Saturday, March 19, 2022

H C Verma solutions, ELECTROMAGNETIC INDUCTION, Chapter-38, EXERCISES, Q41 to Q50, Concepts of Physics, Part-II

Electromagnetic Induction


EXERCISES, Q41 to Q5O


    41.  A wire ab of length l, mass m and resistance R slides on a smooth, thick pair of metallic rails joined at the bottom as shown in figure (38-E16). The plane of the rails makes an angle θ with the horizontal. A vertical magnetic field B exists in the region. If the wire slides on the rails at a constant speed v, show that

B=√(mgRsinθ)/vl²cos²θ.
The figure for Q-41

ANSWER: emf induced in the wire ab,

E =vBl.cosθ

{we will take the component of B that is perpendicular to the plane of movement of wire ab}

Current in the wire ab, 

i =E/R =vBl.cosθ/R

The magnetic force in the wire,

F =ilB.cosθ

  =vBl.cosθ*(lB.cosθ)/R

  =vB²l²cos²θ/R

The direction of this force will be opposite to the movement. 

Weight of the wire, W =mg

Component of the weight along the plane =mg.sinθ 

Since the wire is moving with a uniform speed, the net force on the wire is zero. i.e.

vB²l²cos²θ/R -mgsinθ =0

→B² = mgR.sinθ/vl²cos²θ

→B =√{(mgR.sinθ)/(vl²cos²θ)}

 





    42. Consider the situation shown in figure (38-E17). The wire P₁Q₂ and P₂Q₂ are made to slide on the rails with the same speed of 5 cm/s. Find the electric current in the 19 Ω resistor if (a) both the wires move towards the right and (b) if P₁Q₁ moves towards the left but P₂Q₂ moves towards the right. 
The figure for Q-42

 ANSWER: (a) When both the wires move towards the right with the same speed, the emf induced in each wire will be the same and in the same direction. The emf induced E =vBl, 

→E =0.05*1.0*0.04 V 

   = 0.002 V.

It will be as if two cells are joined in parallel, hence the emf will remain the same = 0.002 V. 

The equivalent resistance of the moving wires that are in parallel,

1/r =1/2 +1/2 = 1

→r = 1 Ω. 

The total resistance of the circuit,

R =1 Ω +19 Ω =20 Ω. 

Hence the current through the 19 Ω resistor,

i =E/R =0.002/20 A 

  = 0.1x10⁻³ A 

  = 0.1 mA.               


(b) When both the wires move parallelly with equal and opposite speeds in a uniform magnetic field, the EMFs induced in each wire are equal but opposite in direction. So the net emf available for the 19 Ω resistor is zero.



 

 

    43.  Suppose the 19 Ω resistor of the previous problem is disconnected. Find the current through P₂Q₂ in the two situations (a) and (b) of that problem.     


ANSWER: (a) When the 19 Ω resistor is disconnected, the points P₁ and P₂ have the same potential. Similarly, the points Q₁ and Q₂ have equal potentials. It is like two cells connected in parallel but there is no circuit completed. So the current through the wire P₂Q₂ will be zero.  


(b) In the second case, when both wires slide in opposite directions, the EMFs induced in the wires are equal and opposite. This situation is as if two cells are connected in series. The emf in the circuit will be the sum of both EMFs, i.e.

E =0.002 +0.002 V =0.004 V 

 The circuit completes through both the wires and the resistance of the circuit 

R =2 Ω +2 Ω =4 Ω 

So the current in the circuit and hence in the wire P₂Q₂,

 i =E/R 

  =0.004/4 A

  =0.001 A 

  =1.0x10⁻³ A

  =1.0 mA.          




 

    44.  Consider the situation shown in the figure (38-E18). The wire PQ has negligible resistance and is made to slide on the three rails with a constant speed of 5 cm/s. Find the current in the 10 Ω resistor when the switch S is thrown to (a) the middle rail (b) the bottom rail.    
The figure for Q-44


ANSWER: (a) When the switch s is thrown to the middle rail, the emf induced in the bottom half of the wire PQ is not contributing to the circuit that is completed through the top and middle rail. The emf induced in the upper half of the wire and contributing to the circuit, 

E =vBl 

   =0.05*1.0*0.02 V  

   =0.001 V 

Resistance in the circuit, R =10 Ω 

Hence the current in the 10 Ω resistor, 

  =E/R 

   =0.001/10 A 

   =0.0001 A

   =0.1 mA.  



(b) When the switch is thrown to the bottom wire, the emf induced in the whole wire PR is contributing to the circuit that is through the top and bottom rails. The emf induced, 

E =vBl 

   =0.05*1.0*0.04 V 

   =0.002 V. 

Resistance R =10 Ω. 

Hence the current in the 10 Ω resistor,

   i = E/R  

     = 0.002/10 A 

     = 0.0002 A 

     = 0.2 mA.                      





