Question

A long circular tube of length $10m$ and radius $0.3m$ carries a current $I$ along its curved surface as shown. A loop created of wire of radius $0.1m$ and a resistance $0.05\mathrm{\Omega }$ has been kept 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 ${}^{\prime }{N}^{\prime }$ is
$\begin{array}{rl}& A.6\\ & B.7\\ & C.8\\ & D.9\end{array}$

Answer 1

Hint: The first step should be to find out the magnetic field inside the circular tube due to the current flowing along its curved surface. Then, from that value of magnetic field, we can calculate the flux linking the wire loop. With the flux, we can calculate the emf induced and then we will get the current induced in it by dividing the emf by the loop resistance. Then, by using the magnetic moment formula we will get the answer.
Formula used:
Magnetic field inside the tube,
$B={\displaystyle \frac{{\mu }_{0}I}{L}}$
Flux linked in a wire loop,
$\varphi =B\pi r^2$
Emf induced,
$\epsilon ={\displaystyle \frac{d\varphi }{dt}}$
Magnetic moment,
$M=\pi I{r}^2$

Complete answer:
We have been given that the length, $L$ and radius, $R$ of the circular tube is $10m$ and $0.3m$ respectively. The current flowing through its curved surface is $I$, which varies as $I=I_{0}cos(300t)$ with $I_0$ being a constant.
Now, using the Ampere’s circuital law, the magnetic field inside the tube can be written as
$B={\displaystyle \frac{{\mu }_{0}I}{L}}$,
And also, the flux linking the wire loop will be
$\varphi =B\pi r^2$
Where $r$ be the radius of the wire loop placed inside the tube.
So,
$\varphi ={\displaystyle \frac{{\mu }_{0}I}{L}}\pi r^2={\displaystyle \frac{{\mu }_0\pi r^2{I}_{0}cos300t}{L}}$
Since, induced emf in a loop is given by,
$\epsilon ={\displaystyle \frac{d\varphi }{dt}}=-{\displaystyle \frac{d}{dt}}\left({\displaystyle \frac{{\mu }_0}{L}}\pi r^2{I}_{0}cos300t\right)$
$⟹\epsilon ={\displaystyle \frac{{\mu }_0\pi r^2{I}_{0}300sin300t}{L}}$
Therefore, current induced in the loop will be,
$i={\displaystyle \frac{\epsilon }{r^{\prime }}}$
Where $r^{\prime }$ is the resistance of the loop.
$\therefore i={\displaystyle \frac{{\mu }_0\pi r^2{I}_{0}300sin300t}{L{r}^{\prime }}}$
Since, magnetic moment of the loop is given by,
$M=\pi I{r}^2$
$\Rightarrow M={\displaystyle \frac{300{\pi }^2{r}^4{\mu }_0{I}_{0}sin300t}{L{r}^{\prime }}}$
Now, let us substitute the given values and taking
${\pi }^2=10$
$\Rightarrow M={\displaystyle \frac{300\times 10\times (0.1{)}^4}{10\times 0.005}}{\mu }_0{I}_{0}sin300t=6{\mu }_0{I}_{0}sin300t$
Comparing this term with the one given in the question, we get
$N=6$

So, the correct answer is “Option A”.

Note:
 Initially, there is no current flowing in the wire loop. The magnetic field produced by the circular tube will induce an emf inside the loop and because of that, there will be current. The value of emf can vary with the position of the loop.