A long circular tube of length 10 m and radi | vedclass

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(B) Given: Length of tube $L = 10 \ m$,radius of tube $r_1 = 0.3 \ m$.Resistance of loop $R = 0.005 \ \Omega$,radius of loop $r_2 = 0.1 \ m$.Current i

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$6$

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(B) Given: Length of tube $L = 10 \ m$,radius of tube $r_1 = 0.3 \ m$.Resistance of loop $R = 0.005 \ \Omega$,radius of loop $r_2 = 0.1 \ m$.Current in the tube $I = I_0 \cos(300t)$,so angular frequency $\omega = 300 \ rad/s$.The magnetic field inside a long solenoid (tube) is $B = \mu_0 n I$,where $n$ is the number of turns per unit length. Assuming a single current sheet equivalent to $n=1/L$ turns per unit length,$B = \frac{\mu_0 I}{L}$.Substituting the values: $B = \frac{\mu_0 I_0 \cos(300t)}{10}$.Magnetic flux through the loop $\Phi = B \cdot A = \frac{\mu_0 I_0 \cos(300t)}{10} \cdot \pi(r_2)^2 = \frac{\mu_0 I_0 \cos(300t)}{10} \cdot \pi(0.1)^2 = \frac{\pi \mu_0 I_0 \cos(300t)}{1000}$.Induced emf $e = -\frac{d\Phi}{dt} = -\frac{d}{dt} \left[ \frac{\pi \mu_0 I_0 \cos(300t)}{1000} \right] = \frac{300 \pi \mu_0 I_0 \sin(300t)}{1000} = 0.3 \pi \mu_0 I_0 \sin(300t)$.Induced current in the loop $i = \frac{e}{R} = \frac{0.3 \pi \mu_0 I_0 \sin(300t)}{0.005} = 60 \pi \mu_0 I_0 \sin(300t)$.Magnetic moment $M = i \cdot A = (60 \pi \mu_0 I_0 \sin(300t)) \cdot (\pi (0.1)^2) = 60 \pi^2 \mu_0 I_0 \sin(300t) \cdot 0.01 = 0.6 \pi^2 \mu_0 I_0 \sin(300t)$.Using $\pi^2 \approx 10$,$M \approx 0.6 \cdot 10 \mu_0 I_0 \sin(300t) = 6 \mu_0 I_0 \sin(300t)$.Comparing with $N \mu_0 I_0 \sin(300t)$,we get $N = 6$.

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