Tuesday, March 8, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, EXERCISES, Q31 to Q40, Concepts of Physics, Part-II

 Electromagnetic Induction


EXERCISES, Q31 to Q4O


    31.  A right-angled triangle abc, made from a metallic wire, moves at a uniform speed v in its plane as shown in figure (38-E10). A uniform magnetic field B exists in the perpendicular direction. Find the emf induced (a) in the loop abc, (b) in the segment bc, (c) in the ac and (d) in the segment ab.    
Figure for Q-31


ANSWER: (a) Since the loop abc moves in its plane and the uniform magnetic field is perpendicular to its plane, the magnetic flux through it remains constant. Since the emf induced in a loop is proportional to the rate of change of flux through it, the emf induced in this loop is zero.  


(b) The emf induced in the segment bc is 

=vB(bc) 

From the right-hand rule, the positive end will be at c.  


(c) Since the segment ac is moving along its length, its exposed portion perpendicular to velocity is zero. Hence the induced emf in the segment ac, 

=vB(ac)*cos90° =0.

Hence the emf induced is zero. 


(d) In the segment ab, the emf induced 

=vB{(ab)*cos(∠abc)} 

=vB(bc). 

From the right-hand rule, the positive end will be at a.               





  

    32.  A copper wire bent in the shape of a semicircle of radius r translates in its plane with a constant velocity v. A uniform magnetic B exists in the direction perpendicular to the plane of the wire. Find the emf induced between the ends of the wire if (a) the velocity is perpendicular to the diameter joining free ends, (b) the velocity is parallel to this diameter.    


ANSWER: (a) Since the velocity is perpendicular to the diameter joining free ends, the projection of the semicircular wire perpendicular to the velocity =2r. 

The emf induced between the ends, 

=vB*2r 

=2rvB.   


(b) As is clear from the figure each segment AB and BC has the same projected length r perpendicular to the velocity. Hence each segment will have equal emf induced =vBr. 
Diagram for Q-32

     Also, the ends A and C will be at the same potential, either positive or negative depending upon the sense of the magnetic field. Since the potentials of A and C are the same, the emf between the ends A and C is zero.             



 


 

    33.  A wire of length 10 cm translates in a direction making an angle of 60° with its length. The plane of motion is perpendicular to a uniform magnetic field of 1.0 T that exists in space. Find the emf induced between the ends of the rod if the speed of translation is 20 cm/s.     


ANSWER: B =1.0 T, l =0.10 m, v =0.20 m/s. Angle between the length and the velocity =60°. Hence the emf induced between the ends, 

E = vBl*sin60°

   =0.2*1.0*0.1*(√3/2) V

   =0.017 V 

   =17x10⁻³ V.                



 


  

    34.  A circular copper ring of radius r translates in its plane with a constant velocity v. A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the ring. Consider different pairs of diametrically opposite points on the ring. (a) Between which pair of points is the emf maximum? What is the value of this maximum emf? (b) Between which pair of points is the emf minimum? What is the value of this minimum emf?      


ANSWER: (a) The emf between a pair of diametrically opposite points is, 

   =vB*2r*sinß, where ß is the angle between the diameter and the velocity. Since sinß is maximum (= 1) for ß =90°, the maximum emf =2vBr, between the ends of the diameter perpendicular to the velocity.  


(b) Minimum emf will be for ß =0° for which sinß =0. Value of this emf =vB*2r*0 =0 (zero). Since ß =0°, this pair of points are located at the ends of the diameter that is parallel to the velocity.          






  

    35.  Figure (38-E11) shows a wire sliding on two parallel, conducting rails placed at a separation l. A magnetic field B exists in a direction perpendicular to the plane of the rails. What force is necessary to keep the wire moving at a constant velocity v?     
Figure for Q-35


ANSWER: The emf developed at the ends of the wire, E =vBL. Since the conducting wires on which it slides are not connected, there is no current in the moving wire, i =0.  

   The force on a current-carrying conductor in a magnetic field, 

F =ilB =zero. If the wire is moving with a uniform velocity, no force is needed. 

             



 

    36.  Figure (38-E12) shows a long U-shaped wire of width l placed in a perpendicular magnetic field B. A wire of length l is slid on the U-shaped wire with a constant velocity v towards the right. The resistance of all the wires is r per unit length. At t = 0, the sliding wire is close to the left edge of the U-shaped wire. Draw an equivalent circuit diagram, showing the induced emf as a battery. Calculate the current in the circuit.       
Figure for Q-36


ANSWER: At time t, let the distance of the moving wire from the left = x. In the next small time dt, distance moved =dx. 

dφ = Bldx, 

emf induced, E =dφ/dt 

→E =Bldx/dt 

      =Blv 

At this time, the length of the wire in the circuit,

        =2l +2x =2l+2vt =2(l+vt)

Given the resistance of wires = r per unit length. Hence the resistance in the circuit,

 R = 2r(l +vt) 

    

 Current in the circuit, 

 i =E/R 

   =Blv/{2r(l+vt)}  

The equivalent circuit diagram can be drawn as below:-
Diagram for Q-36

                     





 

    37.  Consider the situation of the previous problem. (a) Calculate the force needed to keep the sliding wire moving with a constant velocity v. (b) If the force needed just after t =0 is Fₒ, find the time at which the force needed will be Fₒ/2.     


ANSWER: (a) Force on a current-carrying conductor in a magnetic field is,  

F = ilB, 

Equal and opposite force will be needed to move the wire at uniform velocity. 

