A flat circular coil of wire having 400 turns and diameter 6.0 cm carries a current of 7.0 A. It is placed in a magnetic field of with the plane of the coil making an angle of 30° with the magnetic field. What is the magnitude of the magnetic torque on the coil?​

6.8 N.m

Explanation:

The computation of the magnitude of the magnetic torque on the coil is given below:

Given that

n = 400

d =  6.0 cm

Current  is I = 7.0 A

Angle is = 30 degree

Now

We know that

the magnitude of the magnetic torque is

= nIABsin

= (400) (7.0) π ÷ 4 (0.06m)^2 sin(90° - 30°)

As

= (90° - Ф)

=  (400) (7.0) π ÷ 4 (0.06m)^2 sin 60°

= 6.8 N.m

i think they are all correct

explanation:

they seem good

1/0.14nm = 10^10/1.4 a lot