- Home
- Standard 12
- Physics
5.Magnetism and Matter
medium
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
$0$
Solution
(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 $M$.
$B_{ net }=\frac{2 \mu_0 M}{4 \pi d^3}$
Standard 12
Physics
