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.

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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

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Given that two bar magnet placed axially at distance r and force between them is 4.8N.

F = μ0*6M1M2/4π*d^4

Force is inversely proportional to r^4

So the new force will be

F'=F/2^4

Because new distance d will be 2r

F'=4.8/16

=0.3N

Hence the correct answer is option A which is 0.3 N.

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