Question

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

• A. $20A{m}^2$
• B. $25A{m}^2$
• C. $30A{m}^2$
• D. $35A{m}^2$

Answer 1

The correct option is B $25A{m}^2$

We know that, $B={\mu }_{0}H+{\mu }_{0}I$
$\therefore I=\frac{B-{\mu }_{0}H}{{\mu }_0}=\frac{{\mu }_{0}H-{\mu }_{0}H}{{\mu }_0}=(\frac{{\mu }_0}{{\mu }_0}-1)H$
$I=({\mu }_r-1)H$
For a solenoid having $n$ turns/ length and current $i$,
$H=ni$
$I=({\mu }_r-1)ni$
$=(1000-1)500\times 0.5$
$=2.5\times {10}^{5}A{m}^{-1}$
$\therefore$ Magnetic moment, $M=IV$
$=2.5\times {20}^5={10}^{-4}=25A{m}^2$.