Discussion Forum

1)

A long circular tube of length 10 m and radius 0.3 m carries a current l along its curved surface as shown. A wire loop of resistance 0.005 $\Omega$  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 $l=l_{0}\cos 300 t$ , where  $l_{0}$  is constant . If the magnetic moment of the loop  is $N \mu _{0}l_{0} \sin 300t$, then N is 

Answer:

Option A

Explanation:

Take the circular tube as a long solenoid. The wires are closely wound.
Magnetic field inside the solenoid is

  $B= \mu_{0} ni$

 Here, n= number of turns per unit length

 $\therefore$ ni= current per unit length$

 In the given problem,

 $ni= \frac{I}{L}$

 $\therefore$  $B= \frac{\mu_{0} I}{L}$

 Flux passing through the circular coil is 

 $\phi =BS=\left(\frac{\mu_{0}l}{L}\right)(\pi r^{2})$

 Induced emf, $e= -\frac{d \phi}{dt}=-\left(\frac{\mu_{0} \pi r^{2}}{LR}\right). \frac{dl}{dt}$

 magnetic moment   $iA=i \pi r^{2}$

 or    $M=-\left(\frac{\mu_{0} \pi^{2} r^{4}}{LR}\right). \frac{dl}{dt}$....(i)

 Given,  $l=l_{0} \cos 300 t$

$\therefore$  $\frac{dl}{dt}=-300 l_{0} \sin (300 t)$

 Substituting in Eq.(i), we get

 $M=\left(\frac{300 \pi^{2} r^{4}}{LR}\right).\mu_{0}l_{0} \sin 300t$

 $\therefore$  $N= \frac{ 300 \pi^{2} r^{4}}{LR}$

 Substituting the values , we get

 $N= \frac{300(22/7)^{2}(0.1)^{4}}{(10)(0.005)}$=5.926

   or N=6