Two magnetic dipoles are placed in a plane as shown. The magnitude of interaction energy of this system of dipoles is given by

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$\frac{3 μ_{0} m^{2}}{8 π a^{3}}$

$\frac{3 μ_{0} m^{2}}{2 π a^{3}}$

$\frac{μ_{0} m^{2}}{4 π a^{3}}$

$\frac{μ_{0} m^{2}}{2 π a^{3}}$

answer is B.

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Detailed Solution

Interaction energy $= {\vec{m}}_{2} \cdot {\vec{B}}_{12}$

$\begin{aligned} = \frac{μ_{0}}{4 π} {\vec{m}}_{2} \cdot \{\frac{3 {\vec{m}}_{1} \cdot {\vec{r}}_{12}}{r_{12}^{5}} {\vec{r}}_{12} - \frac{{\vec{m}}_{1}}{r_{12}^{3}}\} \\ = \frac{μ_{0}}{4 π} \frac{3 m^{2}}{a^{3}} \cos^{2} 45^{0} = \frac{3 μ_{0} m^{2}}{8 {πa}^{3}} \end{aligned}$

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