MCQ
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
- ASquare of the distance
- BCube of the distance
- CDistance
- ✓Fourth power of the distance
✓
Answer
Correct option: D.
Fourth power of the distance
d
(d) 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}}}$
(d) 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}}}$
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