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Question

An iron rod of volume $$ 10^{-4} m^3 $$ and relative permeability $$ 1000 $$ is placed inside a long solenoid wound with $$ 5 $$ turns/cm. If a current of $$ 0.5 A $$ is passed through the solenoid, then the magnetic moment of the rod is


A
10Am2
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B
15Am2
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C
20Am2
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D
25Am2
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Solution

The correct option is D $$ 25 Am^2$$
We have $$ B = \mu_0 H + \mu_0 I $$

or $$ I = \dfrac { B - \mu_0 H}{\mu_0} $$ or $$ I = \dfrac { \mu H - \mu_0 H }{ \mu_0} = ( \dfrac { \mu}{ \mu_0} - 1)H $$

$$ I  = ( \mu_r - 1)H $$

For solenoid of n-turns per unit length and current i 
$$ H = ni $$

$$ \therefore I = ( \mu_r -1)ni = ( 1000 - 1) \times 500 \times 0.5 $$

$$I  = 2.5 \times 10^5 Am^{-1} $$

$$ \therefore $$ Magnetic moment $$ M = IV $$

$$ M  = 2.5 \times 10^5 \times 10^{-4} = 25 Am $$

Physics

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