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# Force between two identical short bar magnets whose centres are $r$ metre apart is $4.8\,N$ when their axes are in the same line. If the separation is increased to $2r$ metre, the force between then is reduced to?

Last updated date: 21st Jul 2024
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Hint:Let us first understand about the magnetic dipole. A magnetic dipole is the limit of either a closed loop of electric current or a pair of poles when the source's size is reduced to zero while the magnetic moment remains constant.

The difference between two magnetic dipoles, the angle between their centrelines and the Z-axis, and the angle between their centrelines and the X-axis are all expressed by the letters $I,\theta$ and $\varphi$. Force between two magnetic dipoles is:
$F = \dfrac{{{\mu _o} \times 6{M_1}{M_2}}}{{4\pi \times {r^4}}}$
Where, ${M_1}$ and ${M_2}$ are magnetic moments.
$\dfrac{{{F_2}}}{{{F_1}}} = {(\dfrac{{{r_1}}}{{{r_2}}})^4} \\ \Rightarrow \dfrac{{{F_2}}}{{{F_1}}} = {(\dfrac{1}{2})^4} \\ \Rightarrow \dfrac{{{F_2}}}{{{F_1}}} = \dfrac{1}{16}$
$\Rightarrow {F_2} = \dfrac{{{F_1}}}{{16}} \\ \Rightarrow {F_2} = \dfrac{{4.8}}{{16}} \\ \therefore {F_2} = 0.3\,N$
Note:The interaction of one dipole with the magnetic field formed by the other dipole can be interpreted as the repulsion or attraction between two magnetic dipoles. When the magnetic dipole $m$ is associated with $B$, for example, the energy is $- mB$ and the force is in the direction of increasing $B$.