JEE Physics 2026 Question with Solution

Given: A long solenoid has length $100 \, \text{cm}$, turns per unit length $n = 500 \, \text{m}^{-1}$, and current $I(t)=10\sin(\omega t)$ with $\omega = 1000 \, \text{rad/s}$. The circular loop has radius $r = 1 \, \text{cm} = 10^{-2} \, \text{m}$ and resistance $R = 10 \, \Omega$.

Find: The value of $\alpha$ if $i_{\mathrm{rms}} = \frac{\alpha}{\sqrt{2}} \, \mu \text{A}$.

Since the loop is $10 \, \text{cm}$ inside a $100 \, \text{cm}$ long solenoid, it is in the approximately uniform magnetic field region. Therefore, the solution states that motional emf is zero and only the time-varying magnetic field contributes.

The magnetic field inside the solenoid is

$$B(t)=\mu_0 n I(t)$$

So the magnetic flux through the loop is

$$\Phi = B \cdot A = (\mu_0 n I_0 \sin \omega t)(\pi r^2)$$

Substituting the given values,

$$\Phi = (4\pi \times 10^{-7})(500)(10\sin \omega t)(\pi \times (10^{-2})^2)$$ $$\Phi = 2\pi \times 10^{-3} \sin \omega t \times \pi \times 10^{-4} = 2\pi^2 \times 10^{-7} \sin \omega t \; \text{Wb}$$

The induced emf is

$$e = -\frac{d\Phi}{dt} = -2\pi^2 \times 10^{-7} \omega \cos \omega t$$

With $\omega = 1000$,

$$e_0 = 2\pi^2 \times 10^{-4} \; \text{V}$$

Hence the peak current in the loop is

$$i_0 = \frac{e_0}{R} = \frac{2\pi^2 \times 10^{-4}}{10} = 2\pi^2 \times 10^{-5} \; \text{A}$$

Therefore, the rms current is

$$i_{\mathrm{rms}} = \frac{i_0}{\sqrt{2}} = \frac{2\pi^2 \times 10^{-5}}{\sqrt{2}}$$

It is given that

$$i_{\mathrm{rms}} = \frac{\alpha}{\sqrt{2}} \, \mu \text{A} = \frac{\alpha \times 10^{-6}}{\sqrt{2}} \; \text{A}$$

Equating,

$$\alpha \times 10^{-6} = 2\pi^2 \times 10^{-5}$$ $$\alpha = 20\pi^2 \approx 20 \times 9.87 = 197.4$$

Nearest integer is $197$.

Therefore, the correct option is A.

Given: The loop is coaxial with the solenoid and is already $10 \, \text{cm}$ inside a long solenoid.

Find: Why the loop current is determined only by changing magnetic flux due to alternating current in the solenoid.

Deep inside a long solenoid, the magnetic field is uniform over the loop area and directed along the common axis. As the loop slides coaxially in this uniform-field region, the magnetic flux linked with the loop does not change because of motion alone. Hence the motional emf contribution is zero.

Only the alternating current in the solenoid makes the magnetic field time-dependent:

$$B(t)=\mu_0 n I_0 \sin \omega t$$

So the induced emf comes from

$$e = -\frac{d\Phi}{dt}$$

with

$$\Phi = B(t)\,\pi r^2$$

This directly leads to the same result:

$$i_{\mathrm{rms}} = \frac{2\pi^2 \times 10^{-5}}{\sqrt{2}} \; \text{A}$$

and therefore

$$\alpha = 197.4 \approx 197$$

So the correct option remains A.