Question

A circular coil of diameter 10 cm has 10 turns and carries a current of 5A. It is placed in a uniform field of 0.5 T with its plane parallel to the field. The torque on the coil in Nm, is

• A. $\left(6.25\pi \right)\times {10}^{-2}$ N-m
  
• B. $\left(6.25\pi \right)\times {10}^{-3}$ N-m
• C. $\left(6.25\pi \right)\times {10}^{-4}$ N-m
• D. Zero

Answer 1

The correct option is A $\left(6.25\pi \right)\times {10}^{-2}$ N-m


$\theta =90;sin90=1$
T =Bi An
$=0.5\times 5\times \pi r^2\times 10=25\times \pi (5\times {10}^{-2}{)}^2=625\pi \times {10}^{-4}=6.25\pi \times {10}^{-2}Nm$.