Saturday, April 2, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, EXERCISES, Q61 to Q70, Concepts of Physics, Part-II

Electromagnetic Induction


EXERCISES, Q61 to Q7O


    61.  Consider a variation of the previous problem (figure 38-E29). Suppose the circular loop lies in a vertical plane. The rod has a mass m, The rod and the loop have negligible resistances but the wire connecting the O and C has a resistance R. The rod is made to rotate with a uniform angular velocity ⍵ in the clockwise direction by applying a force at the midpoint of OA in a direction perpendicular to it. Find the magnitude of this force when the rod makes an angle θ with the vertical.  



ANSWER: As we have calculated the emf induced in the rod is, 

E = ½B⍵a²

Current in the rod,

i = E/R

  =½B⍵a²/R

Force on the rod due to the current in the magnetic field

F' =iBl

   =(½B⍵a²/R)*B*a

   =½B²⍵a³/R

When the rod makes an angle θ with the vertical, the component of weight in the direction of the applied force F,

W' = mg Cos(90° -θ)

     =mg Sin θ

Diagram for Q-61


The force by the magnetic field due to current, F' will be opposing the motion. Since the rod is made to move with a uniform angular velocity ⍵, Equal and opposite force should be on the rod. Thus,

W' +F =F'

mg Sinθ +F = ½B²⍵a³/R 

→F = ½B²⍵a³/R -mg Sinθ

 This much magnitude of force should be applied.       






    62.  Figure (38-E30) shows a situation similar to the previous problem. All parameters are the same except that a battery of emf Ɛ and a variable resistance R are connected between O and C. The connecting wires have zero resistance. No external force is applied to the rod (except gravity, forces by the magnetic field, and by the pivot). In what way should the resistance R be changed so that the rod may rotate with uniform angular velocity in the clockwise direction? Express your answer in terms of the given quantities and the angle θ made by the rod OA with the horizontal.  
Figure for Q-62


ANSWER: Suppose the rod makes an angle θ with the horizontal and it moves with a uniform angular velocity ⍵. Due to the clockwise movement, the induced emf E will have a positive end at A and negative at O. So in the circuit, the induced emf E and the battery emf Ɛ are connected in series. So the total emf in the circuit =E+Ɛ.

Current in the rod,

i =(E+Ɛ)/R

Force due to this current in the magnetic field will be opposing the movement and the weight component perpendicular to the rod will be balancing it to keep the motion uniform. So,

mg Cosθ =ilB

→mg Cosθ =(E+Ɛ)aB/R

→R =(E+Ɛ)aB/mgCosθ

  =(½B⍵a² +Ɛ)aB/mgCosθ

  =aB(2Ɛ +⍵a²B)/(2mg Cosθ).

    





    63.  A wire of mass m and length l can slide freely on a pair of smooth, vertical rails (figure 38-E31). A magnetic field B exists in the region in the direction perpendicular to the plane of the rails. The rails are connected at the top end by a capacitor of capacitance C. Find the acceleration of the wire neglecting any electric resistance.    
Figure for Q-63


ANSWER: Suppose at any instant the velocity of the wire is v. The emf induced in the wire, 

E =Bvl 

this will be the potential difference across the capacitor. If the charge on the capacitor is Q, then from 

Q =CV 

we have, Q =C*Bvl =CBvl 

Current at any instant in the circuit, 

i =dQ/dt 

  =d(CBvl)/dt 

  =CBl*dv/t

  =CBla, 

where a is the acceleration of the wire.

Force on the wire due to this current in the magnetic field, 

F =iBl

  =(CBla)*B*l 

  =CB²l²a 

This force will be opposing the motion, in the upward direction. There will be the weight of the wire acting downwards, 

W =mg 

The net force on the wire =W-F 

So the acceleration of the wire, 

a =(W-F)/m 

→a =(mg -CB²l²a)/m 

→ma +CB²l²a =mg 

→a =mg/(m +CB²l²). 

                     





    64.  A uniform magnetic field B exists in a cylindrical region, shown dotted in figure (38-E32). The magnetic field increases at a constant rate of dB/dt. Consider a circle of radius r coaxial with a cylindrical region. (a) Find the magnitude of the electric field E at a point on the circumference of the circle. (b) Consider a point P on the side of the square circumscribing the circle. Show that the component of the induced electric field at P along ba is the same as the magnitude found in part (a).    
Figure for Q-64


ANSWER: (a) The work done by the induced electric field E in moving a unit charge completely around a circuit is the induced emf in the circuit. So the induced emf in the given circular path,

Ɛ = ∮E.dl

Since the directions of E and dl are the same on the circular path,

Ɛ = ∮E.dl

   =E*2πr, 

{Due to symmetry E will be the same everywhere on the circle}

→Ɛ =2πrE

Now consider the magnetic flux through the circular area.

φ =B*A =πr²B at any instant.

Induced emf due to the changing magnetic flux is given as 

Ɛ =dφ/dt

  =πr²*dB/dt

Equating the two expressions we get,

2πrE =πr²dB/dt

→E =½r.dB/dt

It will be the magnitude of the electric field at a point on the given circle.


(b) Join the point P to the center of the circle O. Let OP = r'. Now draw a circle with center O and radius r'. It will pass through the point P. Let OP makes an angle θ with the vertical. The direction of the electric field E' at P will be tangential at that point as shown in the diagram below.
Diagram for Q-64

  The magnitude of E' as from (a) above will be,

E' =½r'dB/dt.

  The angle between E' and the side of the square will also be equal to θ. Hence the component of E' along ba,

= E'*Cosθ

=½r'Cosθ.dB/dt

=½rdB/dt.

