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(B) The magnetic moment of each magnet is $M$. Since the two magnets are identical and placed at an angle of $120^{\circ}$,the resultant magnetic mome

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The magnetic moment of each magnet is $M$. Since the two magnets are identical and placed at an angle of $120^{\circ}$,the resultant magnetic moment $M_{net}$ is given by the vector sum of the two individual magnetic moments.$M_{net} = \sqrt{M^2 + M^2 + 2MM \cos(120^{\circ})} = \sqrt{2M^2 + 2M^2(-0.5)} = \sqrt{M^2} = M$.The direction of this resultant magnetic moment $M_{net}$ lies along the angle bisector of the $120^{\circ}$ angle.The point $P$ is located at a distance $d$ from the origin along this angle bisector. Therefore,the point $P$ lies on the axial line of the resultant magnetic moment $M_{net}$.The magnetic field on the axial line of a short bar magnet is given by $B = \frac{\mu_0}{4 \pi} \cdot \frac{2M}{d^3}$.Thus,the magnetic field at point $P$ is $\frac{\mu_0}{4 \pi} \cdot \frac{2M}{d^3}$.

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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