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Question: Force between two identical short bar magnets whose centres are \(r\) metre apart is \(4.8\,N\) when...

Force between two identical short bar magnets whose centres are rr metre apart is 4.8N4.8\,N when their axes are in the same line. If the separation is increased to 2r2r metre, the force between then is reduced to?

Explanation

Solution

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.

Complete step by step answer:
It's a magnetic equivalent of the electric dipole, but it's not a perfect match. A true magnetic monopole, which is the magnetic equivalent of an electric charge, has never been found in nature. When you get farther away from a magnetic source, the magnetic field around it begins to resemble that of a magnetic dipole.

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,θI,\theta and φ\varphi . Force between two magnetic dipoles is:
F=μo×6M1M24π×r4F = \dfrac{{{\mu _o} \times 6{M_1}{M_2}}}{{4\pi \times {r^4}}}
Where, M1{M_1} and M2{M_2} are magnetic moments.
F2F1=(r1r2)4 F2F1=(12)4 F2F1=116\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}
F2=F116 F2=4.816 F2=0.3N\Rightarrow {F_2} = \dfrac{{{F_1}}}{{16}} \\\ \Rightarrow {F_2} = \dfrac{{4.8}}{{16}} \\\ \therefore {F_2} = 0.3\,N

Hence, the force between then is reduced to 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 mm is associated with BB, for example, the energy is mB - mB and the force is in the direction of increasing BB.