Question

A circular coil of 70 turns and radius 5 cm carrying a current of 8 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.5 T. The field lines make an angle of 30 with the normal of the coil then the magnitude of the counter torque that must be applied to prevent the coil from turning is :

• A. 33 N m
• B. 3 N m
• C. 3.3 x ${10}^2$N m
• D. 3.3 x ${10}^{-4}$N m

Answer 1

The correct option is C 3.3 x ${10}^2$N m

Number of turns on the circular coil, $n=70$

Radius of the coil, $r=5.0cm=0.05m$

Area of the coil , $A=\pi r^2=3.14\times (0.05{)}^2=$

Current flowing in the coil, $I=8.0A$

Magnetic field strength, $B=1.5T$

Angle between the field lines and normal with the coil surface,

$\theta ={30}^o$

The coil experiences a torque in the magnetic field. Hence, it turns. The counter torque applied to prevent the coil from turning is given by the relation,

$\tau =nIBAsin\theta$ … (i)

$\tau =70\times 8\times 1.5\times 0.00785\times sin{30}^o$

$\tau =3.3Nm$