 

    45.  The current generator I₉, shown in figure (38-E19), sends a constant current 'i' through the circuit. The wire cd is fixed and ab is made to slide on the smooth, thick rails with a constant velocity v towards the right. Each of these wires has resistance r. Find the current through the wire cd.     
The figure for Q-45


ANSWER: Let the current in the wire cd = i'. The current in the wire ab = i -i'. The potential difference across cd = i'r. The same potential difference will be across the wire ab. Since the wire ab is moving with a velocity v, the induced emf =Bvl. Its polarity will be opposite to the current i-i'. 
Diagram for Q-45

    So the net potential difference across ab =(i-i')r-Blv. Equating the potential difference across both wires we have

i'r =(i-i')r-Blv

→2i'r =ir -Blv

→i' =(ir -Blv)/2r.

  

Watch this solution in the video below↓        




 

    46.  The current generator I₉, shown in figure (38-E20), sends a constant current i through the circuit. The wire ab has a length l and mass m and can slide on the smooth, horizontal rails connected to I₉. The entire system lies in a vertical magnetic field B. Find the velocity of the wire as a function of time.    
The figure for Q-46


ANSWER: The current through the wire ab = i. The force on the wire in the magnetic field B, 

F = ilB, the direction of this force from Flemming's left-hand rule is towards the right. 

Mass of the wire  =m,

Acceleration of the wire, a =F/m 

→a = ilB/m

 If the initial velocity u =0, at t =0 then from, v =u+at, the velocity at time t is, 

v =0 +(ilB/m)t

→v =ilBt/m.  

In the same direction as the force i.e. towards the right away from the generator.            





 

    47.  The system containing the rails and wire of the previous problem is kept vertically in a uniform horizontal magnetic field B that is perpendicular to the plane of the rails (figure 38-E21). It is found that the wire stays in equilibrium. If the wire ab is replaced by another wire of double its mass, how long will it take in falling through a distance equal to its length?    
The figure for Q-47


ANSWER: The direction of the magnetic field is coming out of the plane of the figure, hence the force on the wire will be upwards. If the current = i, then due to the equilibrium, 

ilB =mg,  

 When the mass is doubled, net downward force, 

F = 2mg -ilB =2mg -mg =mg

Downward acceleration, 

a =mg/2m =g/2  

Let time t is taken to travel a distance of l, then from 

s =ut +½at² 

→l = 0 +½(g/2)t² 

→t² =4l/g 

→t = 2√(l/g).          




 

    48.  The rectangular wire-frame, shown in figure (38-E22), has a width d, mass m, resistance R, and a large length. A uniform magnetic field B exists to the left of the frame. A constant force F starts pushing the frame into the magnetic field at t =0. (a) Find the acceleration of the frame when its speed has increased to v. (b) Show that after some time the frame will move with a constant velocity till the whole frame enters into the magnetic field. Find the velocity vₒ. (c) Show that the velocity at time t is given by 

v =vₒ(1 -e-Ft/mvₒ).          
The figure for Q-48


ANSWER: (a) When the speed is v, the emf induced in the width d of the frame, 

E =Bvd 

The current due to this emf 

i =E/R =Bvd/R 

Force on the width d, =iBd =B²d²v/R

It will be opposite to the movement. Hence the net force on the loop 

=F -B²d²v/R 

Acceleration of the loop, 

a = Force/mass 

  =(F -B²d²v/R)/m 

  =(RF -B²d²v)/mR.  


(b) Due to the acceleration speed increases and so does the opposing force. A time comes when the external and opposing forces become equal in magnitude and the net force is zero. At this time there is no acceleration and the frame moves with a constant acquired speed, say vₒ.  

       For this condition, 

   F -B²d²vₒ/R =0, 

→B²d²vₒ =FR 

→vₒ =RF/B²d².   All the entities on RHS are constant, so vₒ is constant. The velocity will remain constant till the whole frame enters the magnetic field. After that same emf will be induced in the opposite width d and net emf will be zero. So no current in the circuit. The frame will move with acceleration under the force F.


(c) As in (a) above, acceleration at time t is given as, 

a =dv/dt = (RF-B²d²v)/mR 

→dv/(RF-B²d²v) =dt/mR

Integrating both sides we get, 

(-1/B²d²) [ln(RF-B²d²v)] =(1/mR)[t]

Putting limits from 0 to v for speed and 0 to t for the time we get, 

(-1/B²d²){ln(RF-B²d²v)/RF}=t/mR

→ln{(RF-B²d²v)/RF}=-B²d²t/mR

→ln{1 -v*B²d²/RF} =-B²d²Ft/RFm 

→ln{1 -v/vₒ} =-Ft/mvₒ

→1 -v/vₒ =e-Ft/mvₒ 

→v/vₒ =1 -e-Ft/mvₒ

→v =vₒ(1 -e-Ft/mvₒ).             

Proved.






