A circular loop of radius a, carrying a current I, is placed in a tow dimensional magnetic field. The centre of the loop coincides with the centre of the filed The strenght of the magnetic field at the pariphery of the loop is B. find the magnetic force on the wire.

Explanation

Let us select a current element on circular loop `Ivecdl`. The magnetic field in radial direction will be perpendicular to current element vector. From left hand rule or right palm rule. we can find the direction of force `dvecF` on the element which comes to be perpendicular into the plane of the paper.
Required force `F=int dF=int idlB sin 90^@=int IdlB`
`=2piaiB` perpendicular to the plane.
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