A current of `4A` enters at the corner 'a' of a square frame of side `10cm` and leaves at opposite corner 'c'. A magnetic field of `B=0*20T` acts in a direction perpendicular to the plane of paper directed outwards. Find the magnitude and direction of the magnetic forces on the four arms of the frame.

Answer

`0*04N` on each arm; force on arms ad and bc is towards right and on arms ab and dc is down wards in plane of frame

Explanation

Here, `I=4A`. This current divides equally in arms abc and adc, i.e., the current through each arm of square is, `I^'=2A`. Length of square `l=0*10m`, `B=0*20T`, `theta=90^@`.
Magnitude of the force on each arm of square frame is
`F=I^'lBsintheta=2xx(0*10)xx(0*20)xxsin90^@`
`=0*04N`
According to Fleming's left hand rule, the forces on arms ad and bc will be towards right and on arms ab and dc will be downwards in the plane of square frame.