## Electromagnetic Induction

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1

## Direction of electric current in loops using Lenz's law

When the North-pole of a bar magnet is moved towards a closed loop like a coil connected to a galvanometer, the magnetic flux through the coil increases. Hence according to Lenz's law, current is induced in the coil in such a direction that it opposes the increase in flux.This is possible only if the current in the coil is in a counter-clockwise direction with respect to an observer situated on the side of the magnet.Note that magnetic moment associated with this current has North polarity towards the North-pole of the approaching magnet. Similarly, if the North pole of the magnet is being withdrawn from the coil, the magnetic flux through the coil will decrease. To counter this decrease in magnetic flux,the induced current in the coil flows in clockwise direction and its South pole faces the receding North-pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in flux.
Three coaxial circular wire loops and a stationary observer are positioned as shown in figure. From the observer's point of view, a current flows counter clockwise in the middle loop, which is moving towards the observer with a velocity . Loops A and B are stationary. This same observer would notice that.
A
Clockwise currents are induced in loops A and B.
B
Counter clockwise currents are induced in loops A and B.
C
A clockwise current is induced in loop A, but a counter clockwise current is induced in loop B.
D
A counter clockwise current is induced in loop A, but a clockwise current is induced in loop B.
Predict the direction of induced current in a metal ring when the ring is moved towards a straight conductor with constant speed . The conductor is carrying current in the direction shown in the figure.
2

## Force required to move a conductor in a magnetic field

where B is the magnetic field,r is the overall resistance of the loop.
Example: A rectangular loop with a sliding connector of length m is situated in uniform and constant magnetic field perpendicular to the plane of loop. Resistance of connector is . Two resistances of and are connected as shown in figure. The external force required to keep the connector moving with a constant velocity perpendicular to and in the plane of the loop is :
Since resistances 3 and 6 are in parallel, tnhe equivalent resistance
Motional emf through the rod is where is the magnetic field, is the length of the rod and is the velocity of the rod.
Current through the rod is
Force on the rod
As shown in figure, a metal rod completes the circuit. The circuit area is perpendicular to a magnetic field with . If the resistance of the total circuit is , how large a force is needed to move the rod as indicated with a constant speed of ?
Lenz's Law (27)
Figure shows a top view of a bar that can slide on two friction less rails. The resistor is , and a magnetic field is directed perpendicularly downward, into the paper. Let .
(a) Calculate the applied force required to move the bar to the right at a constant speed of .
(b) At what rate is energy delivered to the resistor?
3

## Power required to move a conductor in a magnetic field

where B is the magnetic field,
l is the length of the conductor
v is the velocity of the conductor
R is the resistance
The conducting rod shown in Figure has length L and is being pulled along horizontal, frictionless conducting rails at a constant velocity . The rails are connected at one end with a metal strip. A uniform magnetic field , directed out of the page, fills the region in which the rod  moves. Assume that L = 10 cm, v = 5.0 m/s, and B = 1.2 T. What are the (a) magnitude and (b) direction (up or down the page) of the emf induced in the rod? What are the (c) size and (d) direction of the current in the conducting loop? Assume that the resistance of the rod is and that the resistance of the rails and metal strip is negligibly small. (e) At what rate is thermal energy being generated in the rod? (f) What external force on the rod is needed to maintain ? (g) At what rate does this force do work on the rod?
4

## Applications and disadvantages of eddy currents

Eddy currents are used to advantage in certain applications like:
(i) Magnetic braking in trains: Strong electromagnets are situated above the rails in some electrically powered trains. When the electromagnets are activated, the eddy currents induced in the rails oppose the motion of the train. As there are no mechanical linkages, the braking effect is smooth.
(ii) Electromagnetic damping: Certain galvanometers have a fixed core made of nonmagnetic metallic material. When the coil oscillates, the eddy currents generated in the core oppose the motion and bring the coil to rest quickly.
(iii) Induction furnace: Induction furnace can be used to produce high temperatures and can be utilised to prepare alloys, by melting the constituent metals. A high frequency alternating current is passed through a coil which surrounds the metals to be melted. The eddy currents generated in the metals produce high temperatures sufficient to melt it.
(iv) Electric power meters: The shiny metal disc in the electric power meter(analogue type) rotates due to the eddy currents. Electric currents are induced in the disc by magnetic fields produced by sinusoidally varying currents in a coil.
Disadvantages of eddy current include:
1. Heat is lost in the core of transformers due to eddy currents.
2. The value of eddy current is highly sensitive to the value of permeability. This causes difficulty in the functioning of detectors.
Consider a metallic pipe with an inner radius of . If acylindrical bar magnet of radius is droped through the pipe, it takes more time to come down than it takes for a similar unmagnetised cylindrical iron bar dropped through the metallic pipe. Explain.
5

## Mutual inductance of two long co-axial solenoids

A straight solenoid of length m has turns in the primary and turns in the secondary. If the area of cross section is , the mutual inductance will be:

Define the term 'mutual inductance' between the two coils. Obtain the expression for mutual inductance of a pair of long coaxial solenoids each of length and radii and . Total number of turns in the two solenoids are and respectively.
Calculate the mutual inductance between two coils if a current in the primary coil changes the flux by per turn in the secondary coil of turns. Also determine the induced across the ends of the secondary coil in second:
6

## Energy stored in an inductor due to a magnetic field

Example:
A long wire carries a current . Find the energy stored in the magnetic field inside a volume at a distance from the wire.
Solution:
(energy per unit volume) and

Energy

The current (in Ampere) in an inductor is given by , where is in seconds. The self-induced emf in it is . Find the energy stored in the inductor and the power supplied to it at .
The current flowing in the two coils of self-inductance and are increasing at the same rate. If the power supplied to the two coils are equal find the ratio of
The energies stored in the two coils at a given instant.
7

## Emf induced in an AC generator

In an ac generator, mechanical energy is converted to electrical energy by virtue of electromagnetic induction. If coil of turn and area is rotated at revolutions per second in a uniform magnetic field , then the motional emf produced is

where we have assumed that at time t = 0 s, the coil is perpendicular to the field.
An a.c. generator consists of a coil of turns and of area . The coil rotates at an angular speed of rpm in a uniform magnetic field of T. Find the maximum value of the emf induced.