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Two long bar magnets are placed with their axes coinciding in such away that the north pole of the first magnet is 1.0 cm from the south pole of the second. If both the magnets have a pole strength of 40 Am. The force exerted by one magnet on the other is__N

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answer is 1.6.

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Detailed Solution

As the magnets are described as long, only interaction between the nearest poles is consideredGiven : m1=m2=40Am; r=1 cm=1×10−2mForce exerted by two magnetic poles on each other=μ04π.m1m2r2=10−7×40×401×10−22=10−7×16×1021×10−4=16 ×102×10−31=16×10−1=1.6N

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