Friday, February 4, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, EXERCISES, Q1 to Q10, Concepts of Physics, Part-II

Electromagnetic Induction


EXERCISES, Q1 to Q1O


    1.  Calculate the dimensions of 

(a) ∫E.dl 

(b) vBl,   and 

(c) dⲪᵦ/dt. 

The symbols have their usual meanings.     


ANSWER: The dimensions of 

(a) ∫E.dl 

Electric field E =force per unit charge

Its dimensions =[MLT⁻²]/[IT] 

              =[MLI⁻¹T⁻³] 

Dimensions of dl =[L]

Hence dimensions of 

∫E.dl =[MLI⁻¹T⁻³][L] 

     =[ML²I⁻¹T⁻³].    


(b) Dimensions of vBl:- 

Since, the magnetic force F =qvBsinθ

→B =F/(qv.sinθ)

So, vBl =vFl/(qv.sinθ)

          =Fl/(q.sinθ)

Sine is dimensionless due to a ratio. Hence the dimensions of,

vBl = [MLT⁻²][L]/[IT]

     = [ML²I⁻¹T⁻³]. 

 

(c) Dimensions of dⲪᵦ/dt :- 

It is the magnetic flux per unit time. The magnetic flux is BAcosθ. So, 

BAcosθ =FA.tanθ.(qv)⁻¹

Dimensions of flux 

  =[MLT⁻²][L²][I⁻¹T⁻¹][L⁻¹T¹]

  =[ML²I⁻¹T⁻²]

Hence dimensions of dⲪᵦ/dt 

    =[ML²I⁻¹T⁻³]. 

  

   



 

    2.  The flux of magnetic field through a closed conducting loop changes with time according to the equation, 

Φ =at²+bt+c.

(a) Write the SI units of a, b and c. 

(b) If the magnitude of a, b and c are 0.20, 0.40, and 0.60 respectively, find the induced emf at t = 2 s.    


ANSWER: For the given equation to be dimensionally correct, each term should have the same units and dimensions as that of flux. Unit of flux is weber or volt-sec. 

(a) Hence unit of the term at² is also volt-sec. Thus the unit of 'a' is 

 volt-sec/sec² =volt/s. 


Unit of b = Flux/sec =volt-sec/sec 

              = volt. 


Unit of c is the same as the unit of flux, i.e. volt-s (or weber).  


(b) At t =2 s, and magnitudes of a, b, and c as 0.20, 0.40, and 0.60 respectively, the magnitude of induced emf is 

= dΦ/dt

= (2at+b)

= (2*0.20*2 +0.40)

= (0.80 +0.40)

= 1.20 volt 

   

 


 

    3.  (a) The magnetic field in a region varies as shown in figure (38-E1). Calculate the average induced emf in a conducting loop of area 2.0x10⁻³ m² placed perpendicular to the field in each of the 10 ms intervals shown. (b) In which intervals is the emf not constant? Neglect the behavior near the ends of 10 ms intervals.
Figure for Q-3


ANSWER: (a) Area of the loop, A= 2x10⁻³ m². At t = 0, B =0. So the flux is zero. But at t =10 ms, B =0.01 T. Flux now =BA

 =0.01*2x10⁻³ weber 

 =2x10⁻⁵ weber

Hence the change in flux in time Δt =10 ms, 

 ΔΦ = 2x10⁻⁵ weber,

Hence the induced emf,

ξ = -ΔΦ/Δt 

   = -2x10⁻⁵/(10x10⁻³) volt

   = -2x10⁻³ volt

   = -2.0 mV.


At t =20 m, B =0.03 T

So Φ =BA

    =0.03*2x10⁻³ weber

    =6.0x10⁻⁵ weber

In this interval of 10 ms, change in flux, 

ΔΦ =6.0x10⁻⁵ -2.0x10⁻⁵ weber

    =4.0x10⁻⁵ weber

Hence the average induced emf in this interval, 

ξ = -ΔΦ/Δt

  = -4.0x10⁻⁵/(10*10⁻³) volt

  = -4.0x10⁻³ volt

  = -4.0 mV

 

At t =30 ms, again the value of B =0.01 T. Hence flux  = 2.0x10⁻⁵ weber. Change in flux ΔΦ =(2.0 -6.0)x10⁻⁵ weber

       =  -4.0x10⁻⁵ weber.

