Question

A rod of length $L$ and resistance $r$ rotates about one end with angular velocity $\omega$ as shown in the figure, its other end touches a conducting ring of negligible resistance. A resistance $R$ is connected between the centre and periphery. There is a uniform magnetic field $B$ directed as shown in the figure. Find the current in resistance $R$.

• A. $\frac{B\omega L^2}{2r}$
• B. $\frac{B\omega L^2}{2R}$
• C. $\frac{B\omega L^2}{2(R+r)}$
• D. $\frac{B\omega L^2}{R+r}$

Answer 1

The correct option is C $\frac{B\omega L^2}{2(R+r)}$

When a conductor of length $L$ is rotated with uniform angular velocity $\omega$ in a magnetic field $B$, then an emf is induced in that conductor.

$E_{induced}=\frac{B\omega L^2}{2}$

The rod and the resistance can be considered as the circuit shown below,


$\therefore i_{\text{induced}}=\frac{E_{\text{induced}}}{R_{eq}}=\frac{B\omega L^2}{2(R+r)}$

Hence, option $\left(C\right)$ is correct.

Why this Question To understand the concept of induced emf in a rotating conductor placed in a magnetic field.