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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Infinity Learn · Accepted Answer

Force experienced by the square loop $$=FAB$$−$$FCD=i$$.a.μ$$0i2$$$$π$$a−i.a.μ$$0i2$$π$$.2a=$$μ$$0i24$$π$$=10$$−$$7$$×$$(10)2N=10$$−$$5NBy$$ Newton's third law of motion, force exerted by the square loop on $$PQ=10$$−$$5N$$

Force on a current carrying wire placed in a magnetic field

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Force experienced by the square loop
$= F_{AB} - F_{CD} = i . a . \frac{μ_{0} i}{2 πa} - i . a . \frac{μ_{0} i}{2 π . 2 a} = \frac{μ_{0} i^{2}}{4 π} = 10^{- 7} \times {(10)}^{2} N$
$= 10^{- 5} N$
By Newton's third law of motion, force exerted by the square loop on $PQ = 10^{- 5} N$

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Force on a current carrying wire placed in a magnetic field

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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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Force experienced by the square loop
$= F_{AB} - F_{CD} = i . a . \frac{μ_{0} i}{2 πa} - i . a . \frac{μ_{0} i}{2 π . 2 a} = \frac{μ_{0} i^{2}}{4 π} = 10^{- 7} \times {(10)}^{2} N$
$= 10^{- 5} N$
By Newton's third law of motion, force exerted by the square loop on $PQ = 10^{- 5} N$