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Chapter: 12th Physics : Electromagnetic Induction and Alternating Current

Motional emf from Faraday’s law and Energy conservation

Physics : Electromagnetic Induction: Motional emf from Faraday’s law and Energy conservation

Motional emf from Faraday’s law and Energy conservation

Let us consider a rectangular conducting loop of width l in a uniform magnetic field  which is perpendicular to the plane of the loop and is directed inwards. A part of the loop is in the magnetic field while the remaining part is outside the field as shown in Figure 4.10.


When the loop is pulled with a constant velocity  to the right, the area of the portion of the loop within the magnetic field will decrease. Thus, the flux linked with the loop will also decrease. According to Faraday’s law, an electric current is induced in the loop which flows in a direction so as to oppose the pull of the loop.

Let x be the length of the loop which is still within the magnetic field, then its area is lx . The magnetic flux linked with the loop is


As this magnetic flux decreases due to the movement of the loop, the magnitude of the induced emf is given by


Here,  both  B  and  l  are  constants. Therefore,


where v = dx/dt is the velocity of the loop.

This emf is known as motional emf since it is produced due to the movement of the loop in the magnetic field.

From Lenz’s law, it is found that the induced current flows in clockwise direction. If R is the resistance of the loop, then the induced current is given by



Energy conservation

In order to move the loop with a constant velocity  , a constant force that is equal and opposite to the magnetic force, must be applied. Therefore, mechanical work is done to move the loop. Then the rate of doing work or power is


Now, let us find the magnetic force acting on the loop due to its movement in the magnetic field. Let three deflecting forces 1 , 2 and 3 be acting on the three segments of the loop as shown in Figure 4.10. The general equation of such a deflecting force is given by


Forces 2 and 3 are equal in magnitude and opposite in direction and cancel each other. Therefore, the force 1 alone acts on the left segment of the loop in a direction shown in Figure 4.10 and is given by


Here θ is the angle between  and the length vector  for the left segment and is 90º

F1 = il Bsin90º =il B    since sin90º = 1

The applied force  must be equal to  in order to just move the loop with a 1 constant velocity 


(since  and 1 are in opposite direction)

Considering only the magnitudes,

F = F1 = i l B

Substituting for i from equation (4.10)


From equation (4.11), the rate at which the mechanical work is done to pull the loop from the magnetic field or power is given by


When the induced current flows in the loop, Joule heating takes place. The rate at which thermal energy is dissipated in the loop or power dissipated is


This equation is exactly same as the equation (4.13). Thus the mechanical work done in moving the loop appears as thermal energy in the loop.

 

EXAMPLE 4.8

A conducting rod of length 0.5 m falls freely from the top of a building of height 7.2 m at a place in Chennai where the horizontal component of Earth’s magnetic field is 40378.7 nT. If the length of the rod is perpendicular to Earth’s horizontal magnetic field, find the emf induced across the conductor when the rod is about to touch the ground. [Take g = 10 m s-2]

Solution

l = 0.5 m; h = 7.2 m; u = 0 m s-1;

g = 10 m s-2; BH = 40378.7 nT

The final velocity of the rod is


Induced emf when the rod is about to touch the ground, ε = BH lv

= 40, 378.7 ×10−9 × 0.5 ×12

=242 .27 ×10−6V

=242 .27µV

 

EXAMPLE 4.9

A copper rod of length l rotates about one of its ends with an angular velocity ω in a magnetic field B as shown in the figure. The plane of rotation is perpendicular to the field. Find the emf induced between the two ends of the rod.

Solution


Consider a small element of length dx at a distance x from the centre of the circle described by the rod. As this element moves perpendicular to the field with a linear velocity v = xω , the emf developed in the element dx is

d ε = Bvdx = B( xω)dx


This rod is made up of many such elements, moving perpendicular to the field. The emf developed across two ends is



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