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 them is reduced to?

Given,

Two bar magnets placed axially at distance r

Force between them = 4.8N

Solution:

Force between two magnetic dipoles is = F = μ<sub>o</sub>6M1M2/4πd<sup>4</sup>

F is inversely proportional r<sup>4</sup>

Therefore, new force = F’ = F/2<sup>4</sup> (because new d = 2r)

F’ = 4.8/16 = 0.3 N

Therefore, 3N force is reduced between the two identical short bar magnets.

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