Question

A long circular tube of length $10\, m$ and radius $0.3 \,m $ carries a current $I$ along its curved surface as shown. A wire-loop of resistance $0.005$ ohm and of radius $0.1 \,m $ is placed inside the tube with its axis coinciding with the axis of the tube

The current varies as $I=I_{0}\, cos(300t)$ where $I_{0}$ is constant. If the magnetic moment of the loop is $N\mu_{0}I_{0}sin(300t)$, then $N$ is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (D)

According to Amperes circuital law the magnetic field inside the tube is
$B=\frac{\mu_{0}\,I}{L}$ $\dots (i)$
where $L$ is the length of the tube Flux linked with the wire loop is $\phi=B\, \pi\, r^{2}$
where $r$ is the radius of the loop
$\phi=\frac{\mu_{0}I}{L} \pi r^{2}$ (Using (i))
$=\frac{\mu_{0}\pi r^{2}I_{0}\,cos\,300t}{L}$
Induced emf in the loop is
$\varepsilon=-\frac{d \phi}{dt}=-\frac{d}{dt}\left(\frac{\mu_{0}}{L}\pi r^{2}\,I_{0}\,cos\,300t\right)$
$=\frac{\mu_{0}\pi r^{2}I_{0}300\,sin\,300t}{L}$
Induced current in the loop is
$i=\frac{\varepsilon}{R}=\frac{300\,\varepsilon_{0}\,\pi r^{2}\,I_{0}\,sin\,300t}{LR}$
where $R$ is the resistance of the loop
Magnetic moment of the loop $M = i\pi r^{2}$
$=\frac{300\pi^{2}r^{4}\mu_{0} I_{0}\,sin\,300t}{LR}$
Substituting the given values, we get
$M=\frac{300\times10\times\left(0.1\right)^{4}}{10\times0.005} \mu_{0}I_{0}\,sin\,300t$ (Take$ \pi^{2}=10)$
$=6\mu_{0}I_{0}\,sin300t$
$M=N\mu_{0}I_{0}sin 300t$
$\therefore N=6$