Step-by-step Derivation

1. Model the physical setup and interpret the current

The problem describes a long circular tube carrying a current I along its curved surface. For a long tube, this setup is equivalent to an ideal solenoid or a cylindrical current sheet. The magnetic field inside a long solenoid is uniform and parallel to its axis.

The current is given as $I = I_0 \cos(300t)$. The phrase "carries a current I along its curved surface" over a length of $L=10$ m implies that this total current is distributed over the length of the tube. This gives rise to a surface current density (current per unit axial length), $K$, flowing azimuthally (in circles around the tube's axis).

$K(t) = \frac{I(t)}{L} = \frac{I_0 \cos(300t)}{L}$

2. Calculate the magnetic field inside the tube

The magnetic field inside a long solenoid (or a cylindrical current sheet) is given by $B = \mu_0 K$. This field is uniform inside the tube and directed along its axis.

$B(t) = \mu_0 K(t) = \frac{\mu_0 I_0}{L} \cos(300t)$

3. Calculate the magnetic flux through the wire loop

The wire loop is placed coaxially inside the tube. It has a radius of $r = 0.1$ m. The area of the loop is $A_{loop} = \pi r^2$. Since the magnetic field is uniform and parallel to the axis of the loop, the magnetic flux $\Phi_B$ through the loop is:

$\Phi_B(t) = B(t) \cdot A_{loop} = \left( \frac{\mu_0 I_0}{L} \cos(300t) \right) (\pi r^2)$

$\Phi_B(t) = \frac{\mu_0 \pi r^2 I_0}{L} \cos(300t)$

4. Find the induced e.m.f. in the loop

According to Faraday's law of electromagnetic induction, the induced electromotive force (e.m.f.) is the negative time derivative of the magnetic flux:

$\mathcal{E}(t) = -\frac{d\Phi_B}{dt} = -\frac{d}{dt} \left( \frac{\mu_0 \pi r^2 I_0}{L} \cos(300t) \right)$

$\mathcal{E}(t) = -\frac{\mu_0 \pi r^2 I_0}{L} \cdot (-\sin(300t) \cdot 300)$

$\mathcal{E}(t) = \frac{300 \mu_0 \pi r^2 I_0}{L} \sin(300t)$

5. Calculate the induced current in the loop

The loop has a resistance of $R_{loop} = 0.005 \, \Omega$. Using Ohm's law, the induced current $I_{loop}$ is:

$I_{loop}(t) = \frac{\mathcal{E}(t)}{R_{loop}} = \frac{300 \mu_0 \pi r^2 I_0}{L R_{loop}} \sin(300t)$

6. Determine the magnetic moment of the loop

The magnetic moment ($M$) of a current loop is given by the product of the current in the loop and its area:

$M(t) = I_{loop}(t) \cdot A_{loop} = \left( \frac{300 \mu_0 \pi r^2 I_0}{L R_{loop}} \sin(300t) \right) \cdot (\pi r^2)$

$M(t) = \frac{300 \mu_0 \pi^2 r^4 I_0}{L R_{loop}} \sin(300t)$

7. Compare with the given expression to find N

The problem states that the magnetic moment of the loop is $M(t) = N{\mu _0}{I_0}\sin (300t)$. By comparing this with our derived expression, we can find the value of N:

$N{\mu _0}{I_0}\sin (300t) = \frac{300 \mu_0 \pi^2 r^4 I_0}{L R_{loop}} \sin(300t)$

$N = \frac{300 \pi^2 r^4}{L R_{loop}}$

8. Substitute the given values

We are given:

• Length of the tube, $L = 10$ m
• Radius of the loop, $r = 0.1$ m
• Resistance of the loop, $R_{loop} = 0.005 \, \Omega$
• Angular frequency, $\omega = 300$ rad/s (from the argument of cosine)

$N = \frac{300 \cdot \pi^2 \cdot (0.1)^4}{10 \cdot 0.005}$

$N = \frac{300 \cdot \pi^2 \cdot 10^{-4}}{0.05}$

$N = \frac{3 \times 10^2 \cdot \pi^2 \cdot 10^{-4}}{5 \times 10^{-2}} = \frac{3}{5} \pi^2 \frac{10^{-2}}{10^{-2}} = 0.6 \pi^2$

For integer-type questions in JEE, it's common to use the approximation $\pi^2 \approx 10$.

$N = 0.6 \times 10 = 6$

Using a more precise value, $\pi^2 \approx 9.87$, we get $N \approx 0.6 \times 9.87 = 5.922$, which rounds to 6.

Thus, the integer value of N is 6.