Hence the induced emf,

ξ = -ΔΦ/Δt

  =4.0x10⁻⁵/(10x10⁻³) volt

  =4.0x10⁻³ volt

  =4.0 mV


At t = 40 ms, B = 0, hence Φ =0.

Change in flux ΔΦ =0 -2.0x10⁻⁵ weber

  = -2.0x10⁻⁵ weber.

Hence induced emf 

ξ = -ΔΦ/Δt

  = 2.0x10⁻⁵/(10x10⁻³) volt

  =2.0x10⁻³ volt

  =2.0 mV.


(b) Between 10 ms to 20 ms and 20 ms to 30 ms, the magnetic field B does not change linearly. Hence the emf between these two intervals will not be constant. 



      

 

 

    4.  A conducting circular loop having a radius of 5.0 cm, is placed perpendicular to a magnetic field of 0.50 T. It is removed from the field in 0.50 s. Find the average emf produced in the loop during this time. 


ANSWER: Given Δt =0.50 s, and ΔΦ =ΔB*A

=0.50*π*(0.05)² weber

=0.50*7.8x10⁻³ weber

The magnitude of average emf induced.

ξ =ΔΦ/Δt

  =0.50*7.8x10⁻³/0.5 V

  =7.8x10⁻³ V

 


   


 

    5.  A conducting circular loop of area 1 mm² is placed coplanar with a long straight wire at a distance of 20 cm from it. The straight wire carries an electric current that changes from 10 A to zero in 0.1 s. Find the average emf induced in the loop in 0.1 s. 


ANSWER: The magnetic field produced by the current in the wire will be circular in a plane perpendicular to it. So these magnetic fields will pass perpendicularly to the circular loop. Distance of the loop from the wire, 

d =20 cm =0.20 m. 

Current in the wire, i =10 A 

Hence the magnetic field due to the current near the circular loop, 

B = µₒi/(2πd)

  = (µₒ/4π)*2i/d

  = 10⁻⁷*2*10/0.20 T

  = 1x10⁻⁵ T

Area of the loop,

A = 1 mm² = 1x10⁻⁶ m²  

Hence flux at the loop,

  φ =BA

   = 1x10⁻⁵*1x10⁻⁶ weber

   =1x10⁻¹¹ weber

In an interval of time Δt =0.1 s, the current becomes zero. Hence the magnetic field and flux will also become zero. 

   So, Δφ =0 -1x10⁻¹¹ weber

     =-1x10⁻¹¹ weber.

Average emf induced in this time interval will be 

 ξ = -ΔΦ/Δt

   = 1x10⁻¹¹/0.1 V

   =1x10⁻¹⁰ V.

 



 


 

    6.  A square-shaped copper coil has edges of length 50 cm and contains 50 turns. It is placed perpendicular to a 1.0 T magnetic field. It is removed from the magnetic field in 0.25 s and restored to its original place in the next 0.25 s. Find the magnitude of the average emf induced in the loop during (a) its removal, (b) its restoration, and (c) its motion.


ANSWER: Numer of turns, N =50. Area of the cross-section of the coil, A = 0.5*0.5 m² =0.25 m². B =1.0 T. So the flux through one turn, 

Φ =BA  

   =1.0*0.25 weber 

   =0.25 weber. 

In a time interval, Δt =0.25 s the coil is in a place that is without a magnetic field. So the flux is also zero. Hence, 

ΔΦ =0 -0.25 = -0.25 weber. 


(a) Hence the magnitude of the average emf induced in the coil during its removal, 

ξ = N*(ΔΦ/Δt) = 50*(0.25/0.25) V 

   = 50 V.  


(b) During the restoration, the change in flux ΔΦ =0.25 -0 =0.25 weber   

Time interval Δt = 0.25 s. 

Hence the average magnitude of emf induced in the coil during its restoration

ξ = N*(ΔΦ/Δt)

   =50*(0.25/0.25) V

   =50 V.


(c) During the motion of the coil the initial and final strength of the magnetic field is the same and so is the flux through the coil. Hence the change in flux ΔΦ =0, The time interval

Δt =0.25 +0.25 =0.50 s

Hence the magnitude of the average emf developed

ξ = N*(ΔΦ/Δt)

   =50*(0/0.50) V

   = Zero.  



