# (a) State Faraday's law of electromagnetic induction.(b) Derive expression for the emf induced across the ends of a straight conductor of length l moving right angles to uniform magnetic field B with a uniform speed v.(c) Obtain expression for the magnetic energy stored in a solenoid in terms of the magnetic field B, area A and length l of the solenoid through which a current i is passed.

(a)

Faraday's first law of electromagnetic induction:

Whenever a conductor is placed in a changing magnetic field, an electromagnetic force is induced in it (which is called as induced emf). If the conductor is closed the current is induced in it, which is called induced current.

Faraday's second law of electromagnetic induction:

The magnitude of induced electromotive force in a circuit is equal to the rate of change of magnetic flux linked with the coil.

Mathematically induced emf is,

$\varepsilon =\frac{-d\phi}{dt}$

(b)

Here the length of the conductor is 'P' which is moving right angle to the magnetic field 'B' with the uniform speed v.

In the figure, the conductor PQ is free to move.

Consider the rectangular loop PQRS. The magnetic flux through the loop is Blx.

That is $\phi = Blx$ where x is length of RQ

When the conductor moves x is varying. So due to the rate of change of magnetic flux an emf is induced in it.

According to Faraday's law of electromagnetic induction

$\varepsilon =\frac{-d\phi}{dt}$

$\varepsilon =\frac{-d(Blx)}{dt} = Bl\frac{ dx}{dt}= Blv$

This induced emf Blv is motional emf.

(c) We know that

Induced emf $\varepsilon =L\frac{dI}{dt}$                 .......(1)

where L = self inductance

I = current

The rate of work needed in an electric circuit is

$\frac{dW}{dt}=\left | \varepsilon \right |I$                ............(2)

Substituting (1) in (2)  we get

$\frac{dW}{dt}=L\frac{dI}{dt}\times I$

$\frac{dW}{dt}=I\frac{dI}{dt}$

we know that

By integrating we get

$W = \frac{1}{2}LI^{2}$

For a solenoid of  area A and length l, the magnetic field due to the current flowing through is

$B= \mu _{0}nI$

Total flux $\phi= \left (\mu _{0}nI \right )A\left ( nl \right )=\mu_{0}n^{2}AlI$

Self-inductance $=\frac{\phi}{I}= \mu_{0}n^{2}Al$

Therefore energy stored

$W= U_{B}=\frac{1}{2}LI^{2}=\frac{B^2Al}{2\mu_0}$

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