Friday, March 25, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, EXERCISES, Q51 to Q60, Concepts of Physics, Part-II

Electromagnetic Induction


EXERCISES, Q51 to Q6O


    51.  A bicycle is resting on its stand in the east-west direction and the rear wheel is rotated at an angular speed of 100 revolutions per minute. If the length of each stroke is 30.0 cm and the horizontal component of the earth's magnetic field is 2.0x10⁻⁵ T, Find the emf induced between the axis and the outer end of the spoke. Neglect centripetal force acting on the free electrons of the spoke.          


ANSWER: Consider a small length dx at a distance x from the center. Its velocity will be v =ωx. emf induced in this small length due to the earth's horizontal magnetic field, dE =Bv*dx 

 =Bωxdx, 

Hence E =∫dE =Bω∫xdx 

            =Bω[x²/2] 

Taking x from 0 to l, we have, 

E =½Bωl² 

Given, l =30 cm =0.30 m 

ω =100*2π/60 rad/s =10π/3 rad/s, 

B =2.0x10⁻⁵ T, hence

E =½*2.0x10⁻⁵*(10π/3)*(0.30)² V

   =9.4x10⁻⁶ V. 




 


    52.  A conducting disc of radius r rotates with a small but constant angular velocity ⍵ about its axis. A uniform magnetic field B exists parallel to the axis of rotation. Find the motional emf between the center and the periphery of the disc.  


ANSWER: When a conducting rod or flat surface of length l, moves perpendicular to the magnetic field with a velocity v, the emf induced is =Bvl, whatever its width may be. So if we consider a dx wide metallic ring on the moving disc at distance x from the center, the emf induced between the inner circle of radius x and outer circle of radius x+dx is, dE =Bvdx  

→dE =Bωx.dx 

Now, E =∫dE = ∫Bωx.dx 

           =Bω[x²/2] 

           =½Bωr².                     

{Putting the value of x from 0 to r} 

 



 


    53.  Figure (38-E25) shows a conducting disc rotating about its axis in a perpendicular magnetic field B. A resistor of resistance R is connected between the center and the rim. Calculate the current in the resistor. Does it enter the disc or leave it at the center? The radius of the disc is 5.0 cm, angular speed ⍵ = 10 rads/s, B =0.40 T and R =10 Ω. 
The figure for Q-51


ANSWER: As in the previous problem, emf induced between the center and the rim, E =½Bωr². 

r =5 cm =0.05 m, ω =10 rad/s, B =0.4 T, hence, 

E =½*0.4*10*(0.05)² V 

   =0.005 V   

Current in the 10 Ω resistor, i =E/R 

→i =0.005/10 A =0.5x10⁻³ A =0.5 mA.


From Flaming's right hand rule the positive charge will concentrate at the center and the negative at the rim. When the resistor is connected as in the figure, the current will leave the disc at the center.             




 

    54.  The magnetic field in a region is given by B =kBₒy/L where L is a fixed length. A conducting rod of length L lies along the Y-axis between the origin and the point (0, L, 0). If the rod moves with a velocity v =vₒi, find the emf induced between the ends of the rod.            


ANSWER: Consider a very small element dy on the rod at y distance from the origin. The small emf generated in this element,  

dE =Bxv*dy  

    =k(Bₒy/L)xvₒi*dy 

    =j(Bₒvₒ/L)ydy 

Unit vector j shows that the direction of the emf is along Y-axis. Hence the magnitude of full emf developed in the rod is, 

  E =∫dE =(Bₒvₒ/L)∫ydy 

     =(Bₒvₒ/L)[y²/2]

Putting the limit for y between 0 to L, we get 

E =(Bₒvₒ/L)*(L²/2) 

   =½BₒvₒL.                            




 

    55.  Figure (38-E26) shows a straight, long wire carrying a current i and a rod of length l coplanar with the wire and perpendicular to it. The rod moves with a constant velocity v in a direction parallel to the wire. The distance of the wire from the center of the rod is x. Find the motional emf induced in the rod.          
The figure for Q-55


ANSWER: Magnetic field due to the current carrying wire at a distance r from the wire

B =µₒi/2πr.

Consider a small element dr on the rod. The small emf induced in it,  

dE =Bv*dr 

   =(µₒi/2πr)vdr  

    =(µₒiv/2π)(1/r)dr 

Hence the emf induced in the rod, 

E =∫dE =(µₒiv/2π)∫(1/r)dr

   =(µₒiv/2π)[ln r] 

Putting the limits for the rod length, r =(x-l/2) to (x+l/2), we get 

E =(µₒiv/2π)*ln{(x+l/2)/(x-l/2)} 

   =(µₒiv/2π)*ln {(2x+l)/(2x-l)}                



 


    56.  Consider a situation similar to that of the previous problem except that the ends of the rod slide on a pair of thick metallic rails laid parallel to the wire. At one end the rails are connected by a resistor of resistance R. (a) What force is needed to keep the rod sliding at a constant speed v? (b) In this situation what is the current in the resistor R. (c) Find the rate of heat developed in the resistor. (d) Find the power delivered by the external agent exerting the force on the rod.             


