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A parallelogram shaped coil ABCD of side...

A parallelogram shaped coil ABCD of sides `0*3m` and `0*2m` carries a current of `2*0A` as shown in figure. It is placed in a uniform magnetic field `vecB=50T` parallel to AD. Find (i) forces on the sides of the coil and (ii) torque on the coil.

Text Solution

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Here, `l=0*3m`, `b=0*2m`,
`I=2*0A`, `B=50T`
(i) Since magnetic field `vecB` is parallel to sides AD and BC, hence no force acts on these sides.
Force on the side AB is
`F=IlBsintheta=2*0xx0*3xx50xxsin60^@`
`=2xx0*3xx50xxsqrt3//2=25*98N~~26N`
According to Flemings Left Hand rule the direction of the force is normally upwards. Similarly force on the side DC will also be `~~26N`.
Its direction is normally downwards.
(ii) Since force on sides AB and DC are equal, parallel and opposite, they form a couple which exerts a torque. Therefore, torque on coil is
`tau=FxxAD sin 60^@=26xx0*2xxsqrt3//2`
`=4*5Nm`
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