Question

Two identical short bar magnets of magnetic moment $M$ and $3M$ are placed at ${120}^{\circ }$ as shown in the figure. Find the magnetic induction at point $P$ on the angle bisector is

• A. $\frac{\sqrt{7}{\mu }_{0}M}{2\pi d^3}$
• B. $\frac{\sqrt{7}{\mu }_{0}M}{4\pi d^3}$
• C. $\frac{{\mu }_{0}M}{2\pi d^3}$
• D. $\frac{{\mu }_{0}M}{4\pi d^3}$

Answer 1

The correct option is A $\frac{\sqrt{7}{\mu }_{0}M}{2\pi d^3}$

Since both $M$ and $3M$ are making ${120}^{\circ }$, therefore the resultant magnetic moment is,

$M_{res}=\sqrt{M_1^2+M_2^2+2M_1{M}_2\cos \theta }$

$=\sqrt{(M{)}^2+(3M{)}^2+2M\left(3M\right)\cos {120}^{\circ }}$

$=\sqrt{M^2+9M^2+6M^2(-\frac{1}{2})}$

$=\sqrt{7M^2}=\sqrt{7}M.$

Now point $P$ is on the axial line of the resultant magnetic moment.

By symmetry, it is clear that the resultant magnetic field is along the perpendicular bisector.

$\therefore B=\frac{{\mu }_0}{4\pi }\left(\frac{2M}{d^3}\right)$

$B=\frac{{\mu }_0}{4\pi }\left(\frac{2\sqrt{7}M}{d^3}\right)$

$B=\frac{\sqrt{7}{\mu }_{0}M}{2\pi d^3}$

Hence, option $\left(A\right)$ is the correct answer.