Question

A circular coil of $100$ turns, radius $10$cm carries a current of $5$A. It is suspended vertically in a uniform horizontal magnetic field of $0.5$T and the field lines make an angle of ${60}^o$ with the plane of the coil. The magnitude of the torque that make be applied on it to prevent it from turning is?

• A. $2.93$N m
• B. $3.43$N m
• C. $3.93$N m
• D. $4.93$N m

Answer 1

The correct option is C $3.93$N m

Here, $N=100$, $r=10$cm, $=0.10$m, $I=5$A,
$B=0.5$T, $\theta ={90}^o-{60}^o={30}^o$
Area of the coil, $A=\pi r^2=3.14\times (0.1{)}^2$
$\tau =NIBA\sin \theta =100\times 5\times 0.5\times 3.14\times (0.1{)}^2\times \sin {30}^o$
$=3.931$Nm