question_answer

Two short magnets placed along the same axis with their like poles facing each other repel each other with a force which varies inversely as

A)            Square of the distance

B)            Cube of the distance

C)            Distance

D)            Fourth power of the distance

Correct Answer: D

Solution :

                     Both the magnets are placed in the field of one another, hence potential energy of dipole (2) is                    \[{{U}_{2}}=-{{M}_{2}}{{B}_{1}}\cos 0=-{{M}_{2}}{{B}_{1}}={{M}_{2}}\times \frac{{{\mu }_{0}}}{4\pi }.\frac{2{{M}_{1}}}{{{r}^{3}}}\]                    By using \[F=-\frac{dU}{dr}\], Force on magnet (2) is                    \[{{F}_{2}}=-\frac{d{{U}_{2}}}{dr}=-\frac{d}{dr}\left( \frac{{{\mu }_{0}}}{4\pi }.\frac{2{{M}_{1}}{{M}_{2}}}{{{r}^{3}}} \right)=-\frac{{{\mu }_{0}}}{4\pi }.6\frac{{{M}_{1}}{{M}_{2}}}{{{r}^{4}}}\]                    It can be proved \[\left| {{F}_{1}} \right|=\left| {{F}_{2}} \right|=F=\frac{{{\mu }_{0}}}{4\pi }.\frac{6{{M}_{1}}{{M}_{2}}}{{{r}^{4}}}\]            \[\Rightarrow F\propto \frac{1}{{{r}^{4}}}\]