ANSWER: (a) In this case there will be a current in the rod, 

i' = E/R

A current-carrying rod will feel a force in the perpendicular magnetic field. This force

F =i'lB =ElB/R

Here i and l are fixed for uniform v, but B is varying. Let the force on a small length element dr is dF, then

dF =(µₒiv/2πR)ln{(2x+l)/(2x-l)}*(µₒi/2πr)dr

 =(µₒi/2π)²(v/R)ln{(2x+l)/2x-l)}(dr/r)

Force on the whole rod, 

F =∫dF 

  =(µₒi/2π)²(v/R)ln{(2x+l)/(2x-l)}[ln r]

Putting the same limit for r =(x-l/2) to (x+l/2), we get

F =(µₒi/2π)²(v/R)[ln{(2x+l)/(2x-l)]².

=(v/R)[(µₒi/2π) ln{(2x+l)/(2x-l)}]²


This much force is needed to keep the rod moving on the rails with a constant velocity v.


(b) current in the rod is 

i' =E/R

  =(µₒiv/2πR) ln{(2x+l)/(2x-l)}

The expression for E is derived in the previous problem.


(c) The rate of heat developed in the resistor,

H =i'²R

 =[(µₒiv/2πR) ln{(2x+l)/(2x-l)}]²*R

  =(1/R)[(µₒiv/2π) ln{(2x+l)/(2x-l)}]²


(d) Power delivered by the external agent exerting force on the rod,

P = F*v

  =(v/R)[(µₒi/2π) ln{(2x+l)/(2x-l)}]²v

  =(v²/R)[(µₒi/2π) ln{(2x+l)/(2x-l)}]²

  =(1/R)[(µₒiv/2π) ln{(2x+l)/(2x-l)}]²

  = H

So the power required to maintain v is the same as the rate of heat generated. The answer is the same as in (c)

The expression of force F is taken from the part (a). 



 

 


    57.  Figure (38-E27) shows a square frame of wire having a total resistance r placed coplanar with a long, straight wire. The wire carries a current 'i' given by i = iₒsin⍵t. Find (a) the flux of the magnetic field through the square frame. (b) the emf induced in the frame and (c) the heat developed in the frame in the time interval 0 to 20π/⍵.         
The figure for Q-57


ANSWER: (a) Consider a small area strip dA parallel to the wire on the loop. dA is at a distance r from the wire and has a width dr as shown in the diagram below.
Diagram for Q57

Magnetic field due to the long current carrying wire at distance r from it,

B = µₒi/2πr

Magnetic flux on the small strip with area dA,

dφ =B.dA

    =(µₒi/2πr)*(a*dr)

   =(µₒia/2π)(dr/r)

φ =∫dφ

    =(µₒia/2π)∫(dr/r)

    =(µₒia/2π)*[ln r]

Putting the limit of r from b to (a+b), we get

φ  =(µₒia/2π)*ln {(a+b)/b)}

    =(µₒia/2π)*ln {(1 +a/b)}


(b) The emf induced in the frame

E =dφ/dt

  =(µₒa/2π)*ln(1+a/b)*di/dt

  =(µₒa/2π)*ln(1+a/b)*d(iₒsinωt)/dt

  =(µₒa/2π)*ln(1+a/b)*iₒωcosωt 

 =(µₒaiₒω.cos ωt/2π) ln(1+a/b)


(c) Current in the loop at time t

i' =E/r 

Heat developed in a time interval dt,

dH =i'²r dt

  =(E²/r) dt

  =(1/r){(µₒaiₒω/2π) ln(1+a/b)} cos²(ωt)dt

Hence the heat developed in the frame in time t,

H =∫dH

  =(1/r){(µₒaiₒω/2π)ln(1+a/b)}² ∫cos²(ωt)dt

  =(1/r){(µₒaiₒω/2π)ln(1+a/b)}² ∫{½(1+cos2ωt)}dt

  =(1/2r){(µₒaiₒω/2π)ln(1+a/b)}² ∫(1+cos2ωt)dt

  =(1/2r){(µₒaiₒω/2π)ln(1+a/b)}² [t+(1/2ω)sin2ωt)]

  

Putting the value of t from 0 to 20π/ω, we get

H =(1/2r){(µₒaiₒω/2π)ln(1+a/b)}² [20π/ω+(1/2ω)sin40π)]

 =(1/2r){(µₒaiₒω/2π)ln(1+a/b)}² (20π/ω)

 =(10π/ωr){(µₒaiₒω/2π)ln(1+a/b)}²  

 =(10πµₒ²a²iₒ²ω²/4π²ωr) {ln(1+a/b)}²

=(5µₒ²a²iₒ²ω/2πr) ln²(1 +a/b)         



