Thursday, April 7, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, EXERCISES, Q71 to Q80, Concepts of Physics, Part-II

Electromagnetic Induction


EXERCISES, Q71 to Q8O


    71.  A coil having an inductance of 2.0 H and resistance of 20 Ω is connected to a battery of emf 4.0 V. Find (a) the current at the instant 0.20 s after the connection is made and (b) the magnetic field energy at this instant.  


ANSWER: Given L =2.0 H, R =20 Ω. 

Hence 𝝉 =L/R =2.0/20 s =0.1 s, and 

Steady-state current iₒ =4.0/20 A =0.20 A.

(a) So, from i =iₒ(1 -e-t/𝝉)

Current after 0.20 s, 

=0.20*(1-e-0.20/0.10) A

=0.20*(1-e⁻²) A

=0.20*0.86 A

=0.17 A.


(b) The magnetic field energy at this instant 

U =½Li²

  =½*2.0*(0.17)² J

  ≈0.03 J.  





 

    72.  A coil of resistance 40 Ω is connected across a 4.0 V battery. 0.10 s after the battery is connected, the current in the coil is 63 mA. Find the inductance of the coil.   


ANSWER: R =40 Ω, Ɛ =4.0 V. t =0.10 s, i =63 mA =0.063 A.

iₒ =Ɛ/R =4.0/40 A =0.10 A

Hence, i = iₒ(1 -e-t/𝝉)

→0.063 =0.10*(1-e-0.10/𝝉)

→e-0.10/𝝉 =1 -0.63 =0.37

→-0.10/𝝉 =ln 0.37 =-0.99

→𝝉 =0.10

→L/R =0.10

→L =0.10*R =0.10*40 H 

→L =4.0 H.

         




 

    73.  An inductor of inductance 5.0 H, having a negligible resistance, is connected in series with a 100 Ω resistor and a battery of emf 2.0 V. Find the potential difference across the resistor 20 ms after the current is switched on.   


ANSWER: Steady-state current, 

iₒ =2.0/100 A =0.02 A, 

t =20 ms =0.02 s

𝝉 =L/R =5.0/100 s =0.05 s

Hence, i =iₒ(1-e-t/𝝉)

→i =0.02*(1-e-0.02/0.05)

   =0.02*(1-e-0.4)

   =0.0066 A

Hence the potential difference across the resistor,

=iR

=0.0066*100 V

=0.66 V.

          




 

    74.  The time constant of an LR circuit is 40 ms. The circuit is connected at t =0 and the steady-state current is found to be 2.0 A. Find the current at (a) t =10 ms, (b) t =20 ms, (c) t =100 ms and (d) t = 1 s.  


ANSWER: 𝝉 =40 ms =0.040 s

iₒ =2.0 A.

From i =iₒ(1 -e-t/𝝉)

we have,  

(a) At t =10 ms,

t/𝝉 =10/40 =0.25

Hence the current at t =10 ms,

i =2.0*(1 -e-0.25) A

  =0.44 A. 


(b) At t = 20 ms,

t/𝝉 =20/40 =0.5

Hence the current at t =20 ms,

i =2.0*(1-e-0.5) A

  =0.79 A.


(c) At t = 100 ms,

t/𝝉 =100/40 =2.5

Hence the current at t =100 ms,

i =2.0*(1 -e-2.5) A

  =1.8 A.


(d) At t =1 s,

 t/𝝉 =1/0.04 =25

Hence the current at t =1 s,

i =2.0*(1 -e⁻²⁵) A

  =2.0 A.  




 

    75.  An LR circuit has L = 1.0 H and R =20 Ω. It is connected across an emf of 2.0 V at t =0. Find di/dt at (a) t =100 ms, (b) t =200 ms and (c) t =1.0 s.  


ANSWER: Steady-state current, 

iₒ =V/R =2.0/20 A =0.10 A 

𝝉 =L/R =1.0/20 s =0.05 s

Hence the current at time t is,

i =iₒ(1 -e-t/𝝉)

So, di/dt =iₒ{-e-t/𝝉*(-1/𝝉)}

→di/dt =iₒe-t/𝝉/𝝉

          =0.10 e-t/0.05/0.05 A/s

        =2 e-20t A/s


(a) at t =100 ms =0.10 s,

di/dt = 2*e-20*0.10 A/s

       = 2*e-2

      = 0.27 A/s


(b) at t = 200 ms =0.20 s

di/dt =2*e-20*0.20 A/s

      =2*e-4 A/s

     =0.036 A/s


(c) at t = 1.0 s

di/dt =2*e-20*1.0 A/s

      = 2/e20 A/s

     = 4.1x10⁻⁹ A/s.

 




 

    76.  What are the values of the self-induced emf in the circuit of the previous problem at the times indicated therein?  


ANSWER: The magnitude of self-induced emf is given by, 

Ɛ = L*di/dt

For the previous problem

   di/dt =2 e-20t A/s


(a) At t =100 ms, the calculated di/dt =0.27 A/s

Hence the value of self-induced emf  at t =100 ms, 

 Ɛ =L*di/dt 

   =1.0*0.27 V

   =0.27 V.


(b) At t = 200 ms, the calculated value of di/dt =0.036 A/s.

Hence the value of self-induced emf, Ɛ

Ɛ =L*di/dt

  =1.0*0.036 V

  =0.036 V.


(c) At t = 1.0 s, the calculated value of di/dt =4.1x10⁻⁹ A/s

Hence the value of self-induced emf,

 Ɛ =L*di/dt

   =1.0*4.1x10⁻⁹ V

   =4.1x10⁻⁹ V.  

