JEE Physics 2023 Question with Solution

Given: A metallic rod of length $L$ rotates with angular speed $\omega$ in a uniform magnetic field $B$ about one end.

Find: The induced emf across the rod.

For an element of the rod at a distance $x$ from the axis, the linear speed is

$$v = \omega x$$

The differential emf across a small element $dx$ is

$$d\epsilon = Bv \, dx$$

Substituting $v = \omega x$,

$$d\epsilon = B\omega x \, dx$$

Now integrate from $x = 0$ to $x = L$:

$$\epsilon = \int_0^L B\omega x \, dx$$ $$\epsilon = B\omega \int_0^L x \, dx$$ $$\epsilon = B\omega \left[ \frac{x^2}{2} \right]_0^L$$ $$\epsilon = \frac{1}{2} B L^2 \omega$$

Therefore, the induced emf is $\frac{1}{2} B L^2 \omega$, so the correct option is C.

Given: A rod of length $L$ rotates about one end with angular speed $\omega$ in uniform magnetic field $B$.

Find: The induced emf.

For a rod rotating about one end in a uniform magnetic field normal to the plane of rotation, the standard result is

$$\epsilon = \frac{1}{2} B L^2 \omega$$

Therefore, the correct option is C.