loop is at a distance x from the center of the coil. In the beginning, the sliding contact of the rheostat is at the left end and then onwards it is moved towards the right at a constant speed v. Find the emf induced in the small circular loop at the instant (a) the contact begins to slide and (b) it has slid through half the length of the rheostat. 
Figure for Q-20


ANSWER: The magnetic field at the small circular loop, 

B =µₒiNa²/{2(a²+x²)³ˊ²} 

The magnetic flux at the smaller loop, 

φ =BA 

   =µₒiNa²*πa'²/{2(a²+x²)³ˊ²} 

emf induced in this loop, 

E =dφ/dt 

  ={½πµₒNa²a'²/(a²+x²)³ˊ²}di/dt  

At time t, the resistance of the rheostat, 

r' =R-Rvt/L 

The total resistance of the circuit =r'+r

  =R+r-Rvt/L

The current in the circuit,

i = Ɛ/(r+R-Rvt/L)

di/dt =ƐRv/{L(r+R-Rvt/L)²}

   

Hence the emf in the small loop,

E ={½πµₒNa²a'²/(a²+x²)³ˊ²}*ƐRv/{L(r+R-Rvt/L)²}


(a) At the time t =0 when the contact begins to slide, 

E ={½πµₒNa²a'²/(a²+x²)³ˊ²}*ƐRv/{L(r+R)²}

E =½πµₒNa²a'²ƐRv/{L(a²+x²)³ˊ²(r+R)²}


(b) When the contact slid through half the length, the time

t =½L/v 

Now emf,

E ={½πµₒNa²a'²/(a²+x²)³ˊ²}*ƐRv/{L(r+R-R/2)²}

E =½πµₒNa²a'²ƐRv/{L(a²+x²)³ˊ²(r+½R)²}

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CHAPTER- 34- Magnetic Field

CHAPTER- 29- Electric Field and Potential











CHAPTER- 28- Heat Transfer

OBJECTIVE -I







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CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


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CHAPTER- 24-Kinetic Theory of Gases







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CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




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