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In the question number 73, the given coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 5T in the horizontal direction exists such that initially the axis of the coil is in direction of the field. The coil rotates through an angle of `60^(@)` under the influence of magnetic field. The magnitude of torque on the coil in the final position isA. 25 N mB. `25 sqrt3 N m`C. 40 N mD. `40 sqrt3 N m`

Answer» Correct Answer - B
Torque `|vec(tau)|=|vecm xx vecB|=m B sin theta`
Here, m =`"10 A m"^(2), B = 5 T`
Now initially `theta=0^(@)`
Thus, initial torquie, `tau_(i)=0`
In final position `theta=60^(@)`
`therefore" "tau_(f)=m B sin 60^(@)=10xx5xx(sqrt3)/(2)=25sqrt3"N m"`


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