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Magnetic scalar potential due to a magnetic dipole at a point on its axis situated it a distance of 20 cm from the centre of is found to be 1.2 × 10^–5 Tm. Find the magnetic dipole moment of the dipole is

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1.8 Am²

2.8 Am²

5.8 Am²

4.8 Am²

answer is D.

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Detailed Solution

$\begin{array}{l} V = \frac{μ_{0}}{4 π} \frac{Mcos θ}{r^{2}} \Rightarrow 10 \\ 1 .2 \times 10^{- 5} = \frac{10^{- 7} \times M \times 1}{(0 .2)^{2}}; M = 4 .8 {Am}^{2} \end{array}$

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