Question

Two long bare magnets are placed with their axes coinciding in such a way that the north pole of the first magnet is 2.0 cm from the south pole of the second. If both the magnets have a pole strength of 10 Am, find the force exerted by one magnet of the other.

Answer 1

Given:
Pole strength = m₁ = m₂ = 10 Am
Distance between the north pole of the first magnet and the south pole of the second magnet, r = 2 cm = 0.02 m
We know,
Force $\left(F\right)$ exerted by two magnetic poles on each other is given by
$$\begin{gathered} F=\frac{{\mathrm{\mu }}_0}{4\mathrm{\pi }}\frac{m_1{m}_2}{r^2} \\ =\frac{4\mathrm{\pi }\times {10}^{-7}\times {10}^2}{4\mathrm{\pi }\times 4\times {10}^{-4}} \\ =2.5\times {10}^{-2}\mathrm{N} \end{gathered}$$