Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a cu

Question

Suppose a long solenoid of 100 cm length, radius 2 cm having 500 turns per unit length, carries a current $I=10 \sin (\omega \mathrm{t}) \mathrm{A}$, where $\omega=1000 \mathrm{rad} . / \mathrm{s}$. A circular conducting loop $(B)$ of radius 1 cm coaxially slided through the solenoid at a speed $v=1 \mathrm{~cm} / \mathrm{s}$. The r.m.s. current through the loop when the coil $B$ is inserted 10 cm inside the solenoid is $\alpha / \sqrt{2} \mu \mathrm{~A}$. The value of $\alpha$ is $\_\_\_\_$ .
[Resistance of the loop $=10 \Omega$ ]

Professor PC Thomas Classes · Answer

The magnetic field inside the solenoid is $B = \mu_0 n I = \mu_0 (500) (10 \sin \omega t) = 5000 \mu_0 \sin \omega t$. The flux through the loop is $\Phi = B A_{loop} = (5000 \mu_0 \sin \omega t) (\pi 	imes 0.01^2) = 0.5 \pi \mu_0 \sin \omega t$. The induced EMF is $E = -\frac{d\Phi}{dt} = -0.5 \pi \mu_0 \omega \cos \omega t$. The amplitude of the EMF is $E_0 = 0.5 \pi (4\pi 	imes 10^{-7}) (1000) = 2\pi^2 	imes 10^{-4} \approx 19.7 	imes 10^{-4} 	ext{ V}$. The RMS EMF is $E_{rms} = \frac{E_0}{\sqrt{2}}$. The RMS current is $I_{rms} = \frac{E_{rms}}{R} = \frac{19.7 	imes 10^{-4}}{10 \sqrt{2}} = \frac{197 	imes 10^{-6}}{\sqrt{2}} 	ext{ A} = \frac{197}{\sqrt{2}} \mu	ext{A}$. Thus, $\alpha = 197$.