Acircular coil 17.7 cm in diameter and containing 18 loops lies flat on the gound. the earth's magnetic field at this location has magnitude 5.5e-5 t and points into the earth at an angle of 66.0 degrees below the horizontal line and aimed due north. determine the torque if a 7.4 a clockwise current passes through the coil.

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

the answer is a)

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