Answer:
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 [tex]\theta[/tex] = 30 degree
Now
We know that
the magnitude of the magnetic torque is
= nIABsin[tex]\theta[/tex]
= (400) (7.0) π ÷ 4 (0.06m)^2 sin(90° - 30°)
As
[tex]\theta[/tex] = (90° - Ф)
= (400) (7.0) π ÷ 4 (0.06m)^2 sin 60°
= 6.8 N.m
