Question

A metallic rod of length $l$is rotated with a frequency $\nu $ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $l$, about an axis passing through the centre and perpendicular to the plane of the ring. A constant uniform magnetic field \[{\mathbf{B}}\] parallel to the axis is present everywhere. Using Lorentz force, explain how emf. is induced between the centre and the metallic ring and hence obtain the expression for it.

Answer 1

Hint:We already know that when the rod rotates, the free electron/free electrons in the rod move towards the outer end due to Lorentz force and get distributed over the ring. This produces an emf across the ends of the rod. After a certain value of emf there is no more flow of electrons and a steady state is reached.

Formula Used:We know from our knowledge of Lorentz force that:
\[dE = B\dfrac{{dA}}{{dt}}\]
Here, B is the uniform magnetic field and A represents the area swept by the infinitesimal change in length $dx$.

Complete step by step solution:
Let a metallic rod of length $l$, rotating with angular velocity in a uniform magnetic field\[B\], the plane of rotation being perpendicular to the magnetic field. Let an external length, \[dx\]at a distance \[x\]from the centre have a linear velocity\[v\].
Now, the area is equal to $A = vdxdt$.
Then, substituting the value of $A$in the above mathematical formula we have:
\[dE = B\dfrac{{dA}}{{dt}} = Bvdx\]
We, also know that from the formula for angular velocity, we have:
$v = x\omega $
Putting this value instead of the velocity in the above equation, we have:
\[dE = Bx\omega dx\]
To obtain the expression for the induced emf we have to integrate it over the limits 0 to $l$. Thus,
\[E = \int\limits_0^l {Bx\omega dx} \]
Integrating the above equation, we get;
$E = B\omega \dfrac{{{l^2}}}{2}$

This is the final mathematical expression for induced emf.

Note:All points on the rod are moving perpendicular to the magnetic field. Hence, all elementary small elements of the rod induce a small potential difference and the net potential difference in the rod is the integration of the potential differences along the rod.