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
B
15Am2
C
20Am2
D
25Am2

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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