Electromagnetic Induction , Faraday’s Law of Electromagnetic Induction

Electromagnetic Induction:

Whenever the magnetic flux linked with a coil (or a closed circuit) changes with time, an emf and a current are induced in the circuit. This phenomenon first observed by Faraday is called electromagnetic induction.

Magnetic flux :

Magnetic flux associated with any area is given by the expression

$\large d\phi_B = \vec{B}.\vec{ds}$

$\large \phi_B = \int \vec{B}.\vec{ds}$

Here $\vec{B}$ is the magnetic field and $\vec{ds}$ is area element. Area is taken as a vector quantity and its direction is perpendicular to surface

Magnetic Flux associated with plane coil having N turns in a uniform magnetic field $\vec{B}$  is equal to

$\large \phi_B = N ( \vec{B}.\vec{A} )$

$\large \phi_B = N B A cos\theta$

Here N is the number of turns in the loop

B = magnetic field strength

A = area of the loop

θ = angle between area vector and magnetic field.

Whenever the magnetic flux linked with a coil (or a closed circuit) changes with time, an emf and a current are induced in the circuit.

Induced emf can be given by the expression

$\displaystyle \xi = -\frac{d\phi_B}{dt}$

As such Faraday’s law in itself is complete to tell the magnitude and polarity of induced emf, it is the Lenz’s rule that is commonly used to determine the polarity of induced emf or direction of induced current.

Magnetic flux associated with a loop can be changed by various method . By changing N by changing B by changing A or by changing the angle between A and B

Example: A rectangular loop of length l and breadth b is situated in a uniform magnetic field of induction B with its plane perpendicular to the field as shown in the figure.

Calculate the rate of change of magnetic flux and the induced e.m.f. if the loop is rotated with constant angular velocity ω

(a) about an axis passing through the centre and perpendicular to the loop,

(b) about an axis passing through the centre parallel to the breadth through angle 180°

Solution : (a)

The flux linked through the loop at any instant is

φ = BA cosθ.

= Blb cos θ = Blb = constant

Rate of change of magnetic flux is

$\displaystyle = \frac{d\phi}{dt} = \frac{d}{dt}(B l b)$

$\displaystyle \frac{d\phi}{dt} = 0$

Hence Indiced e.m.f  , ξ = 0

(b)

When the loop is rotated through 180° about an axis passing through centre and parallel to breadth then the change in magnetic flux

dφ = − Blb − (Blb) = − 2Blb.

The induced e.m.f.

$\displaystyle \xi = \frac{d\phi}{dt} = \frac{2 B l b}{\pi/\omega}$

$\displaystyle \xi = \frac{2 B l b \omega}{\pi}$

Exercise 1: Referring to the previous illustration what will be the induced e.m.f. at any instant if the loop is rotated about an axis passing through the centre and parallel to the length?