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Two identical current carrying coaxial l...

Two identical current carrying coaxial loops, carry current I in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as C,

A

`oint_(c)vecB*vec(dl)=pm2mu_(0)I`

B

the value of `oint_(c)vecB*vec(dl)` is independent of sense of C.

C

there may be a point on C where `vecBandvec(dl)` are perpendicular.

D

`vecB` vanishes everywhere on C.

Text Solution

Verified by Experts

The correct Answer is:
B, C
Current in both loops are opposite to each other, so according to Ampere.s circuital law,
`ointvecB*vec(dl)=mu_(0)sumI`
= `mu_(0)(I-I)=0`
As the magnetic field inside the loop is perpendicular to the direction of plane of loop,
so, `ointBdl=0or|vecB*vec(dl)|cos90^(@)=0`
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