Question

If at a distance of $40 \, cm$ at an axial position of a dipole, the "magnetic potential" (analogous to electric potential) is $2.4\times 10^{- 5} \, J \, A \, m^{- 1}$ , then the magnetic moment of the dipole is

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

Answer 1

Answer: (C)

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