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

Answer:

Torque is $6.41\times 10^{- 5} N-m$

Solution:

As per the question;

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

Magnetic Field of the earth, B = $5.50\times 10^{- 5}T$

Angle, $\theta = 54.0^{\circ}$

Current, I = 6.30 A, clockwise

(a) Now,

Torque on the coil is given by:

$\tau = AN(\vec{l}\times \vec{B}) = ANBlsin\theta$

where

A = cross-sectional area

$A = \pi(\frac{d}{2})^{2} = \pi(\frac{0.18}{2})^{2} = 0.0254 m^{2}$

Now,

$\tau = 9\times 6.30\times (5.50\times 10^{- 5})\times 0.0254\times sin54^{\circ} = 6.41\times 10^{- 5} N-m$