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

Options

A. 33 Nm
B. 3.3 Nm
C. `3.3 xx 10^(-2)Nm`
D. `3.3 xx 10^(-4)Nm`

Answer

B

Explanation

`N=70, r=5cm = 5xx10^(-2)m, I=8A`
`B=1.5T, theta=30^(@)`
The counter torque to prevent the coil from turning will be equal and opposite to the torque acting on the coil,
`therefore" "tau=NIAB sin theta = NI pi r^(2)B sin 30^(@)`
`=70xx8xx3.14xx(5xx10^(-2))^(2)xx1.5xx(1)/(2)="3.297 N m"`
`~~3.3" N m"`