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A small circular flexible loop of wire of radius r carries a current I. It is placed in a uniform magnetic field B. The tension in the loop will be doubled if ______ . |
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Answer» I is doubled `F=mBsintheta` but MAGNETIC moment `m=Ipir^(2)` `thereforeF=Ipir^(2)Bsintheta""...(1)` Suppose RADIUS `r_(1)`, magnetic FIELD `B_(1)` and current `I_(1)` then force acting is `F_(1)` and when radius `r_(2)` magnetic field `B_(1)` and current `I_(2)` then the force acting is `F_(2)`. `therefore` From EQUATION (1), If `F_(2)=2F_(1)`, then `F_(1)/F_(2)=(r_(1)^(2)I_(1)B_(1))/(r_(2)^(2)I_(2)B_(2))` becomes `2(r_(1)^(2)I_(1)B_(1))=r_(2)^(2)I_(2)B_(2)""...(2)` Now only if `r_(2)=2r_(1)`, then equation (2) is not solved, so option ( C) is wrong and if `B_(2)=B_(1)/2` is done then also equation (2) is not solved so option (B) is wrong too. If `B_(1)=2B_(1)andI_(2)=2I_(1)` then also equation (2) is not solved so option (D) is also wrong. But if r and B are kept constant and only `I_(2)=2I_(1)` then F can be doubled. so option (A) is TRUE. |
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