Electromagnetic Induction Part 2

Concept:The loop inside a solenoid experiences a time-varying magnetic field. Even though the loop moves, its motion along the solenoid’s axis does not change the flux because the field is uniform inside. Induced emf is due only to the time‑variation of current.Explanation:The magnetic field inside a long solenoid is given by $B = \mu_0 n I$.Here $n = 500$ turns/m, $I = 10 \sin(1000 t)$ A, and $\mu_0 = 4\pi \times 10^{-7}$ T·m/A.The circular loop has radius $r = 1\ \text{cm} = 0.01\ \text{m}$, area $A = \pi r^2$.The magnetic flux through the loop is $\Phi = B A = \mu_0 n I \pi r^2 = \mu_0 n \pi r^2 \cdot 10 \sin(1000 t)$.Using Faraday’s law, induced emf $\varepsilon = -\frac{d\Phi}{dt} = -\mu_0 n \pi r^2 \cdot 10 \cdot 1000 \cos(1000 t)$.The magnitude of induced current is $i = \frac{|\varepsilon|}{R}$, where $R = 10\ \Omega$.So $i = \frac{10^4 \mu_0 n \pi r^2}{R} \cos(1000 t)$.Peak current $i_0 = \frac{10^4 \times (4\pi \times 10^{-7}) \times 500 \times \pi \times (0.01)^2}{10}$ A.Simplify: $i_0 = 197 \times 10^{-6}$ A = $197\ \mu$A.For a sinusoidal current, rms value is $i_{\text{rms}} = \frac{i_0}{\sqrt{2}} = \frac{197}{\sqrt{2}}\ \mu$A.Given $i_{\text{rms}} = \frac{\alpha}{\sqrt{2}}\ \mu$A, so $\alpha = 197$.Answer:$\alpha = 197$ (Option D).