Question

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

• A. $\frac{\mu_{0}}{4 \pi} \cdot \frac{M}{d^{3}}$
• B. $\frac{\mu_{0}}{4 \pi} \cdot \frac{2 M}{d^{3}}$
• C. $\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \sqrt{2} M}{d^{3}}$
• D. Zero

Answer 1

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 $M$.
$\frac{2 \mu_{0}}{4 \pi}\left(\frac{M}{d^{3}}\right)$

$B_{\text{net} }=\frac{2 \mu_{0} M}{4 \pi d^{3}}$