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A long current carrying wire , carrying current I_(1) such that I_(1) is flowing out from the plane of paper is placed at O. A steady state current I_(2) is flowingin the loop ABCD |
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Answer» the net FORCE is zero Force on `AB` . ` dF xxBxx sin 0@)= 0` Force on CD `: Similarly the magnetic field due to current `I_(1)` is along ` DC` . Because ` theta = 180^(@)` here , therefore force on ` DC` is zero . Force on `BC ` : Consider a small element `dl`. ` dF = I_(2) dl B_(1) sin 90^(@) rArr dF = I_(2) dl B_(1) ` By Fleming's left hand rule , the direction of this force is perpendicular to the plane of the paper directed outwards . Force on ` AD` : ` dF = I_(2) dl B_(1) sin 90^(@) rArr dF = I_(2) dl B_(1) ` By Fleming's left hand rule , the direction of this force is perpendicular to theplane of paper directed inwards. Since the current elements are located symmetrical to current ` I_(1) , therfore force on `BC` will cancel out the effect of force on ` AD`. rArr Net force on loop ` ABCD` is zero. ` NEt Torque on the loop : Theforce on ` BC and AD` will create a torque on ` ABCD` in clockwise direction about ` OO` as seen by the observer at `O`.   
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