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. $10\, Am ^{2}$
• B. $15\, Am ^{2}$
• C. $20\, Am ^{2}$
• D. $25\, Am ^{2}$

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 $i$
$H=ni$
$\left.\therefore I={(} \mu_{r}-1\right) n i=(1000-1) \times 500 \times 0.5$
$I=2.5 \times 10^{5} Am ^{-1}$
$\therefore $ Magnetic moment, $M=I V$
$M=2.5 \times 10^{5} \times 10^{-4}=25\, Am$