If at a distance of 40 cm at an axial position of a dipole, the...

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

Options

A $28.6Am^{2}$
B $32.2Am^{2}$
C $38.4Am^{2}$
DNone of these

Correct answer

C. $38.4Am^{2}$

Step-by-step solution

Here, $r = 40cm = 0.4m$
$\theta = 0 °$ (an axial line)
$M = ?$

As $V = \dfrac{{\mu}_{0}}{4\pi}\dfrac{Mcos\theta}{r^{2}}$
$⇒ 2.4 \times {10}^{- 5} = {10}^{- 7} \times \dfrac{M \times 1}{\left(0.4\right)^{2}}$
$⇒ M = 38.4Am^{2}$

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