JEE Physics 2023 Question with Solution

Given: Number of turns $N = 100$, area of each turn $A = 14 \times 10^{-2} \, m^2$, magnetic field $B = 3.0 \, \text{T}$, rotation speed $= 360$ rev/min.

Find: Maximum emf induced in the generator.

For an ac generator, the maximum emf is

$$\xi_{\max} = NAB\omega$$

The angular speed is obtained from rotational frequency:

$$f = \frac{360}{60} = 6 \, \text{s}^{-1}$$

Therefore,

$$\omega = 2\pi f = 2\pi \times 6 = 12\pi$$

Now substitute in the emf expression:

$$\xi_{\max} = 100 \times 14 \times 10^{-2} \times 3 \times 12\pi$$

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

$$\xi_{\max} = 100 \times 14 \times 10^{-2} \times 3 \times 12 \times \frac{22}{7}$$ $$\xi_{\max} = 1584 \, \text{V}$$

Therefore, the maximum emf produced is $1584 \, \text{V}$.

Given: The coil rotates at $360$ revolutions per minute.

Find: First convert this to angular speed, then evaluate $\xi_{\max}$.

Since $1$ minute $= 60$ seconds,

$$f = \frac{360}{60} = 6 \, \text{rev/s}$$

Hence,

$$\omega = 2\pi f = 2\pi \times 6 = 12\pi \, \text{rad/s}$$

Now,

$$\xi_{\max} = NAB\omega$$ $$= 100 \times 0.14 \times 3 \times 12\pi$$ $$= 14 \times 3 \times 12\pi$$ $$= 504\pi$$

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

$$504\pi = 504 \times \frac{22}{7} = 1584$$

Therefore, the required numerical value is 1584.