Suppose a long solenoid of 100 cm length, ra | vedclass

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(A) The magnetic field inside the long solenoid is given by $B = \mu_0 n I$, where $n = 500 \ turns/cm = 50000 \ turns/m$.Given $I = 10 \sin(\omega t)

Vedclass · Accepted Answer

$197$

Vedclass · Answer

(A) The magnetic field inside the long solenoid is given by $B = \mu_0 n I$, where $n = 500 \ turns/cm = 50000 \ turns/m$.Given $I = 10 \sin(\omega t)$, so $B = \mu_0 n (10 \sin(\omega t))$.The magnetic flux through the loop of radius $r = 1 \ cm = 0.01 \ m$ is $\phi = B \cdot A = \mu_0 n (10 \sin(\omega t)) \cdot (\pi r^2)$.The induced $EMF$ is $\varepsilon = -\frac{d\phi}{dt} = -\mu_0 n \pi r^2 (10 \omega \cos(\omega t))$.The magnitude of induced current is $i = \frac{|\varepsilon|}{R} = \frac{\mu_0 n \pi r^2 (10 \omega \cos(\omega t))}{R}$.The peak current is $i_0 = \frac{\mu_0 n \pi r^2 (10 \omega)}{R}$.Substituting values: $\mu_0 = 4\pi 	imes 10^{-7} \ T\cdot m/A$, $n = 5 	imes 10^4 \ m^{-1}$, $r = 10^{-2} \ m$, $\omega = 10^3 \ rad/s$, $R = 10 \ \Omega$.$i_0 = \frac{(4\pi 	imes 10^{-7}) 	imes (5 	imes 10^4) 	imes \pi 	imes (10^{-2})^2 	imes 10 	imes 10^3}{10} = 20 \pi^2 	imes 10^{-6} \ A$.$i_0 = 20 	imes (9.8696) 	imes 10^{-6} \ A \approx 197.39 \ \mu A$.The r.m.s. current is $i_{rms} = \frac{i_0}{\sqrt{2}} = \frac{197.39}{\sqrt{2}} \ \mu A$.Thus, $\alpha \approx 197$.

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