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 $0.5A$ is passed through the solenoid, then the magnetic moment of the rod is

• A. $10A{m}^2$
• B. $15A{m}^2$
• C. $20A{m}^2$
• D. $25A{m}^2$

Answer 1

Answer: (D)

We have, $B={\mu }_{0}H+{\mu }_{0}I$
or $I=\frac{B-{\mu }_{0}H}{{\mu }_0}$
or $I=\frac{\mu H-{\mu }_{0}H}{{\mu }_0}$
$=\left(\frac{\mu }{{\mu }_0}-1\right)H$
$I=\left({\mu }_r-1\right)H$
For a solenoid of n -turns per unit length and current $i$
$H=ni$
$\therefore I=({\mu }_r-1)ni=(1000-1)\times 500\times 0.5$
$I=2.5\times 10^{5}A{m}^{-1}$
$\therefore$ Magnetic moment, $M=IV$
$M=2.5\times 10^5\times 10^{-4}=25Am$