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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 $$0.005 \Omega$$ 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 $$I=I_{0}\cos(300t)$$ where $$I_{0}$$ is constant. lf the magnetic moment of the loop is $$N\mu_{0}I_{0}\sin(300t)$$, then ` $$N$$' is

28731_d35a65032f5f406aa018043df40cf156.png


A
6
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B
7
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C
8
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D
9
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Solution

The correct option is A $$6$$
According to Ampere's circuital law the magnetic filed inside the tube is 
$$B \, = \, \dfrac{\mu_0I}{L}$$
where L is the lenght of the tube.
Flux linked the wire loop is $$\phi \, = \, B\pi r^2$$
where r is the radius of the loop
$$\phi \, = \, \dfrac{\mu_0I}{L} \pi r^2$$
$$= \, \dfrac{\mu_0\pi r^2 I_0 \, cos \, 300t}{L}$$
Induced emf in the loop is 
$$\varepsilon \, = \, \dfrac{d\phi}{dt} \, = \, - \, \dfrac{d}{dt} \, \left(\frac{\mu_0}{L} \pi r^2I_0 \, cos \, 300t \right)$$
$$= \, \dfrac{\mu_0 \pi r^2 I_0 300 \, sin \, 300t}{L}$$
Induced current in the loop is
$$i \, = \, \dfrac{\varepsilon}{R} \, = \, \dfrac{300\mu_0 \pi r^2 I_0 \, sin \, 300t}{LR}$$
where R is the resistance of the loop Magnetic moment of the loop $$M \, = \, i\pi r^2$$
$$= \, \dfrac{300 \pi^2 r^4 \mu_0 I_0 \, sin \, 300t}{LR}$$
Substitutiing the given value, we get
$$m \, = \, \dfrac{300 \, \times \, 10 \, \times \, (0.1)^4}{10 \, \times \, 0.005} \, \mu_0 I_0 \, sin \, 300t \, (Take \, \pi^2 \, = \, 10)$$
$$= \, 6\mu_0I_0 \, sin \, 300t$$
$$M \, = \, N\mu_0I_0 \, sin \, 300t$$
$$\therefore \, N = \, 6$$

Physics

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