Magnetic Moment of Wire Loop Inside Long Current Tube

To find the value of $N$, we analyze the electromagnetic induction in the wire-loop step-by-step.

1. Magnetic Field Inside the Long Tube

The tube acts as a long solenoid carrying a surface current $I$ around its circumference. The magnetic field $B(t)$ inside such a uniform tube along its axis is given by:
$B(t) = \mu_0 \left(\frac{I(t)}{L}\right)$

Given that $I(t) = I_0 \cos(300t)$ and length $L = 10\text{ m}$:
$B(t) = \frac{\mu_0 I_0}{L} \cos(300t)$

2. Magnetic Flux Linked with the Wire-Loop

The loop has a radius $r = 0.1\text{ m}$ and its plane is perpendicular to the axis of the tube. The magnetic flux $\Phi(t)$ through the area of the loop $A = \pi r^2$ is:
$\Phi(t) = B(t) \cdot A = \left(\frac{\mu_0 I_0}{L} \cos(300t)\right) (\pi r^2) = \frac{\pi r^2 \mu_0 I_0}{L} \cos(300t)$

3. Induced EMF and Current in the Loop

By Faraday's Law of Electromagnetic Induction, the magnitude of the induced electromotive force (EMF) $\mathcal{E}(t)$ in the loop is:
$\mathcal{E}(t) = \left| -\frac{d\Phi(t)}{dt} \right| = \frac{\pi r^2 \mu_0 I_0}{L} \cdot 300 \sin(300t)$

Using Ohm's law, the induced current $i(t)$ flowing through the loop of resistance $R = 0.005\ \Omega$ is:
$i(t) = \frac{\mathcal{E}(t)}{R} = \frac{300 \pi r^2 \mu_0 I_0}{L R} \sin(300t)$

4. Magnetic Moment of the Loop

The magnetic moment $M(t)$ of the wire-loop is the product of the induced current $i(t)$ and the area of the loop $A = \pi r^2$:
$M(t) = i(t) \cdot (\pi r^2) = \left( \frac{300 \pi r^2 \mu_0 I_0}{L R} \sin(300t) \right) (\pi r^2) = \frac{300 \pi^2 r^4}{L R} \mu_0 I_0 \sin(300t)$

5. Calculation of $N$

Comparing this with the given expression $M(t) = N \mu_0 I_0 \sin(300t)$, we obtain:
$N = \frac{300 \pi^2 r^4}{L R}$

Substituting the given numerical values:

• $r = 0.1\text{ m} \implies r^4 = 10^{-4}\text{ m}^4$
• $L = 10\text{ m}$
• $R = 0.005\ \Omega = 5 \times 10^{-3}\ \Omega$
• Taking $\pi^2 \approx 10$

$N = \frac{300 \times 10 \times 10^{-4}}{10 \times (5 \times 10^{-3})} = \frac{0.3}{0.05} = 6$

Thus, the value of $N$ is $6$.