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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.
Solution
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 (F) exerted by two magnetic poles on each other is given by
`F = u_0/(4pi) = (m_1m_2)/r^2`
= `(4pi xx 10^-7 xx 10^2)/(4pi xx 4 xx 10^-4)`
= `2.5 xx 10^-2 "N"`
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