          




 

    77.  An inductor coil of inductance 20 mH having resistance 10 Ω is joined to an ideal battery of emf 5.0 V. Find the rate of change of the induced emf at (a) t =0, (b) t = 10 ms, and (c) t =1.0 s.  


ANSWER: L =20 mH =0.02 H, 

R =10 Ω, 𝝉 =L/R =0.02/10 =0.002 s

Steady-state current,

 iₒ =E/R =5.0/10 =0.5 A

At any instant t/𝝉 = t/0.002 =500t

Hence the current at any instant,

i =iₒ(1 -e-t/𝝉)

  =0.5*(1 -e⁻⁵⁰⁰t)

Hence di/dt = 0.5*{-e-500t*(-500)}

 →di/dt =250e-500t


So the self-induced emf at time t is,

Ɛ =-Ldi/dt

  =-0.02*250 e-500t

  =-5 e-500t

The rate of change of the induced emf at time t,

dƐ/dt =-5*(-500) e-500t

    =2500 e-500t


(a) at t =0,

dƐ/dt =2500*e⁰ V/s

     =2500 V/s. 

     =2.5x10³ V/s.


(b) At t =10 ms =0.01 s

  dƐ/dt =2500*e-500*0.01 V/s

       =2500*e⁻⁵ V/s

       ≈17 V/s


(c) At t =1.0 s,

dƐ/dt =2500*e⁻⁵⁰⁰ V/s

       ≈ 0 V/s.

 




 

    78.  An LR circuit contains an inductor of 500 mH, a resistor of 25.0 Ω, and an emf of 5.0 V in series. Find the potential difference across the resistor at t = (a) 20.0 ms, (b) 100 ms and (c) 1.00 s.  


ANSWER: L =500 mH =0.50 H, 

R =25 Ω, 𝝉 =L/R =0.5/25 s =0.02 s

Steady-state current, 

iₒ =5.0/25 A =0.20 A

If i is the current at any time t, then

The potential difference across the resistor =iR

  =R*iₒ(1 -e-t/𝝉)

  =25*0.20*(1-e-50t)

  =5(1 -e-50t)


(a) At t = 20.0 ms =0.02 s

P.D. =5*(1-e⁻¹) V =3.16 V.


(b) At t =100 ms =0.10 s

P.D. =5*(1-e⁻⁵) V =4.97 V


(c) At t =1.00 s

P.D. =5*(1-e⁻⁵⁰) V

      =5.00 V. 




 

    79.  An inductor coil of resistance 10 Ω and inductance 120 mH is connected across a battery of emf 6.0 V and internal resistance 2 Ω. Find the charge that flows through the inductor in (a) 10 ms, (b) 20 ms, and (c) 100 ms after the connections are made.  


ANSWER: R =10 Ω, r =2 Ω,

L =120 mH =0.12 H 

𝝉 =L/(R+r) =0.12/12 s =0.01 s

Ɛ =6.0 V

Hence steady-state current 

iₒ =6.0/(10+2) A =0.5 A.

The current at any time t is,

i =iₒ(1 -e-t/𝝉)

The flow of charge in a small time dt is,

dQ =idt

→dQ =iₒ(1 -e-t/𝝉)dt

Hence Q =∫dQ

    =∫iₒ(1 -e-t/𝝉)dt

   =iₒ[t -e-t/𝝉/(-1/𝝉)]

   =iₒ[t +𝝉e-t/𝝉]

Keeping the limits of integration between t =0 to t =t we get,

 Q =iₒ{t+𝝉e-t/𝝉 -𝝉}

  =0.5{t+0.01 e-100t -0.01}

  =5x10⁻³{100t+e-100t -1}


(a) Up till t=10 ms =0.01 s, 

100t =100*0.01 s =1.0 s 

Charge flows through the inductor, 

  =5x10⁻³{1+e⁻¹-1} C

  =5x10⁻³/e C

  =1.8x10⁻³ C

  =1.8 mC.


(b) For t =20 ms =0.02 s

100t =100*0.02 s =2 s

Charge flows through the inductor in 20 ms, 

Q =5x10⁻³{2 +e⁻² -1} C

   =5x10⁻³*1.14 C

   =5.7x10⁻³ C

   =5.7 mC.


(c) For t =100 ms =0.10 s 

100t =100*0.10 s =10 s

Hence the flow of charge through the inductor in 100 ms, 

Q =5x10⁻³{10+e⁻¹⁰-1} C

  =5x10⁻³{9+e⁻¹⁰} C

  =5x10⁻³*9.00 C

  =45x10⁻³ C

  =45 mC.

 

 

        




 

    80.  An inductor coil of the inductance of 17 mH is constructed from copper wire of a length of 100 m and a cross-sectional area of 1 mm². Calculate the time constant of the circuit if this inductor is joined across an ideal battery. The resistivity of copper =1.7x 10⁻⁸ Ω-m.   


ANSWER: Inductance, L =17 mH =0.017 H

Length of the wire, l =100 m, 

The cross-sectional area of the wire, A =1 mm² = 1x10⁻⁶ m².  

Given resistivity,

⍴ =1.7x10⁻⁸ Ω-m

The relation between the resistance of a wire R and resistivity is,

⍴ =R*(A/l)

→R =⍴l/A

    =1.7x10⁻⁸*100/1x10⁻⁶ Ω

   =1.7 Ω

Hence the time constant of the circuit,

𝜏 =L/R

  =0.017/1.7 s

  =0.01 s

  =10 ms.

 

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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 (11-20)

Click here for → Exercises (21-30)

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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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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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