   


 

    7.  Suppose the resistance of the coil in the previous problem is 25 𝛀. Assume that the coil moves with uniform velocity during its removal and restoration. Find the thermal energy developed in the coil during (a) its removal, (b) its restoration, and (c) its motion.  


ANSWER: Resistance of the coil, R =25 Ω. 

(a) Energy dissipated during its removal,

U = Power*Time

   =(ξ²/R)*Δt

   = (50²/25)*0.25 J 

   = 25 J. 


(b) Energy dissipated during its restoration is again 

U = (ξ²/R)*Δt

   = (50²/25)*0.25 J

   = 25 J.


(c) The energy dissipated during its motion will be the sum of energy dissipated during removal and restoration because it is a scalar quantity. So, 

U = 25 +25 J =50 J.        


   


 

    8.  A conducting loop of area 5.0 cm² is placed in a magnetic field which varies sinusoidally with time as B = Bₒsin ωt where Bₒ =0.20 T and ω =300 s⁻¹. The normal to the coil makes an angle of 60° with the field. Find (a) maximum emf induced in the coil. (b) the emf induced at τ =(π/900) s and(c) the emf induced at t =(π/600) s. 

 

ANSWER: (a) Angle between normal to the coil and the field, ß =60°.  Hence  emf developed at time t, 

ξ =dΦ/dt 

  =d(BAcosß)/dt 

  =A.cosß*d(BₒSin ωt)/dt

  =A.cos 60°*Bₒω*Cos ωt

  =(5x10⁻⁴)*½*0.20*300*Cos (300t)

  = 0.015*Cos(300t)

The maximum value of Cos (300t) will be =1, hence maximum emf developed,

ξₘₐₓ =0.015 V.

       

  (b) emf induced at t =(π/900) s 

ξ = 0.015*Cos (300π/900) 

  =0.015*Cos(π/3) 

  =0.015*½ V 

  =7.5x10⁻³ V. 


(c) Emf induced at t =(π/600) s 

ξ = 0.015*Cos (300π/600) V 

   =0.015*Cos(π/2) V

  =0.015*0 V =zero.        



 


 

    9.  Figure (38-E2) shows a conducting square loop placed parallel to the pole-faces of a ring magnet. The pole-faces have an area of 1 cm² each and the field between the poles is 0.10 T. The wires making the loop are all outside the magnetic field. If the magnet is removed in 1.0 s, what is the average emf induced in the loop? 

Figure for Q-9

ANSWER: Area of pole faces, A =1 cm² each =1.0x10⁻⁴ m². Field between the poles, B =0.10 T. Since the plane of the loop is perpendicular to the magnetic field, the flux Φ =BA 

→Φ =0.10*1.0x10⁻⁴ weber 

      =1.0x10⁻⁵ weber

After removing, the flux inside the loop is zero in time Δt =1.0 s. Hence the average emf developed is

ξ =ΔΦ/Δt

   =1.0x10⁻⁵/1.0 V 

   = 10x10⁻⁶ V 

   = 10 µV.    




 

    10.  A conducting square loop with having edge of length 2.0 cm is rotated through 180° about a diagonal in 0.20 s. A magnetic field B exists in the region which is perpendicular to the loop in its initial position. If the average induced emf during the rotation is 20 mV. Find the magnitude of the magnetic field.


ANSWER: Area of the square loop, A =(0.02)² m² = 4x10⁻⁴ m². 

Initial flux Φ =BA.cosß 

→Φ =B*4x10⁻⁴*cos0°

     = 4x10⁻⁴B weber 

Final flux Φ' =B*4x10⁻⁴*cos180° 

→Φ' =-4x10⁻⁴B weber

Change in flux ΔΦ =-Φ' -Φ

→ΔΦ =-8x10⁻⁴B weber


Δt = 0.20 s

Hence the induced emf ξ = -ΔΦ/Δt

→20x10⁻³ = 8x10⁻⁴B/0.20 

→B =½*10 T =5 T.       


 

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Links to the Chapters




CHAPTER- 38- Electromagnetic Induction


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

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Click here for → OBJECTIVE-II

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Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

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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)

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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


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CHAPTER- 3 - Kinematics - Rest and Motion

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Click here for EXERCISES (Question number 31 to 40)


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CHAPTER- 2 - "Physics and Mathematics"

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