Here the force required to move the wire at uniform velocity at time t is given as, 

F =(lB)*Blv/{2r(l+vt)} 

   =B²l²v/{2r(l+vt)}.


(b) Initial force at time t = 0, 

Fₒ =B²l²v/(2rl)

If at time t = t' the force becomes Fₒ/2, then

B²l²v/{2r(l+vt')} =½B²l²v/(2rl)

→1/(l+vt') = 1/2l

→l+vt' =2l

→vt' =l

→t' = l/v.


 



 

    38.  Consider the situation shown in figure (38-E13). The wire PQ has mass m, resistance r, and can slide on the smooth, horizontal parallel rails separated by a distance l. The resistance of the rails is negligible. A uniform magnetic field b exists in the rectangular region and a resistance R connects the rails outside the field region. At t =0, the wire PQ is pushed towards the right with a speed vₒ. Find (a) the current in the loop at an instant when the speed of the wire PQ is v, (b) the acceleration of the wire at this instant, (c) the velocity v as a function of x and (d) the maximum distance the wire will move. 
Figure for Q-38


ANSWER: (a) Initially, the emf induced in the wire PQ = vₒBl.

Resistance of the circuit =R+r

Hence the current induced in the circuit, 

i =vₒBl/(R+r).

So, when the speed of the wire is v, the current induced in the loop is,

  =vBl/(R+r).


(b) Due to the current of the wire in the magnetic field, an opposing force will act on the wire,

F =ilB

  =vBl*Bl/(R+r)

  =vB²l²/(R+r)

(Towards left)

Acceleration of the wire, a =F/m

  =vB²l²/{m(R+r)}.

Directed towards left.


(c) The velocity of the wire after time t,

v =vₒ -at

  =vₒ -vB²l²t/{m(R+r)} 

  =vₒ -B²l²(vt)/{m(R+r)}

  =vₒ -B²l²x/{m(R+r)}.


(d) For the maximum distance that wire will move (xₘₐₓ), v = 0.

→B²l²xₘₐₓ/{m(R+r)} =vₒ

→xₘₐₓ =mvₒ(R+r)/B²l². 

 



 

 

    39.  A rectangular frame of wire abcd has dimensions 32 cm x 8.0 cm and a total resistance of 2.0 Ω. It is pulled out of a magnetic field B = 0.020 T by applying a force of 3.2x10⁻⁵ N (figure 38-E14). It is found that the frame moves at a constant speed. Find (a) this constant speed, (b) the emf induced in the loop, (c) the potential difference between points a and b, and (d) the potential difference between points c and d.     
Figure for Q-39


ANSWER: (a) Suppose the current in the loop is = i. Only the part 'ab' of the loop is in the magnetic field and it is perpendicular to the motion. The emf will be induced only in this part. This emf,

E =vBl, where v is the constant speed. 

Current in the loop, 

i = E/R 

  =vBl/R 

Since the loop is moving with a constant speed, the magnitude of the magnetic force on the loop will be equal to the applied force but opposite in direction. So, magnetic force, 

F =3.2x10⁻⁵ N. 

But the force on a current-carrying conductor in a magnetic field,

F =ilB =vBl*lB/R =vB²l²/R

→3.2x10⁻⁵ =v*(0.02)²*(0.08)²*/2

→v =2*3.2x10⁻⁵/{(0.02)²*(0.08)²} 

   =25 m/s. 


(b) The emf induced in the loop is the same as the emf induced in part ab. It is calculated in above part (a) as,

E = vBl 

  =25*0.02*0.08 V 

  =0.04 V 

  =4x10⁻² V.   


(c) Total resistance of the loop =2 Ω. 

Resistance of the part adcb, (excluding part ab) =(72/80)*2 Ω.

→R' = 1.8 Ω. 

Current in the loop, i =vBl/R 

→i =25*0.02*0.08/2 A 

→i =0.02 A 

Hence the potential difference between points a and b due to this current =iR' 

   =0.02*1.8 V 

   =0.036 V

   =3.6x10⁻² V.

     

(d) The potential difference between points c and d =iR" 

  =(0.02 A)*{(8/80)*2 Ω} 

  =0.02*0.2 V 

  =0.004 V 

  =4.0x10⁻³ V. 

                                   





  

    40.  Figure (38-E15) shows a metallic wire of resistance 0.20 Ω sliding on a horizontal, U-shaped metallic rail. The separation between the parallel arms is 20 cm. An electric current of 2.0 µA passes through the wire when it is slid at a rate of 20 cm/s. If the horizontal component of the earth's magnetic field is 3.0x10⁻⁵ T, calculate the dip at the place.    
Figure for Q-40


ANSWER: Speed v =20 cm/s =0.20 m/s. 

l = 20 cm =0.20 m 

R =0.20 Ω 

BH =3.0x10⁻⁵ T 

i =2.0x10⁻⁶ A

Since the wire moves horizontally, the emf induced in it is due to the earth's vertical component of the magnetic field. 


emf induced, E =Bvvl

→E =Bv*0.20*0.20 V 

   =Bv*4x10⁻² V

Current induced =E/R

  =Bv*4x10⁻²/0.20 A

Equating it with the given current,

0.20Bv = 2x10⁻⁶  

→Bv =1.0x10⁻⁵ T

Hence the dip at the place, θ is given as,

 tan θ =Bv/BH

      =1.0x10⁻⁵/3.0x10⁻⁵

      =1/3

→θ = tan⁻¹(1/3).               

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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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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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CHAPTER- 6 - Friction

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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


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Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

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CHAPTER- 3 - Kinematics - Rest and Motion

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CHAPTER- 2 - "Physics and Mathematics"

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