Q. 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) $10\ Am^{2}$
B) $15\ Am^{2}$
C) $20\ Am^{2}$
D) $25\ Am^{2}$
Explanation
(d) We have, $B = \mu_{0}H + \mu_{0}I$
or $I = \frac{B - \mu_{0}H}{\mu_{0}}$ or $I = \frac{\mu H - \mu_{0}H}{\mu_{0}} = \left\lbrack \frac{\mu}{\mu_{0}} - 1 \right\rbrack H$
$I = (\mu_{r} - 1)H$
For a solenoid of $n$-turns per unit length and current $i$ $H = ni$
$\therefore I = \left( \mu_{r} - 1 \right)ni = (1000 - 1) \times 500 \times 0.5$
$I = 2.5 \times 10^{5}\ Am^{- 1}$
$\therefore$ Magnetic moment $M = IV$
$M = 2.5 \times 10^{5} \times 10^{- 4} = 25\ Am^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.