Answer:
Explanation:
Diameter of the circular coil d = 17.7 cm = 0.177m
Current in the coil, I = 7.4 A
Number of loops in the coil, N = 18
Magnetic field, B = 5.5 x 10-5 T
Angle below line, (θ) = 66°
Write the expression for the torque in the coil
Torque = N x I x B x A x sin (θ)
Where A is the area of the coil and θ° is the angle between the magnetic field and the coil face.
The cross- sectional area of the coil = (pie)(d/2)²
A = (π)(0.177/2)² = 0.0246 m²
The Earth’s magnetic field points into the earth at an angle 66 below the line. The line points towards the north
Hence to find the angle between the magnetic field and the coil face we need to subtract the given angle by 90°
Theta = 90 – 66 = 24°
Substitute the value to find out torque
Torque = 18 x 7.4 x 5.5 x 10⁻⁵ x 0.0246 x sin(24) = 7.33 x 10⁻⁵ N-m
The torque on the coil is 7.33 x 10⁻⁵ N-m
