Answer:
Torque, [tex]\tau=0.1669\ N-m[/tex]
Explanation:
It is given that,
Radius of the circular loop, r = 0.7 cm = 0.007 m
Number of turns, N = 520
Current in the loop, I = 3.9 A
The axis of the loop makes an angle of 57 degrees with a magnetic field.
Magnetic field, B = 0.982 T
We need to find the magnitude of torque on the loop. It is given by :
[tex]\tau=\mu\times B[/tex]
[tex]\tau=NIABsin(90-57)[/tex]
[tex]\tau=520\times 3.9\ A\times \pi (0.007\ m)^2\times 0.982\ T\ cos(57)[/tex]
[tex]\tau=0.1669\ N-m[/tex]
[tex]\tau=0.167\ N-m[/tex]
So, the magnitude of torque is 0.1669 N-m. Hence, this is the required solution.
