Question

A coil of 50 turns is situated in a magnetic field $B=0.25weber/m^2$ as shown in figure. A current of 2A is flowing in the coil. Torque acting on the coil will be

• A. 0.15 N ​
• B. 0.3 N
• C. 0.45 N
• D. 0.6 N ​

Answer 1

The correct option is B 0.3 N

Since plane of the coil is parallel to magnetic field. So $\theta ={90}^{\circ }$
Hence $\tau =NBiAsin{90}^{\circ }=NBiA=50\times 0.25\times 2\times (12\times {10}^{-2}\times 10\times {10}^{-2})=0.3N$.