 

    49.  Figure (38-E23) shows a smooth pair of thick metallic rails connected across a battery of emf Ɛ having a negligible internal resistance. A wire ab of length l and resistance r can slide smoothly on the rails. The entire system lies in a horizontal plane and is immersed in a uniform vertical magnetic field B. At an instant t, the wire is given a small velocity v towards the right. (a) Find the current in it at this instant. What is the direction of the current? (b) What is the force acting on the wire at this instant? (c) Show that after some time the wire ab will slide with a constant velocity. Find this velocity. 
The figure for Q-49


ANSWER: (a) Induced emf in the moving wire =Bvl. 

Its direction will be positive at b and negative at a. So the net emf in the circuit 

=Ɛ -Bvl

So the current in the wire,

i = (Ɛ -Blv)/r.

 Since the velocity given is small, the direction of the current will be from b to a.


(b) The force acting on the wire ab,

F =ilB

  = {(Ɛ -Blv)/r}*lB

  = lB(Ɛ -Blv)/r. 

Towards the right.


(c) Under the force the wire will accelerate towards the right and due to this the opposing emf Blv will also increase. A time will come when the net emf in the circuit becomes zero and hence the current. Suppose the velocity at this instant is vₒ, then

Ɛ =Blvₒ

→vₒ = Ɛ/Bl.

 




  

    50.  A conducting wire ab of length l, resistance r and mass m starts sliding at t =0 down a smooth, vertical, thick pair of connected rails as shown in figure(38-E24). A uniform magnetic field B exists in the space in a direction perpendicular to the plane of the rails. (a) Write the induced emf in the loop at an instant t when the speed of the wire is v. (b) What would be the magnitude and direction of the induced current in the wire? (c) Find the downward acceleration of the wire at this instant. (d) After sufficient time, the wire starts moving with a constant velocity. Find the velocity vₘ. (e) Find the velocity of the wire as a function of time. (f) Find the displacement of the wire as a function of time. (g) Show that the rate of heat developed in the wire is equal to the rate at which the gravitational potential energy is decreased after a steady-state is reached.      
The figure for Q-50


ANSWER: (a) Induced emf in the loop is due to the movement of wire ab in the magnetic field, and it is equal to, 

E =Bvl. 


(b) Magnitude of the current in the wire

i =E/r =Bvl/r. 

 The positive charge in the moving wire will accumulate at the end b and the negative at a. So the current will be from b to a.  


(c) Upward force on the wire due to the induced emf (opposing the movement)

=ilB 

=(Bvl/r)*lB 

=vB²l²/r

Downward force 

=weight 

=mg

Net downward force, 

F = mg -vB²l²/r 

Downward acceleration of the wire at this instant, 

a = F/m 

  =(mg -vB²l²/r)/m 

  =g -vB²l²/mr.   


(d) Due to the acceleration v increases and a time comes when the opposing force vB²l²/r becomes equal to the weight mg. Now the net force on the wire is zero and it moves with a constant velocity vₘ. In this situation, 

vₘB²l²/r =mg

→vₘ =mgr/B²l².


(e) Acceleration of the wire,

dv/dt =a =g -vB²l²/mr

→dv/dt =g -vg(B²l²/mgr)

→dv/dt =g -gv/vₘ =(gvₘ -gv)/vₘ 

→dv/(gvₘ -gv) =dt/vₘ 

Integrating both sides we get,

(-1/g)[ln(gvₘ -gv)] =[t]/vₘ

Putting the limits for velocity from 0 to v and for time 0 to t, we have

{ln(gvₘ -gv) -ln(gvₘ)} =-gt/vₘ

→ln{(vₘ -v)/vₘ} =-gt/vₘ

→(vₘ -v)/vₘ = e-gt/vₘ

→vₘ -v = vₘe-gt/vₘ 

→v = vₘ -vₘe-gt/vₘ

      = vₘ(1 -e-gt/vₘ).


(f) v =dx/dt 

→dx = vdt

→X =∫dx =∫vdt

     =∫vₘ(1 -e-gt/vₘ)dt

    =∫(vₘ -vₘe-gt/vₘ)dt

    =[vₘt + vₘ(vₘ/g)e-gt/vₘ)]

     Put limits of t from 0 to t. 

    =vₘt +(vₘ²/g)e-gt/vₘ)-vₘ²/g
    =vₘt -(vₘ²/g)(1-e-gt/vₘ).
 
(g) Rate of heat developed is the power consumed by the wire resistance,
P =V*i
  =Blv*Blv/r
  =B²l²v²/r
When the steady-state is achieved,
 v =vₘ  
So, P =B²l²vₘ²/r 
     =(B²l²/r)(mgr/B²l²)² 
    =m²g²r/B²l² 

Rate of decrease of gravitational potential energy, 
dU/dt =d(mgx)dt =mg.dx/dt =mgvₘ
  =mg(mgr/B²l²)
  =m²g²r/B²l². 
Thus P =dU/dt. 

Hence proved.

  

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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 (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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For more practice on problems on friction solve these- "New Questions on Friction".

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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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