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A closely wound solenoid of 100 turns...

A closely wound solenoid of 100 turns and area of cross section `5xx10^(-4) m^(2)` carries current of 2A it is placed in such a way that its horizontal axis is at `30^(@)` with the direction of a uniform magnetic field of intensity 0.2 T what is the torque experienced by the solenoid in the magnetic field ?

A

0.1 Nm

B

0.2 Nm

C

0.01 Nm

D

0.02 Nm

Text Solution

Verified by Experts

The correct Answer is:
C
`vec(tau)=vec(M)xxvec(B) or tau=M sin theta`
But for the solenoid M=nIA =`100xx12xx5xx10^(-4)`
`therefore M =1000xx10^(-4)=10^(-1) Am^(2)`
and `tau =10^(-1)xx0.2xxsin 30^(@)`
`=1/10xx2/10xx1/2=1/100=0.01 Nm`
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