Question

The figure shows two short bar magnets of equal length $l$ having magnetic moments ${\overrightarrow{M}}_1$ and ${\overrightarrow{M}}_2$ respectively, with their centres of separation being $r$. If $l<<r$, what would be the magnitude of force on the magnet $2$ by $1$ ?

• A. $\frac{6{\mu }_0}{4\pi }\frac{M_1{M}_2}{r^3}$
• B. $\frac{3{\mu }_0}{4\pi }\frac{M_1{M}_2}{r^3}$
• C. $\frac{{\mu }_0}{4\pi }\frac{M_1{M}_2}{r^4}$
• D. $\frac{6{\mu }_0}{4\pi }\frac{M_1{M}_2}{r^4}$

Answer 1

The correct option is D $\frac{6{\mu }_0}{4\pi }\frac{M_1{M}_2}{r^4}$


Magnetic field $\overrightarrow{B_1}$ at the centre of magnet $2$ due to $\overrightarrow{M_1}$ be,

$\overrightarrow{B_1}=\frac{{\mu }_0}{4\pi }\frac{2\overrightarrow{M_1}}{r^3}(\because l<<r)$

Potential energy of the magnet $2$,

$U_2=-\overrightarrow{M_2}.\overrightarrow{B_1}$

$U_2=-M_2\times \frac{{\mu }_0}{4\pi }\times \frac{2M_1}{r^3}(\text{As}{\overrightarrow{B}}_1\parallel {\overrightarrow{M}}_2)$

Using the relation between force and potential energy gradient,

$F_2=-\frac{dU}{dr}=-\frac{d}{dr}(-\frac{2{\mu }_0}{4\pi }\frac{M_1{M}_2}{r^3})$

$F_2=-\frac{6{\mu }_0}{4\pi }\frac{M_1{M}_2}{r^4}$

$\therefore |F_2|=\frac{6{\mu }_0}{4\pi }\frac{M_1{M}_2}{r^4}$

And negative sign is showing attractive nature of net force.

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