A long circular tube of length \(10 \mathrm{~m}\) and radius \(0.3 \mathrm{~m}\) carries a current I along its curved surface as shown. A wire loop of resistance \(0.005 \Omega\) and of radius \(0.1 \mathrm{~m}\) is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as \(I=I_0 \cos 300 t\), where \(I_0\) is constant. If the magnetic moment of the loop is \(N \mu_0 I_0 \sin 300 t\), then \(N\) is

Take the circular tube as a long solenoid. The wires are closely wound. Magnetic field inside the solenoid is
\(
B=\mu_0 n i
\)
Here, \(n=\) number of turns per unit length
\(\therefore n i=\) current per unit length
In the given problem,
\(
\begin{aligned}
& n i=\frac{I}{L} \\
\therefore \quad B & =\frac{\mu_0 I}{L}
\end{aligned}
\)
Flux passing through the circular coil is
\(
\phi=B S=\left(\frac{\mu_0 I}{L}\right)\left(\pi r^2\right)
\)
Induced emf, \(e=-\frac{d \phi}{d t}=-\left(\frac{\mu_0 \pi r^2}{L R}\right) \cdot \frac{d I}{d t}\)
Induced current,
\(
i=\frac{e}{R}=-\left(\frac{\mu_0 \pi r^2}{L R}\right) \frac{d I}{d t}
\)
Magnetic moment \(i A=i \pi r^2\)
or \(\quad M=-\left(\frac{\mu_0 \pi^2 r^4}{L R}\right) \cdot \frac{d I}{d t}\)
Given, \(\quad I=I_0 \cos 300 t\)
\(
\therefore \quad \frac{d I}{d t}=-300 I_0 \sin (300 t)
\)

Substituting in Eq. (i), we get
\(
\begin{aligned}
M & =\left(\frac{300 \pi^2 r^4}{L R}\right) \mu_0 I_0 \sin 300 t \\
\therefore \quad N & =\frac{300 \pi^2 r^4}{L R}
\end{aligned}
\)
Substituting the values, we get
\(
\begin{aligned}
N & =\frac{300(22 / 7)^2(0.1)^4}{(10)(0.005)} \\
& =5.926
\end{aligned}
\)
or \(\quad N \simeq 6\)
Analysis of Question
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