JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Electromagnetic Induction

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 μ_{0} I_{0} sin (300 t)

, then 'N' is

Answer: 6

Solution

Magnetic field inside a solenoid:

B = μ_{0} ni = \frac{μ _{0} I}{L}

Magnetic flux through the loop:

Φ = B \cdot π r^{2} = \frac{μ _{0} I π r ^{2}}{L}

Induced EMF:

E = - \frac{d Φ}{d t} = - \frac{μ _{0} π r ^{2}}{L} \frac{d I}{d t} = - \frac{μ _{0} π r ^{2}}{L} \cdot (- 300 I_{0} sin 300 t)

Induced current:

i = \frac{E}{R} = \frac{300 μ _{0} π r ^{2} I _{0} s i n 300 t}{L R}

Magnetic moment:

M = i \cdot π r^{2} = \frac{300 μ _{0} π ^{2} r ^{4} I _{0} s i n 300 t}{L R}

Substituting values (L=10 m, r=0.1 m, R=0.005 Ω):

M = \frac{300 μ _{0} π ^{2} ( 0.1 ) ^{4} I _{0} sin 300 t}{10 \times 0.005} = \frac{300 μ _{0} π ^{2} \times 1 0 ^{- 4} I _{0} sin 300 t}{0.05} = 6 μ_{0} I_{0} sin 300 t

Therefore, N = 6.

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Integer answer

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JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Electromagnetic Induction

Solution

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