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. $\frac{{\mu }_0}{4\pi }\cdot \frac{2M}{d^3}$
• 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 $M$.
$\frac{2{\mu }_0}{4\pi }\left(\frac{M}{d^3}\right)$

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