Two short bar magnets of magnetic moments 'M' | vedclass

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Question

(A) Let the square have vertices at $(0,0), (d,0), (d,d),$ and $(0,d)$. Let the magnets be placed at $(0,0)$ and $(d,d)$.For the magnet at $(0,0)$,the

Vedclass · Accepted Answer

$\frac{\mu_0}{4 \pi} \frac{M}{d^3}$

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(A) Let the square have vertices at $(0,0), (d,0), (d,d),$ and $(0,d)$. Let the magnets be placed at $(0,0)$ and $(d,d)$.For the magnet at $(0,0)$,the corner $(d,0)$ is on its axial line. The magnetic field $B_1$ at $(d,0)$ is $B_1 = \frac{\mu_0}{4 \pi} \frac{2M}{d^3}$ (directed along the axis).For the magnet at $(d,d)$,the corner $(d,0)$ is on its equatorial line. The magnetic field $B_2$ at $(d,0)$ is $B_2 = \frac{\mu_0}{4 \pi} \frac{M}{d^3}$ (directed opposite to the axis).Since the like poles are in the same direction,the fields $B_1$ and $B_2$ are in opposite directions.The net magnetic induction is $B_{	ext{net}} = B_1 - B_2 = \frac{\mu_0}{4 \pi} \frac{2M}{d^3} - \frac{\mu_0}{4 \pi} \frac{M}{d^3} = \frac{\mu_0}{4 \pi} \frac{M}{d^3}$.

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