Question

A $1.0\,m$ long metallic rod is rotated with an angular frequency of $400\,rad{\kern 1pt} {s^{ - 1}}$ about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of $0.5\,T$ parallel to the axis exists everywhere. Calculate the emf developed between the center and the ring.

Answer 1

In order to solve this question we need to understand faraday’s law of magnetic induction. According to Faraday's law, if a magnetic flux through a loop changes then emf would be developed in the loop and due to this current flows in loop. Flux could be changed through a loop by three factors, first is by changing magnetic field, second is by changing area under loop and third is by changing orientation of the loop in magnetic field.

Complete step by step answer:
When we vary the area under the loop by providing velocity to the loop, emf induced in the loop is known as motional emf. So when a rod of length “l” rotated in magnetic field “B” with angular frequency $\omega$ about its one end, then the emf induced in loop is given by,
$e = \dfrac{1}{2}B{l^2}\omega$
According to question,
$B = 0.5T$
$\Rightarrow l = 1.0\,m$
$\Rightarrow \omega = 400{\kern 1pt} rad{\kern 1pt} {\sec ^{ - 1}}$
So putting the values we get,
$e = \dfrac{1}{2}(0.5){(1.0)^2}(400)$
$\therefore e = 100{\kern 1pt} {\kern 1pt} V$

So the emf developed in loop is given by, $E = 100\,V$.

Note: It should be remembered that current induced in loop flows in a direction which is determined by Lenz law. Lenz law states that induced current would flow in circuit in such a direction so as to oppose the cause of its generation. For example, when magnetic flux increases through a loop, then it would induce in such a direction such that it opposes the increment in magnetic flux.