JEE Physics 2023 Question with Solution

Given: side of square loop = $2.0 \, \text{cm}$, turns per centimeter in solenoid = $50$, current amplitude = $2.5 \, \text{A}$, angular frequency = $700 \, \text{rad/s}^{-1}$.

Find: the value of $x$ in the induced emf amplitude $x \times 10^{-4} \, \text{V}$.

For a long solenoid,

$$B_0 = \mu_0 n I_0$$

where $n = 50 \times 100 = 5000 \, \text{m}^{-1}$.

So,

$$B_0 = 4\pi \times 10^{-7} \times 5000 \times 2.5 = 5\pi \times 10^{-3} \, \text{T}$$

Area of the square loop is

$$A = (2.0 \times 10^{-2})^2 = 4 \times 10^{-4} \, \text{m}^2$$

The amplitude of induced emf is

$$\mathcal{E}_0 = A \omega B_0$$

Substituting the values,

$$\mathcal{E}_0 = 4 \times 10^{-4} \times 700 \times 5\pi \times 10^{-3}$$ $$\mathcal{E}_0 = 44 \times 10^{-4} \, \text{V}$$

Therefore, the value of $x$ is $44$.

The solution concludes $44$, which disagrees with the answer key $2.07$. As per the extracted working, the answer is $44$.

Given: the loop is coaxial with the solenoid, so the magnetic field passes normally through the loop.

Find: induced emf amplitude.

Magnetic flux through the loop is

$$\Phi = BA$$

If the current varies sinusoidally, then the magnetic field also varies sinusoidally, and the induced emf amplitude becomes

$$\mathcal{E}_0 = \omega \Phi_0 = \omega A B_0$$

Using

$$B_0 = \mu_0 n I_0$$

with

$$n = 5000 \, \text{m}^{-1}, \quad I_0 = 2.5 \, \text{A}$$

we get

$$B_0 = 5\pi \times 10^{-3} \, \text{T}$$

Now,

$$A = 4 \times 10^{-4} \, \text{m}^2$$

Hence,

$$\mathcal{E}_0 = 700 \times 4 \times 10^{-4} \times 5\pi \times 10^{-3}$$

which gives

$$\mathcal{E}_0 = 44 \times 10^{-4} \, \text{V}$$

Therefore, $x = 44$.