Tuesday, April 12, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, EXERCISES, Q81 to Q90, Concepts of Physics, Part-II

Electromagnetic Induction


EXERCISES, Q81 to Q9O


    81.  An LR circuit having a time constant of 50 ms is connected with an ideal battery of emf Ɛ. Find the time elapsed before (a) the current reaches half its maximum value, (b) the power dissipated in heat reaches half its maximum value, and (c) the magnetic field energy stored in the circuit reaches half its maximum value. 


ANSWER: Given L/R =50 ms =0.05 s,

(a) Since i =iₒ(1 -e-tR/L) 

For i =iₒ/2 

iₒ/2 =iₒ(1 -e-t/0.05) 

→1 -e-20t = 0.5 

→e-20t =0.5 

→-20t =ln 0.5 =-0.693 

→t = 0.035 s =35 ms.  


(b) The power dissipated in heat at an instant is,

P = i²R

  =iₒ²R(1 -e-tR/L)²

  =iₒ²R(1 -e-20t)², 

while the maximum power dissipated in heat is,   

Pₒ = iₒ²R  

The given condition is,

P = Pₒ/2. So we have,

  iₒ²R(1 -e-20t)² =iₒ²R/2 

→(1 -e-20t)² = 1/2  

→ 1-e-20t =土1/√2  ------(i)

→e-20t = 1 土1/√2 =1.707 or 0.292 

→-20t = ln 1.707 or ln 0.292

→-20t = 0.53 or -1.23 

Since for a positive value of t, the value of -20t ≠ 0.53, hence we have 

-20t = -1.23 

→t =0.061 s =61 ms.

 

(c) The magnetic field energy stored in the circuit at any time t is given as, 

U =½Li², 

The maximum value of this stored energy is when the current is maximum i.e. = iₒ. So the maximum magnetic field energy stored in the circuit,

Uₒ =½Liₒ² 

The given condition is, 

U = ½Uₒ 

→½Li² =½*½Liₒ² 

→i² =½iₒ² 

→iₒ²(1 -e-20t)² =½iₒ² 

→1 -e-20t = ±1/√2

It is the same equation (i) in the (b) part of the solution above. Hence after we solve this equation we again get t = 61 ms.           





 

    82.  A coil having an inductance L and a resistance R is connected to a battery of emf Ɛ. Find the time taken for the magnetic energy stored in the circuit to change from one-fourth of the steady-state value to half of the steady-state value.  


ANSWER: At steady-state the magnetic energy stored in the circuit, 

Uₒ =½Liₒ² 

If t is the time to reach the magnetic field enrgy value one fourth the steady-state value and the current in the circuit at this time is i, then 

¼Uₒ =½Li² 

→¼*½Liₒ² =½Liₒ²(1 -e-tR/L)² 

→1 -e-tR/L =±(1/2) 

→e-tR/L = 3/2 or 1/2

→-tR/L =ln (3/2) or ln (1/2)

(ln 3/2) will have positive value hence we can take only (ln 1/2) on the RHS. So,

-tR/L =ln 1/2 =ln 1 -ln 2 =-ln 2

→t =𝜏 ln 2

 Now let t' be the time to reach the magnetic energy value half the steady-state and the current is i'. Then

U' =½Li'² 

But U' =½Uₒ

→½Liₒ²(1 -et'R/L)² =½*½Liₒ²

→1 -et'R/L =±1/√2

→e-t'R/L =1±(1/√2) 

→e-t'R/L =(√2+1)/√2 or (√2-1)/√2

→-t'R/L =ln(√2+1)/√2 or ln(√2-1)/√2

Only the later value on the RHS is negative hence,

-t'/𝜏 =ln (√2-1)/√2

→t' =𝜏 ln √2/(√2-1)

Hence the required time is =t' -t

  =𝜏 ln √2/(√2-1) - 𝜏 ln 2

  =𝜏 ln √2/2(√2-1)

  =𝜏 ln {1/(2 -√2). 

       




 

    83.  A solenoid having inductance 4.0 H and resistance 10 Ω is connected to a 4.0 V battery at t =0. Find (a) the time constant, (b) the time elapsed before the current reaches 0.63 of its steady-state value, (c) the power delivered by the battery at this instant, and (d) the power dissipated in Joule heating at this instant.      


ANSWER: (a) Given, L = 4.0 H,

 R =10 Ω 

Hence the time-constant

𝜏 =L/R =4.0/10 s =0.40 s.


(b) We know that in one time constant the current reaches 0.63 of its steady-state value. 

Hence the required time, t = 𝜏 =0.40 s.

Otherwise, we can solve it by

063 iₒ =iₒ(1-e-t/𝜏)

→e-t/𝜏 =1 -0.63 =0.37 

→-t/0.40 =ln 0.37 = -1

→t =0.40 s 


(c) In steady-state, iₒ =Ɛ/R

→iₒ =4.0/10 A =0.40 A

The power delivered by the battery at the instant when i =0.63 iₒ is

P =Ɛi

   =0.63Ɛiₒ

   =0.63*4.0*0.40 W

   =1.0 W.


