Wednesday, April 13, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, EXERCISES, Q91 to Q98, Concepts of Physics, Part-II

Electromagnetic Induction


EXERCISES, Q91 to Q98


    91.  A current of 1.0 A is established in a tightly wound solenoid of a radius of 2 cm having 1000 turns/meter. Find the magnetic energy stored in each meter of the solenoid.    


ANSWER: Number of turns of the solenoid, n =1000 turns/meter. Current, i =1.0 A, radius of solenoid, r =2 cm =0.02 m.  

Hence the magnetic field inside it is, 

B =µₒni 

 =4πx10⁻⁷*1000*1.0 T 

  =4πx10⁻⁴ T

Energy density inside the solenoid, 

u =B²/2µₒ 

  =16π²x10⁻⁸/(2*4πx10⁻⁷) 

  =0.2π J/m³  

The volume of each meter of the coil, 

v =πr²l 

   =π*(0.02)²*1 m³ 

   =4πx10⁻⁴ m³

Hence the magnetic energy stored in each meter of the solenoid, 

=u*v

=0.2π*4πx10⁻⁴ J

=7.9x10⁻⁴ J.

         




 

    92.  Consider a small cube of volume 1 mm³ at the center of a circular loop of radius 10 cm carrying a current of 4 A. Find the magnetic energy stored inside the cube.   


ANSWER: In comparison to the sizes of the loop and the cube at the center, the magnetic field at the cube will be fairly uniform. The magnetic field at the center, 

B =µₒi/2a 

Hence the energy density at the center, 

u =B²/2µₒ 

   =µₒ²i²/(4a²*2µₒ) 

   =µₒi²/(8a²) 

  =4πx10⁻⁷*4²/(8*0.10²) J/m³

  =8πx10⁻⁵ J/m³

The volume of the cube,

v =1 mm³ =1x10⁻⁹ m³

Hence the magnetic energy stored in the cube,

U =u*v

   =8πx10⁻⁵*1x10⁻⁹ J 

   =8πx10⁻¹⁴ J. 


      



    

    93.  A long wire carries a current of 4.00 A. Find the energy stored in the magnetic field inside a volume of 1.00 mm³ at a distance of 10.0 cm from the wire.    


ANSWER: The magnetic field at a distance of 10.0 cm from the long current-carrying wire, 

B =µₒi/(2πd) 

 Magnetic energy density at this place, 

u =B²/2µₒ 

  =µₒ²i²/(4π²d²*2µₒ) 

  =µₒi²/(8π²d²) 

  =4πx10⁻⁷*4²/(8π²*0.1²) J/m³ 

  =2.55x10⁻⁵ J/m³  

Hence the magnetic energy stored in a volume of 1.00 mm³ is, 

U =2.55x10⁻⁵*1x10⁻⁹ J 

   =2.55x10⁻¹⁴ J. 

             




 

    94.  The mutual inductance between two coils is 2.5 H. If the current in one coil is changed at the rate of 1 A/s what will be the emf induced in the other coil?    


ANSWER: Given, M =2.5 H. 

Rate of change of current in the coil, 

di/dt =1 A/s. 

emf induced in the other coil, 

Ɛ =-M*di/dt 

  =-2.5*1 V 

  =-2.5 V.

So the emf induced in the other coil will be 2.5 V such that it will oppose the change of flux in the first coil.          




 

    95.  Find the mutual inductance between the straight wire and the square loop of the figure (38-E27).   
The figure for Q-95


ANSWER: We have calculated the emf induced in the coil in the solution of Q-57(b) of this chapter as, 

Ɛ =(µₒaiₒ⍵.cos⍵t/2π) ln(1+a/b)

(Link - Exercises, Q51 to 60)

If M is the mutual inductance between them, then it's magnitude 

Ɛ =M*di/dt

(µₒaiₒ⍵.cos⍵t/2π) ln(1+a/b) =M.d(iₒsin⍵t)/dt

→(µₒaiₒ⍵.cos⍵t/2π) ln(1+a/b) =M.iₒ⍵cos⍵t 

→(µₒa/2π)*ln (1 +a/b) =M

→M = (µₒa/2π)*ln (1 +a/b)

 

      




 

    96.  Find the mutual inductance between the circular coil and the loop shown in figure (38-E8).   
The figure for Q-96


ANSWER: As we have solved in Q-20 of this exercise, emf-induced

Ɛ ={½πµₒNa²a'²/(a²+x²)3/2}*di/dt

(Link - Exercises, Q11 to Q20)

If M is the mutual inductance between the coils, then

Ɛ =M*di/dt 

Comparing the above two expressions, 

 M =½πµₒNa²a'²/(a²+x²)3/2.

       



 

    97.  A solenoid of a length of 20 cm, area of cross-section of 4.0 cm², and having 4000 turns is placed inside another solenoid of 2000 turns having a cross-sectional area of 8.0 cm² and length of 10 cm. Find the mutual inductance between the solenoids.    


ANSWER: Given, N =4000 turns. Length, l =20 cm =0.20 m, hence, 

n =N/l =4000/0.20 =20000 turns/m. 

Area of cross-section, A =4.0 cm² =4.0x10⁻⁴ m². 

So the magnetic field inside this solenoid, 

B =µₒni.   

The number of turns in this solenoid =2000

So the magnetic flux through the outer solenoid 

φ =BAN 

   =µₒni*4.0x10⁻⁴*2000 weber

   =4πx10⁻⁷*20000*0.8i weber 

   =2.0x10⁻²i weber 

Hence the emf induced in the outer coil,

Ɛ =dφ/dt

→Ɛ  =2.0x10⁻²*di/dt

But, Ɛ = M*di/dt

Hence the mutual inductance,

M =2.0x10⁻² H.           





 

    98.  The current in a long solenoid of radius R and having n turns per unit length is given by i =iₒ sin ωt. A coil having N turns is wound around it near the center. Find (a) the induced emf in the coil and (b) the mutual inductance between the solenoid and the coil.   


ANSWER: The magnetic field inside the long solenoid, 

B =µₒni 

The magnetic flux,

φ =µₒni*πR²*N   ----------(i)

   =µₒπnNR²iₒSin ωt 

(a) Hence the emf induced, 

Ɛ =dφ/dt 

  =πµₒiₒnNωR² Cosωt 


(b) From (i), 

φ =πµₒnNR²i 

Hence the emf induced in the other coil,

Ɛ =dφ/dt

  =πµₒnNR²*di/dt

But if M is the mutual inductance,

Ɛ =M*di/dt, 

hence, M =πµₒnNR².            

 

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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

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)

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

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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