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Q. The force between two short bar magnets with magnetic moments $M_{1}$ and $M_{2}$ whose centres are $r$ metre apart is $0.8\, N$ when their axes are in the same line. If the separation is increased to $2\, r$, then force between them is reduced to

Magnetism and Matter

Solution:

$F=\frac{\mu_{0}}{4 \pi} \cdot \frac{6 M_{1} M_{2}}{r^{4}}$
i.e., $\propto \frac{1}{r^{4}} $
$\therefore F'=\frac{F}{16}=\frac{8}{16}=0.5 \,N$