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 $I=I_0 \cos (300 t)$ where $I_0$ is constant. If the magnetic moment of the loop is $N \mu_0 I_0 \sin (300 t)$, then ' $N$ ' is
IIT 2011, Diffcult
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$l =10 m, r _1=0.3 \ m$
$R =0.005 \Omega, r _2=0.1 \ m$
$I = I _0 \cos (300 t )$
So we get $\omega=300$
The magnetic field of the tube, $B=\frac{\mu_0 NI }{ L }$
In the problem $N =1, \quad L=10 m, \quad I = I _0 \cos (300 t$ )
So, $B=\frac{\mu_0 I_0 \cos (300 t)}{10}$
$Q=\frac{\mu_0 I_0 \cos (300 t)}{10} \times \pi(0.1)^2$
$\text { Induced emf, } e=-\frac{d Q}{d t}=300 \mu_0 \pi I_0 \sin (300 t) \times 10^{-3}$
$e=0.3 \mu_0 \pi I_0 \sin (300 t)$
The current in the ring $i =\frac{ e }{ R }$
$i =\frac{0.3 \pi \mu_0 I _0 \sin (300 t)}{0.005}$
$\text { or } i =60 \mu_0 \pi I _0 \sin (300 t)$
$\text { Magnetic moment, } M = iA = i \pi r ^2$
$M =60 \mu_0 \pi I _0 \sin (300 t) \times \pi(0.1)^2$
$=0.6 \mu_0 \pi^2 I _0 \sin (300 t) A$
$=\left(0.6 \times \pi^2\right) \mu_0 I _0 \sin (300 t)$
$=6 \mu_0 I _0 \sin (300 t)$
So, the value of $N =6$
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