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 ______ .

Options

A. I is doubled
B. B is halved
C. r is doubled
D. Both B and I are doubled

Answer

A

Explanation

The force acting on the ring.
`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.