(d) The power dissipated in joule heating at this time is,

P' =i²R

   =(0.63iₒ)²R

   =(0.63*0.40)²*10 W

   =0.64 W.


     




 

    84.  The magnetic field at a point inside a 2.0 mH inductor coil becomes 0.80 of its maximum value in 20 µs when the inductor is joined to a battery. Find the resistance of the circuit.       


ANSWER: Let the resistance of the coil =R and the maximum value of the current in it is iₒ. Hence the maximum value of the magnetic field inside the coil,

Bₒ =µₒniₒ 

Let the current in the coil be 'i' when the value of the magnetic field inside it reaches 0.80Bₒ. So, 

0.80Bₒ =µₒni 

→0.80µₒniₒ =µₒni 

→0.80iₒ =iₒ(1-e-tR/L) 

→0.80 =1 -e-20x10⁻⁶R/0.002 

→e-0.01R =1 -0.80 =0.20 

→-0.01R =ln 0.20 =-1.61 

→R =1.61/0.01 Ω =161 Ω 

         




 

    85.  An LR circuit with emf Ɛ is connected at t =0. (a) Find the charge Q which flows through the battery from 0 to t. (b) Calculate the work done by the battery during this period. (c) Find the heat developed during this period. (d) Find the magnetic field energy stored in the circuit at time t. (e) Verify that the results in the three parts above are consistence with energy conservation.        


ANSWER: (a) Current at time t after the connection, 

i =iₒ(1 -e-tR/L) 

In a small time interval dt, the small amount of charge flow, 

dQ =i dt  

→dQ =iₒ(1 -e-tR/L)dt 

Hence total charge flow 

Q =∫dQ

  =iₒ∫(1-e-tR/L)dt

  =iₒ[t +(L/R)e-tR/L]

For the limit of time from 0 to t,

Q =iₒ{t+(L/R)e-tR/L-L/R}

   =(Ɛ/R){t -(L/R)(1 -e-tR/L)}

   =(Ɛ/R){t -(L/R)(1 -x)}, 

where x = e-tR/L.


(b) W =Work-done by the battery during this period = Total charge flow in this period X emf

=(Ɛ/R){t -(L/R)(1-x)}*Ɛ

=(Ɛ²/R){t -(L/R)(1-x)},

Where x =e-tR/L


(c) Heat developed during this period 

H = ∫i²Rdt

 =iₒ²R∫(1 -e-tR/L)²dt

 =iₒ²R∫(1-2e-tR/L +e-2tR/L)dt 

=iₒ²R[t +(2L/R)e-tR/L -(L/2R)e-2tR/L]

  Putting the limit of time between 0 to t we get,

 H =(Ɛ²/R²)R{t+(2L/R)x-(L/2R)x²-(2L/R) +(L/2R)}

=(Ɛ²/R){t -(L/2R)(x²-1+4-4x)}

=(Ɛ²/R){t -(L/2R)(3-4x+x²)}.


(d) Magnetic field energy stored in the circuit at time t,

U =½Li²

  =½Liₒ²(1-e-tR/L)²

 =½L(Ɛ²/R²)(1-e-tR/L)² +e-2tR/L)

=(LƐ²/2R²)(1 -x)².


(e) From the law of conservation of energy, the total work done by the battery should be equal to the sum of heat developed during this period and the magnetic field energy stored in the circuit. i.e.

W =H +U

Let us verify it.

H+U =(Ɛ²/R){t-(L/2R)(3-4x+x²)} +(LƐ²/2R²)(1-x)²}

 =(Ɛ²/R){t-(L/2R)(3-4x+x²)+(L/2R)(1-2x+x²)}

 =(Ɛ²/R){t-(L/2R)(3-4x+x²-1+2x-x²)}

 =(Ɛ²/R){t-(L/2R)(2-2x)}

 =(Ɛ²/R){t-(L/R)(1-x)}

 = W. 

Hence verified.     



 

 

   


    86.  An inductor of inductance 2.00 H is joined in series with a resistor of resistance 200 Ω and a battery of emf 2.00 V. At t = 10 ms, find (a) the current in the circuit. (b) the power delivered by the battery, (c) the power dissipated in heating the resistor, and (d) the rate at which energy is being stored in the magnetic field.        


ANSWER: (a) L =2.00 H, R =200 Ω,

𝜏 =L/R =2.00/200 s =0.01 s

Ɛ =2.0 V, So iₒ =Ɛ/R =2.0/200 A =0.01 A

t =10 ms =0.010 s, t/𝜏 = 1.

Hence the current at time t is,

i =iₒ(1-e-t/𝜏)

  =0.01(1 -e⁻¹) A

  =0.0063 A

  =6.3 mA.


(b) The power delivered by the battery at t =10 ms is

 =Ɛi

 =(2.00 V)*(6.3 mA)

 =12.6 mW.


(c) The power dissipated in heating the resistor, 

=i²R

=(6.3x10⁻³)²*200 W

≈80x10⁻⁴ W

=8.0 mW.


