Question

A circular coil 18.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this location has magnitude 5.50×10−5T and points into the Earth at an angle of 54.0 ∘ below a line pointing due north. A 6.30-A clockwise current passes through the coil. Part A Determine the torque on the coil. Express your answer using three significant figures and include the appropriate units.

Answer 1

Torque is
`6.41* 10^(- 5) N-m`

Solution:

As per the question;

Diameter of the circular, d = 18.0 cm = 0.18 m

Magnetic Field of the earth, B =
`5.50* 10^(- 5)T`

Angle,
`\theta = 54.0^(\circ)`

Current, I = 6.30 A, clockwise

(a) Now,

Torque on the coil is given by:

`\tau = AN(\vec{l}* \vec{B}) = ANBlsin\theta`

where

A = cross-sectional area

`A = \pi((d)/(2))^(2) = \pi((0.18)/(2))^(2) = 0.0254 m^(2)`

Now,

`\tau = 9* 6.30* (5.50* 10^(- 5))* 0.0254* sin54^(\circ) = 6.41* 10^(- 5) N-m`