Explanation

Inside a long solenoid, the magnetic field is uniform and given by

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

where $n = 500\,m^{-1}$ and $I(t) = 10\sin(\omega t)$.

The loop is inserted $10\,cm$ inside the solenoid. Since the loop radius is only $1\,cm$ and edge effects are negligible, the entire loop lies completely within the uniform magnetic field region.

Hence the magnetic flux linked with the loop is

$$ \Phi(t) = B(t)\,A $$

where $A = \pi r^2 = \pi(10^{-2})^2$.

Although the loop is moving, its area inside the magnetic field remains constant. Therefore, the motional emf term is zero.

Using Faraday’s law,

$$ \mathcal{E} = \left|\frac{d\Phi}{dt}\right| = A\left|\frac{dB}{dt}\right| $$

Now,

$$ \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} = \mu_0 n (10\omega\cos\omega t) $$

Thus the induced emf is sinusoidal with amplitude

$$ \mathcal{E}_0 = \mu_0 n A (10\omega) $$

Substituting $\mu_0 = 4\pi\times10^{-7}$, $n = 500$, $r = 10^{-2}\,m$, $\omega = 1000$,

$$ \mathcal{E}_0 = 1.97\times10^{-3}\,V $$

Hence,

$$ \mathcal{E}_{rms} = \frac{\mathcal{E}_0}{\sqrt{2}} $$

Using $I_{rms} = \dfrac{\mathcal{E}_{rms}}{R}$ with $R = 10\Omega$,

$$ I_{rms} = \frac{1.97\times10^{-3}}{10\sqrt{2}} = \frac{197}{\sqrt{2}}\,\mu A $$

Therefore,

$$ \alpha = 197 $$

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