Tuesday, March 1, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, EXERCISES, Q21 to Q30, Concepts of Physics, Part-II

Electromagnetic Induction


EXERCISES, Q21 to Q3O


    21.  A circular coil of radius 2.00 cm has 50 turns. A uniform magnetic field B =0.200 T exists in the space in a direction parallel to the axis of the loop. The coil is now rotated about a diameter through an angle of 60°. The operation takes 0.100 s. (a) Find the average emf induced in the coil. (b) If the coil is a closed one (with the two ends joined together) and has a resistance of 4.00 Ω. Calculate the net charge crossing a cross-section of the wire of the coil.    


ANSWER: (a) Initial magnetic flux, 

  φ =n*BA 

   =50*0.20*π(0.02)² weber 

   =0.004π weber 

Final magnetic flux through the coil, 

  φ' =nB*A cos60° 

   =0.004π*½ weber

   =0.004π/2 weber

Change in flux, Δφ =0.004π -0.004π/2 weber

→Δφ =0.004π/2 weber

     =0.002π weber

Δt =0.100 s 

Average emf induced = Δφ/Δt

   =0.002π/0.100 V

   =0.0628 V

   =6.28x10⁻² V.


(b) R =4.00 Ω

Net charge crossing a cross-section of the wire of the coil =i*Δt

   =(E/R)*Δt

   =(6.28x10⁻²/4)*0.100 C

   =0.00157 C

   =1.57x10⁻³ C. 

 




 

    22.  A closed coil having 100 turns is rotated in a uniform magnetic field B = 4.0x10⁻⁴ T about a diameter that is perpendicular to the field. The angular velocity of rotation is 300 revolutions per minute. The area of the coil is 25 cm² and its resistance is 4.0 Ω. Find (a) the average emf developed in half a turn from a position where the coil is perpendicular to the magnetic field. (b) the average emf in a full turn and (c)the net charge displaced in part (a). 


ANSWER: N =100, B =4.0x10⁻⁴ T, Angular speed =300 revolutions per minute. Area, A = 25 cm², Resistance, R =4.0 Ω. 

  (a) The angle covered in half a turn,

  ß = 180°, 

     Time to cover half turn,

 Δt =½*(60/300) s

     =0.10 s

Δφ =NBA(cos0° -cos180°)

     =100*4.0x10⁻⁴*25x10⁻⁴*(1+1) web

    =2.0x10⁻⁴ weber

Average emf developed,

E =Δφ/Δt =2.0x10⁻⁴/0.10 V  

   =2x10⁻³ V.  


(b) Since the magnetic flux at the initial and final position in a complete turn will be the same, the change in flux Δφ =0. Hence the average emf developed Δφ/Δt will be zero. In fact, the average emf in each half-turn will be equal and opposite hence the net average emf = zero.


(c) Net charge displaced in part (a) is.

Q = i*Δt

   = (E/R)*Δt 

   = (2x10⁻³/4.0)*0.10 C  

   = 0.50x10⁻⁴ C

   = 5.0x10⁻⁵ C.        






 

    23.  A coil of radius 10 cm and resistance 40 Ω has 1000 turns. It is placed with its plane vertical and its axis parallel to the magnetic meridian. The coil is connected to a galvanometer and is rotated about the vertical diameter through an angle of 180°. Find the charge which flows through the galvanometer. If the horizontal component of the earth's magnetic field is BH = 3.0x10⁻⁵ T.  


ANSWER: If the time-taken in this rotation =Δt.

    Emf devevolepd in this rotation,

 E = Δφ/Δt

 =NAB(cos0° -cos180°)/Δt

 =1000*π(0.1)²*3x10⁻⁵*2/Δt 

 =6πx10⁻⁴/Δt

The charge that flows through the galvanometer,

 Q =i*Δt

  =(E/R)*Δt

  =6πx10⁻⁴/40 C

  =0.47x10⁻⁴ C

  =4.7x10⁻⁵ C.

 



 

    24.  A circular coil of one turn of radius 5.0 cm is rotated about a diameter with a constant angular speed of 80 revolutions per minute. A uniform magnetic field B =0.010 T exists in a direction perpendicular to the axis of rotation. Find (a) the maximum emf induced, (b) the average emf induced in the coil over a long period, and (c) the average of the squares of emf induced over a long period.  


ANSWER: (a) A =area of circular coil, 

  =π*(0.05)² m² 

  =2.5πx10⁻³ m². 

  B =0.010 T  

 Ѡ =dß/dt =80 rev/minute 

   =80/60 rev/s 

   =4/3 rev/s

  =4*2π/3 rad/s

  =8π/3 rads/s.  

emf developed in a small time dt, 

  E =dφ/dt

→E =d(BACosß)/dt 

    =-BA*sinß*dß/dt     

    =-BAѠ*sinß. 

 The maximum value of sinß =1

Hence the maximum value of emf induced,  

 Em = BAѠ

  =0.01*2.5πx10⁻³*8π/3 V

  =6.6x10⁻⁴ V.

          

  (b) Since the direction of the induced emf changes every half revolution, the average emf induced in each complete revolution will be zero, and hence over a long time, it will also be zero.  


(c) Average of the squares of emf induced over a long period, say between 0 to T, 

=(1/T)∫E²dt

=(1/T)∫B²A²Ѡ²sin²ß.dt 

=(B²A²Ѡ²/2T)∫{1-cos(2Ѡt)}dt

=(B²A²Ѡ²/2T)[t -{sin(2Ѡt)}/2Ѡ]

=(B²A²Ѡ²/2T)[T-{sin(2ѠT)}/2Ѡ]

{Putting the limit t =0 to t =T}

=(B²A²Ѡ²/2T)[T-sin(2ß)2Ѡ]

For a long period let us put ß =2π for a full revolution.