{Because r'Cosθ =r}

It is the same as derived in (a).

 





    65.  The current in an ideal, long solenoid is varied at a uniform rate of 0.01 A/s. The solenoid has 2000 turns/m and its radius is 6.0 cm. (a) Consider a circle of radius 1.0 cm inside the solenoid with its axis coinciding with the axis of the solenoid. Write the change in the magnetic flux through the circle in 2.0 seconds. (b) Find the electric field induced at a point on the circumference of the circle. (c) Find the electric field induced at a point outside the solenoid at a distance 8.0 cm from its axis.      



ANSWER: Magnetic field inside a long solenoid, 

B =µₒni 

(a) Consider a circle of radius r inside the solenoid with its axis coinciding with the axis of the solenoid. Flux through this circle, 

φ =BA =µₒni*πr² 

   =µₒπnr²i 

Rate of change of flux through this circle,

dφ/dt =µₒπnr²di/dt 

Here, n =2000 turns/m, r =0.01 m, di/dt =0.01 A/s

→dφ/dt=(4πx10⁻⁷)π*2000*1x10⁻⁴*0.01 weber/s

       =7.9x10⁻⁹ Weber/s 

Hence the change in magnetic flux in 2 s,

    =(7.9x10⁻⁹ Weber/s)*(2 s) 

   =1.6x10⁻⁸ Weber.  


(b) Let the electric field induced at a point on the circle =E. Then, 

E*2πr =dφ/dt {=emf induced}

→E =(µₒπnr²di/dt)/(2πr) 

  =½µₒnrdi/dt

  =½(4πx10⁻⁷)*2000*0.01*0.01 V/m 

 =4πx10⁻⁸ V/m 

 =1.2x10⁻⁷ V/m. 


(c) Consider a circle of radius 8.0 cm with its axis coinciding with the axis of the solenoid. Flux changing through this circle is the same as the flux changing through the solenoid. Hence, 

dφ/dt =µₒπnr²di/dt 

  =(4πx10⁻⁷)*π*2000*(0.06)²*0.01 W/s

 =0.288π²x10⁻⁷ weber/s 

If E' is the magnitude of the electric field at a point on the 8.0 cm radius circle outside the solenoid, then

E'*2π*0.08 =dφ/dt 

→E' =0.288π²x10⁻⁷/(0.16π) V/m 

    =1.8πx10⁻⁷ V/m 

   =5.6x10⁻⁷ V/m.  

                 





    66. An average emf of 20 V is induced in an inductor when the current in it is changed from 2.5 A in one direction to the same value in the opposite direction in 0.1 s. Find the self-inductance of the inductor.    

ANSWER: Induced emf Ɛ =L*di/dt 

Ɛ =20 V, di/dt ={2.5 -(-2.5)}/0.1 A/s

→di/dt =50 A/s

Hence, 20 =L*50

→L =20/50 H

     =0.40 H.

 





    67.  A magnetic flux of 8x10⁻⁴ weber is linked with each turn of a 200-turn coil when there is an electric current of 4 A in it. Calculate the self-inductance of the coil.  

ANSWER: The magnetic flux linked with each turn, 

φ =8x10⁻⁴ Weber, 

Number of turns, N =200, 

i =4 A. If the sef-inductance of the coil is L, then

  Nφ =Li

→L =Nφ/i 

  =200*8x10⁻⁴/4 H 

  =4x10⁻² H.           





    68.  The current in a solenoid of 240 turns, having a length of 12 cm and a radius of 2 cm, changes at a rate of 0.8 A/s. Find the emf induced in it.    

ANSWER: B =µₒni, A =πr², di/dt =0.8 A/s, r =2 cm =0.02 m.

n =240/0.12 turns/m =2000 turns/m. 

Magnetic field inside the solenoid,

  B=µₒni. 

Magnetic flux through each turn of the solenoid, 

φ' =BA =µₒπnr²i      

Magnetic flux through N turns of the solenoid,

φ =µₒπnNr²i

emf induced =dφ/dt 

 =µₒπnNr²di/dt 

=(4πx10⁻⁷)*π*2000*240*(0.02)²*0.8 V 

=6.0x10⁻⁴ V.  



   




    69.  Find the value of t/𝞽 for which the current in an LR circuit builds up to (a) 90%, (b) 99%, and (c) 99.9% of the steady-state value.   

 

ANSWER: If iₒ is the steady-state current, then current at time t is,  

i =iₒ(1 -e-t/𝞃)

→1 -e-t/𝞃 =i/iₒ

→e-t/𝞃 = 1 -i/iₒ

→t/𝞃 = -ln(1 -i/iₒ)

(a) For the current builds up to 90%,

i/iₒ =0.90

So, t/𝞃 =-ln(1-0.90) =-ln (0.1)

         = 2.3


(b) For the current builds up to 99%,

i/iₒ =0.99

So, t/𝞃 =-ln(1-0.99) =-ln(0.01)

          =4.6


(c) For the current builds up to 99.9%,

i/iₒ =0.999

So, t/𝞃 =-ln(1-0.999) =-ln(0.001)

        =6.9 

       





    70.  An inductor coil carries a steady-state current of 2.0 A when connected across an ideal battery of emf 4.0 V. If its inductance is 1.0 H, find the time constant of the circuit.   


ANSWER: iₒ =2.0 A, Ɛ =4.0 V, Hence the resistance of the coil,

R =Ɛ/iₒ =4.0/2.0 Ω =2.0 Ω

Given L =1.0 H

So the time constant, 

𝝉 =L/R =1.0/2.0 s =0.5 s.

  

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Links to the Chapters






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

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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

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CHAPTER- 6 - Friction

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

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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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