Question

A closely wound solenoid of $800$ turns and area of cross section $2.5\times {10}^{-4}{m}^2$ carries a current of $3A$, is free to turn about the vertical direction and a uniform magnetic field of $0.25T$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of ${30}^{\circ }$ with the direction of applied force?

Answer 1

Given number of turns of solenoid= 800,current I=3 A, magnetic field B=0.25 T,$\theta$=30

We know

$\tau =\text{BINAcos}\theta =0.25\times 3\times 800\times 2.5\times {10}^{-4}\times cos30=231.37\times {10}^{-4}=0.023\text{N m}$