JEE Physics 2023 Question with Solution

Given: rod length $l = 20 \, \text{cm} = 0.20 \, \text{m}$, magnetic field $B = 0.2 \, \text{T}$, rotation speed $210 \, \text{rpm}$.

Find: the emf developed between the centre and the ring.

For a rod of length $l$ rotating with angular speed $\omega$ in a uniform magnetic field parallel to the axis, the motional emf is

$$\mathcal{E} = \frac{1}{2} B \omega l^2$$

Convert angular speed into SI units:

$$\omega = 210 \times \frac{2\pi}{60} = 7\pi \; \text{rad s}^{-1}$$

Using $\pi = \dfrac{22}{7}$,

$$\omega = 22 \; \text{rad s}^{-1}$$

Substitute the values:

$$\mathcal{E} = \frac{1}{2} \times 0.2 \times 22 \times (0.20)^2$$ $$\mathcal{E} = 0.1 \times 22 \times 0.04 = 0.088 \; \text{V}$$

Convert into millivolts:

$$0.088 \; \text{V} = 88 \; \text{mV}$$

Therefore, the emf developed between the centre and the ring is $88 \, \text{mV}$.

Given: $B = 0.2 \, \text{T}$, $l = 0.20 \, \text{m}$, and $\omega = 22 \; \text{rad s}^{-1}$.

Find: the induced emf.

The shortcut is to first convert $210 \, \text{rpm}$ directly to $22 \; \text{rad s}^{-1}$ using the given value of $\pi$, and then use

$$\mathcal{E} = \frac{1}{2} B \omega l^2$$

So,

$$\mathcal{E} = \frac{1}{2} \times 0.2 \times 22 \times (0.20)^2 = 0.088 \; \text{V} = 88 \; \text{mV}$$

This works because the magnetic field is uniform and parallel to the axis of rotation. Therefore, the required answer is $88 \, \text{mV}$.