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Magnetism

Section: Physics

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Question : 39 of 46

Marks: +1, -0

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 Ω 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)

I_0

is constant. If the magnetic moment of the loop is

N\mu_0 I_0\sin(300t)

, then N is ______.

[JEE Adv 2011 P1]

Your Answer:

Solution:

The flux through the ring is

ϕ =

B\pi r^2

Assuming the cylinder as a solenoid with close winding, we have

B =

\frac{\mu_0 I}{L}

Therefore,

ϕ =

\left(\frac{\mu_0 I_0}{L}\right) \pi r^2 \cos 300t

The induced emf is

ε =

-\frac{d\phi}{dt}

=

300 \left(\frac{\mu_0 I_0}{L}\right) \pi r^2 \sin 300t

Therefore, the current induced is

i =

\frac{\varepsilon}{R}

=

\left(\frac{\pi r^2 300}{RL}\right) \mu_0 I_0 \sin 300t

The magnetic moment is

M = Current × Area of loop

Therefore,

m =

\left(\frac{(3.14)^2 \times (0.1)^2 \times 300}{0.005 \times 10}\right) \mu_0 I_0 \sin 300t

=

6\mu_0 I_0 \sin 300t

Hence, N = 6.

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