Question

The magnitude of torque experienced by a square coil of side 12 cm which consists of 25 turns and carries a current 10 A suspended vertically and the normal to the plane of coil makes an angle of 30 with the direction of a uniform horizontal magnetic field of magnitude 0.9 T is:

• A. 1.6 Nm
• B. 1.2 Nm
• C. 1.4 Nm
• D. 1.8 Nm

Answer 1

The correct option is A 1.6 Nm

Here,

$N=25,I=10A,B=0.9T,\theta ={30}^0$

$A=a^2=12\times {10}^{-2}\times 12\times {10}^{-2}=144\times {10}^{-4}{m}^2$

$\tau$= $NIA\sin \theta$

$\therefore \tau =25\times 10\times 144\times {10}^{-4}\times 0.9\times \sin {30}^0=1.6Nm$