Question

The magnetic potential due to a magnetic dipole at a point on its axis distant $40 \, cm$ from its centre is found to be $2.4\times 10^{- 5} \, J \, A \, m^{- 1}$ .The magnetic moment of the dipole will be

• A. $28.6Am^{2}$
• B. $32.2 \, A \, m^{2}$
• C. $38.4 \, A \, m^{2}$
• D. None of these

Answer 1

Answer: (C)

Here, $r=40 \, cm=0.4 \, m$
$\theta =0^\circ $ (an axial line)
$M=?$
As $V=\frac{\mu _{0}}{4 \pi }\frac{M \, c o s \theta }{r^{2}}$
$2.4 \times 10^{-5}=10^{-7} \frac{M \times 1}{(0.4)^2}$
$M=38.4$ $Am^{2}$