Two identical short bar magnets are placed at | vedclass

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Question

(b)

Since two equal vectors $M$ are inclined at $120^{\circ}$, their resultant will also be $M$ and along its angular bisector. So point $P$ is on

Vedclass · Accepted Answer

$\frac{\mu_0}{4 \pi} \cdot \frac{2 M}{d^3}$

Vedclass · Answer

(b)

Since two equal vectors $M$ are inclined at $120^{\circ}$, their resultant will also be $M$ and along its angular bisector. So point $P$ is on axial line of resultant moment $M$.

$B_{ net }=\frac{2 \mu_0 M}{4 \pi d^3}$

Two identical short bar magnets are placed at $120^{\circ}$ as shown in the figure. The magnetic moment of each magnet is $M$. Then the magnetic field at the point $P$ on the angle bisector is given by

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