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

Answer 1

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

We have, $B={\mu }_{0}H+{\mu }_{0}IOrI=\frac{B-{\mu }_{0}H}{{\mu }_0}orI=\frac{\mu H-{\mu }_{0}H}{{\mu }_0}=(\frac{\mu }{{\mu }_0}-1)HI=({\mu }_r-1)H$
For a solenoid of n-turns per unit length and current i
$H=ni\therefore I=({\mu }_r-1)ni=(1000-1)\times 500\times 0.5I=2.5\times {10}^{5}A{m}^{-1}$
$\therefore$ Magnetic moment M=IV
$M=2.5\times {10}^5\times {10}^{-4}=25A{m}^2$