Question

A long circular tube of length $10m$ and radius $0.3m$ carries a current $I$ along its curved surface as shown. A wire-loop of resistance $0.005$ ohm and of radius $0.1m$ 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 $\varphi =B\pi r^2$
where $r$ is the radius of the loop
$\varphi =\frac{{\mu }_{0}I}{L}\pi r^2$ (Using (i))
$=\frac{{\mu }_0\pi r^2{I}_{0}cos300t}{L}$
Induced emf in the loop is
$\epsilon =-\frac{d\varphi }{dt}=-\frac{d}{dt}\left(\frac{{\mu }_0}{L}\pi r^2{I}_{0}cos300t\right)$
$=\frac{{\mu }_0\pi r^2{I}_{0}300sin300t}{L}$
Induced current in the loop is
$i=\frac{\epsilon }{R}=\frac{300{\epsilon }_0\pi r^2{I}_{0}sin300t}{LR}$
where $R$ is the resistance of the loop
Magnetic moment of the loop $M=i\pi r^2$
$=\frac{300{\pi }^2{r}^4{\mu }_0{I}_{0}sin300t}{LR}$
Substituting the given values, we get
$M=\frac{300\times 10\times {\left(0.1\right)}^4}{10\times 0.005}{\mu }_0{I}_{0}sin300t$ (Take${\pi }^2=10)$
$=6{\mu }_0{I}_{0}sin300t$
$M=N{\mu }_0{I}_{0}sin300t$
$\therefore N=6$