 

 

 


    58.  A rectangular metallic loop of length l and width b is placed coplanar with a long wire carrying current i (Figure 38-E28). The loop is moved perpendicular to the wire with a speed v in the plane containing the wire and the loop. Calculate the emf induced in the loop when the rear end of the loop is at a distance 'a' from the wire. Solve by using Faraday's law for the flux through the loop and also by replacing different segments with equivalent batteries. 
The figure for Q-58

            


ANSWER: (a) consider a dr wide strip on the loop parallel to the current carrying a long wire at a distance r from the wire. Area of the strip, dA =bdr

The magnetic field at distance r from the wire B =µₒi/2πr

Magnetic flux at the strip

dφ =B.dA

   =(µₒi/2π)(bdr/r)

   =(µₒib/2π)(dr/r)

Hence total flux on the loop 

φ=∫dφ

  =(µₒib/2π) [ln r]

  =(µₒib/2π) ln{(x+l)/x}

  =(µₒib/2π) ln(1+l/x) 

for r = x to x+l

i.e. when the rear end of the loop is at a distance x from the wire.

From Faraday's law emf induced in the loop = -dφ/dt 

  =-(µₒib/2π){1/(1+l/x)}(-l/x²)dx/dt

 = µₒiblv/{2πx(l+x)}  

{since v =dx/dt}  

When the rear end of the loop is at a distance x =a from the wire, emf induced in the loop  

  =µₒiblv/{2πa(l+a)}.  


Let us solve this problem by considering different segments of the loop. The emf induced in the two sides of the loop having length l is zero because the projection of length on a plane perpendicular to the velocity is zero. 

Now the emf induced in a side of loop having length b and parallel to the wire is  

 =Bvb 

When this loop segment is at a distance  x from the wire, the emf induced, 

  =(µₒi/2πx)vb 

  =µₒivb/2πx 

So when the rear end of the loop is at a distance 'a' from the wire, the emf induced in the rear segment of the width b = µₒivb/2πa 

and in the farther segment, emf 

  =µₒivb/2π(l+a) 

Since these two emf will have the same direction in the rods, they will oppose each other in the circuit of the loop. Hence the net emf in the loop 

 =µₒivb/2πa -µₒivb/2π(l+a) 

=(µₒivb/2π){1/a -1/(l+a)} 

 =µₒiblv/{2πa(l+a)}.  

It is the same as derived by Faraday's law above.                



 


    59.  Figure (38-E29) shows a conducting circular loop of radius 'a' placed in a uniform, perpendicular magnetic field B. A thick metal rod OA is pivoted at the center O. The other end of the rod touches the loop at A. The center O and fixed-point C on the loop are connected by a wire OC of resistance R. A force is applied at the middle point of the rod OA perpendicularly so that the rod rotates clockwise at a uniform angular velocity ⍵. Find the force.       
The figure for Q-59


ANSWER: Let us find the emf induced in the rod for a uniform angular velocity ω. We can either calculate it by calculus as in earlier problems or by the average speed of the rod. Since the tangential speed of the rod is zero at the center and ωa at the loop, the average speed of the rod =½ωa. Hence the emf induced in the rod 

=B(½ωa)a 

=½Bωa². 

Current through the rod, i =(½Bωa²)/R

Hence the force on the rod due to this current, 

    =iBa 

   =(½Bωa²)Ba/R 

   =½B²a³ω/R. 

This is the resisting force on the rod due to the magnetic field and opposite to the direction of motion. Hence to maintain a uniform angular velocity net force should be zero on the rod. So the applied force also should have the same magnitude. Thus 

F =½B²a³ω/R.           

 




 

    60.  Consider the situation shown in the figure of the previous problem. Suppose the wire connecting O and C has zero resistance but the circular loop has a resistance R uniformly distributed along its length. The rod OA is made to rotate with a uniform angular speed ⍵ as shown in the figure. Find the current in the rod when ∠AOC =90°.              


ANSWER: In this case, there will be two resistances connected in parallel in the circuit. Since the resistance R is uniformly distributed in the loop, one resistance will be for a quarter of the loop

r' =R/4, and the other resistance will be for the rest of the loop which is three-quarters. So, 

r" =3R/4. 

If the equivalent resistance of these two parallel resistances is r, then 

1/r =1/r' +1/r" =4/R +4/3R 

→1/r =(4/R)(1 +1/3) =(4/R)*4/3

→1/r =16/3R 

→r =3R/16. 

So the current in the rod 

 =(induced emf)/Resistance 

 =(½Bωa²)/r 

 =(½Bωa²)/(3R/16) 

 =8Bωa²/3R.             

    

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

 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 No comments:
Labels: , , , ,
Subscribe to: Posts (Atom)