Two short magnets of equal dipole moments M a | vedclass

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(C) The two magnets are joined as shown in the figure. The magnetic field at point $P$ due to the magnet along the $Y$-axis $(M_1)$ is on its axial li

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$\frac{M_0}{\sqrt{5}}$

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(C) The two magnets are joined as shown in the figure. The magnetic field at point $P$ due to the magnet along the $Y$-axis $(M_1)$ is on its axial line, while the magnetic field due to the magnet along the $X$-axis $(M_2)$ is on its equatorial line.For a short magnet of dipole moment $M$, the axial magnetic field at distance $R$ is $B_{	ext{axial}} = \frac{\mu_0}{4 \pi} \frac{2M}{R^3}$ and the equatorial magnetic field at distance $R$ is $B_{	ext{equatorial}} = \frac{\mu_0}{4 \pi} \frac{M}{R^3}$.Since these fields are perpendicular to each other at point $P$, the net magnetic field is:$B_{	ext{net}} = \sqrt{B_{	ext{axial}}^2 + B_{	ext{equatorial}}^2} = \sqrt{\left(\frac{\mu_0}{4 \pi} \frac{2M}{R^3}ight)^2 + \left(\frac{\mu_0}{4 \pi} \frac{M}{R^3}ight)^2}$$B_{	ext{net}} = \frac{\mu_0}{4 \pi} \frac{M}{R^3} \sqrt{2^2 + 1^2} = \frac{\mu_0}{4 \pi} \frac{\sqrt{5}M}{R^3}$Given that $B_{	ext{net}} = \frac{\mu_0}{4 \pi} \frac{M_0}{R^3}$, we equate the two expressions:$\frac{\mu_0}{4 \pi} \frac{\sqrt{5}M}{R^3} = \frac{\mu_0}{4 \pi} \frac{M_0}{R^3}$$\sqrt{5}M = M_0 \implies M = \frac{M_0}{\sqrt{5}}$

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