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 ${60}^0$ with the plane of the coil. The magnitude of the torque that must be applied on it to prevent it from turning is going to be

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

Answer 1

Answer: (C)

$3.93$ N m

Given that,

Number of turns, $N=100$

Radius of coil, $r=10cm=0.1m$

Current in the coil, $i=5A$

Magnetic moment of the coil, $M=Ni\left(\pi r^2\right)$

Magnetic field, $B=0.5T$

$\theta ={90}^o-{60}^o={30}^o$

Torque that must be applied to the coil, $\tau =Ni\left(\pi r^2\right)\times B\times sin\theta =100\times 5\times (3.14\times {0.1}^2)\times 0.5\times sin{30}^o=3.93Nm$

So, correct option is $\left(C\right)$