=B²A²Ѡ²/2  ----- (i)

=(0.01)²*(2.5πx10⁻³)²*(8π/3)²/2 V²

=2.2x10⁻⁷ V².


One interesting thing:-

Maximum emf, Em =BAѠ

Em² =B²A²Ѡ²

Minimum emf, E =0,  →E² =0

Average of these two squares,

  =(B²A²Ѡ² -0)/2

  =B²A²Ѡ²/2

It is the same as (i). 



 

 

    25.  Suppose the ends of the coil in the previous problem are connected to a resistance of 100 Ω. Neglecting the resistance of the coil, find the heat produced in the circuit in one minute.   


ANSWER: Heat produced in a time interval dt =i²Rdt 

Heat produced over a time T, 

H =∫i²Rdt

  =∫(E/R)²Rdt

  =(1/R)∫E²dt

  =(1/R)∫{BAѠsin(Ѡt)}²dt

  =(B²A²Ѡ²/R)½∫(1-cos2Ѡt)dt

  =(B²A²Ѡ²/2R)[t -(sin2Ѡt)/2Ѡ]

 =(B²A²Ѡ²/2R)[T -(sin2ѠT)/2Ѡ]

(After putting the limits)

T =60 s, 2ѠT =2*60*8π/3 =160*2π 

Hence sin2ѠT =0. 
Now,  

H =B²A²Ѡ²T/2R
=0.01²*(2.5πx10⁻³)²(8π/3)²*60/200 J
=1.3x10⁻⁷ J.   

   



 

    26.  Figure (38-E9) shows a circular wheel of radius 10.0 cm whose upper half shown dark in the figure, is made of iron and the lower half of wood. The two junctions are joined by an iron rod. A uniform magnetic B of magnitude 2.00x10⁻⁴ T exists in the space above the central line as suggested by the figure. The wheel is set into pure rolling on the horizontal surface. If it takes 2.00 seconds for the iron part to come down and the wooden part to go up, find the average emf induced during this period.  
The figure for Q-26


ANSWER: In 2 seconds the iron part comes down, at this moment there is no magnetic field through the iron part. Hence the magnetic flux through the iron part, φ = 0.  

  Initial magnetic flux, φ' =BA 

→φ' =2x10⁻⁴*½π(0.10)² 

     =3.14x10⁻⁶ weber

Δt =2 s.

Hence the average emf developed =Δφ/Δt

  =(3.14x10⁻⁶ -0)/2 V  

  =1.57x10⁻⁶ V. 





 

    27.  A 20 cm long conducting rod is set into pure translation with a uniform velocity of 10 cm/s perpendiculars to its length. A uniform magnetic field of magnitude 0.10 T exists in a direction perpendicular to the plane of motion. (a) Find the average magnetic force on the free electrons of the rod. (b) For what electric field inside the rod, the electric force on a free electron will balance the magnetic force? How is the electric field created? (c) Find the motional emf between the ends of the rod.    


ANSWER: (a) Average magnetic force on a free electron of the rod, F =evB, 

→F =1.6x10⁻¹⁹*0.10*0.10 N

  =1.6x10⁻²¹ N.


(b) When this magnetic force is balanced by the electric force due to the electric field inside the rod, the electric force

F =1.6x10⁻²¹ N. 

If the electric field inside the rod =E, then,  

eE =F

→E =F/e 

   =1.6x10⁻²¹/1.6x10⁻¹⁹ V/m 

   =1.0x10⁻² V/m. 

This electric field is created due to the accumulation of opposite charges at the ends of the rod.

(c) The motional emf between the ends of the rod,  

V =vBl   

  =0.10*0.10*0.20 volts 

  =2x10⁻³ Volts.      





  

    28.  A metallic meter stick moves with a velocity of 2 m/s in a direction perpendicular to its length and perpendicular to a uniform magnetic field of magnitude 0.20 T. Find the emf induced between the ends of the stick.  


ANSWER: Length of the meter stick,

   l =1 m.   

Velocity v =2 m/s, 

Magnetic field, B=0.20 T   

The emf induced between the ends of the stick =Bvl 

   =0.20*2*1 V 

   =0.4 V.             





 

    29.  A 10 m wide spacecraft moves through the interstellar space at a speed of 3x10⁷ m/s. A magnetic field B =3x10⁻¹⁰ T exists in the space in a direction perpendicular to the plane of motion. Treating the spacecraft as a conductor, calculate the emf induced across its width.  


ANSWER: B =3x10⁻¹⁰ T, v =3x10⁷ m/s

Width, l =10 m

The emf induced across its width =Bvl

   =3x10⁻¹⁰*3x10⁷*10 V  

   =9x10⁻² V =0.09 V.

     




 

    30.  The two rails of a railway track, insulated from each other and from the ground, are connected to a millivoltmeter. What will be the reading of the millivoltmeter when a train travels on the track at a speed of 180 km/h? The vertical component of the earth's magnetic field is 0.2x10⁻⁴ T and the rails are separated by 1 m.  


ANSWER: The emf will be induced across the width of the train and since the separated railway tracks are in contact with the train through the wheels, the emf will exist between the railway track.  

emf =Bvl 

  =0.2x10⁻⁴*(180x10³/3600)*1 V

  =1.0x10⁻³ V  

  =1.0 mV.       

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

 Buy Home Furnishing

Click here for all links → kktutor.blogspot.com 

===<<<O>>>===


My Channel on YouTube  →  SimplePhysics with KK


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"

Posted by at
Labels: , , , ,

No comments:

Post a Comment

Subscribe to: Post Comments (Atom)