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Q.

A square loop ABCD of side length 2m is placed mean a very long straight conductor PQ carrying a current of 10 A as shown in figure. The square loop is carrying a current of 10 A and the conductor PQ lies in the plane of the loop ABCD. Then the magnetic force exerted by the square loop on the straight conductor is

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Detailed Solution

Force experienced by the square loop =FAB−FCD=i.a.μ0i2πa−i.a.μ0i2π.2a=μ0i24π=10−7×(10)2N=10−5NBy Newton's third law of motion, force exerted by the square loop on PQ=10−5N

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