(d) The energy stored in the magnetic field at time t is

U =½Li²

 =½Liₒ²(1-e-t/𝜏)²

Hence the rate of energy storage, 

dU/dt=½Liₒ²*2(1-e-t/𝜏)*(-e-t/𝜏)(-1/𝜏)

  =(Liₒ²/𝜏)(1-e⁻¹)*e⁻¹

{Since t/𝜏 =1}

 =(2*0.01²/0.01)(0.63)(0.37) W/s

 =0.0046 W/s

 =4.6 mW/s.

       





 

    87.  Two coils A and B have inductances of 1.0 H and 2.0 H respectively. The resistance of each coil is 10 Ω. Each coil is connected to an ideal battery of emf 2.0 V at t =0. Let iA and iB be the currents in the two circuits at time t. Find the ratio iA/iB at (a) t = 100 ms, (b) t =200 ms and (c) t =1 s.      


ANSWER: For the coil A 

  𝜏 =L/H =1.0/10 s =0.10 s

For coil B, 𝜏' =2.0/10 s =0.20 s

Steady-state current for each coil, 

iₒ =2.0/10 A =0.20 A

Current at any time t in the coil A,

iA =iₒ(1 -e-t/𝜏) =iₒ(1 -e-10t)

and in the coil B,

iB =iₒ(1 -e-t/𝜏') =iₒ(1 -e-5t)

Hence at any time t, the ratio of the currents in coils

iA/iB =(1 -e-10t)/(1 -e-5t)


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

10t =10*0.10 =1 and 5t =5*0.10 =0.50

Hence the ratio,

iA/iB =(1-e⁻¹)/(1-e⁻⁰῾⁵)

      =0.63 A/0.39 A 

      =1.6


(b) At t =200 ms =0.20 s

10t =10*0.20 =2 and 5t =5*0.20 =1.0

Hence the ratio

 iA/iB =(1-e⁻²)/(1-e⁻¹)

         =0.86 A/0.63 A

         ≈1.4


(c) At t =1 s, 10t =10 and 5t =5

Hence the ratio

iA/iB =(1-e⁻¹º)/(1-e⁻⁵)

        =0.999 A/0.993 A

       ≈ 1.

 




 

    88.  The current in a discharging LR circuit without the battery drops from 2.0 A to 1.0 A in 0.10 s. (a) Find the time constant of the circuit. (b) If the inductance of the circuit is 4.0 H, what is its resistance?       


ANSWER: For a discharging LR circuit, the current at time t is given as,

i =iₒe⁻t/𝜏, where tau is the time constant.


(a) Given iₒ =2.0 A, 

at t =0.10 s, i =1.0 A. So

1.0 =2.0 e-0.10/𝜏

→e-1/10𝜏 =0.5

→-1/10𝜏 =ln 0.5 = -0.69

→6.9𝜏 =1

→𝜏 =1/6.9 =0.14 s.


(b) Given L =4.0 H. So

𝜏 =L/R 

→0.14 =4.0/R

→R =4.0/0.14 Ω

      =28.6 Ω


       




  

    89.  A constant current exists in an inductor-coil connected to a battery. The coil is short-circuited and the battery is removed. Show that the charge flew through the coil after the short-circuiting is the same as that which flows in one time constant before short-circuiting.     


ANSWER: Charge flown after short-circuiting

Q =∫dQ

  =∫idt

  =iₒ∫e-t/𝜏 dt

  =iₒ[e-t/𝜏]*(-𝜏)

  =-iₒ𝜏[e-t/𝜏]

For the total charge flown through the coil after short-circuiting, we take the limit of time from t =0 to t =∞.

So, Q =-iₒ𝜏{0 -1} =iₒ𝜏

Before short-circuiting, a steady current iₒ is flowing through the coil. Hence in one time constant t =𝜏, the charge flown through the coil

Q' =iₒ*t

    =iₒ*𝜏

    =iₒ𝜏

    =Q.

 Proved.    





    90.  Consider the circuit shown in figure (38-E33). (a) Find the current through the battery a long time after the switch S is closed. (b) Suppose the switch is again opened at t =0. what is the time constant of the discharging circuit? (b) Find the current through the inductor after one time constant. 
Diagram for Q-90

 

ANSWER: (a) A long time after the switch is closed, the steady-state is reached. In this case, the current iₒ through the battery is the same as the current through the equivalent resistance R. For the parallel connection

1/R =1/R₁ +1/R₂ 

→R =R₁R₂/(R₁+R₂) 

Hence,

iₒ =Ɛ/R 

   =Ɛ(R₁+R₂)/R₁R₂ 


(b) When the switch is again opened, the discharging circuit completes through the upper loop leaving the battery branch. In this circuit, both resistors are in series. So total resistance of the circuit, R =R₁+R₂. Inductance in the circuit is L. Hence the time constant of the discharging circuit,

𝜏 =L/R

  =L/(R₁+R₂)


(c) Suppose the steady-state current through the inductor is i'. 

So, i'R₁ =Ɛ

→i' =Ɛ/R₁ 

It will be the initial current at time t =0 for the discharging circuit. Hence after one time constant, i.e. t =𝜏, the current through the inductor is

i =i' e-t/𝜏

  =(Ɛ/R₁)e⁻¹

  =Ɛ/(R₁e).

              

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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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