A circular coil 11.0 cm in diameter and containing nine loops lies flat on the ground. The...

Question:

A circular coil 11.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this location has magnitude {eq}\rm 4.50 \times 10^{-5} \ T {/eq} and points into the Earth at an angle of 51.0 degrees below a line pointing due north. A 7.20 A clockwise current passes through the coil.

(a) Determine the torque on the coil.

(b) Which edge of the coil rises up: north, east, south, or west?

Magnetic Torque:

When a current loop is immersed in a magnetic field, a torque can be induced such that there will be a net rotation of the loop such that the loop's normal vector will align to the magnetic field. This is a direct result of the magnetic force on the loop, which depends on the field strength and the current flowing through the loop, given as:

{eq}\displaystyle \vec{F} = I\vec{L}\ \times\ \vec{B} {/eq}

The resulting torque due to the magnetic field will then be:

{eq}\displaystyle \tau = N\mu B sin\theta {/eq}

where:

• {eq}\displaystyle \mu = IA {/eq} is the magnetic moment, which is essentially the current flowing multiplied by the cross-sectional area of the loop.
• N is the number of turns of the loop
• B is the magnetic field strength
• {eq}\displaystyle \theta {/eq} is the angle between the normal vector of the loop and the field direction.

Given:

• {eq}N= 9 {/eq} is the number of turns
• {eq}I=7.2 \ A {/eq} is the current on the loop
• {eq}d=11 \ cm = 0.11 \ m {/eq} is the diameter of the loop
• {eq}B=4.5\ \times\ 10^{-5}\ T {/eq} is the magnetic field strength
• {eq}\theta =51^\circ {/eq} is the direction of the magnetic field

a) We determine the magnetic torque on the loop through the equation:

{eq}\displaystyle \tau = N\mu B sin\theta {/eq}

or:

{eq}\displaystyle \tau = NIA B sin\theta {/eq}

The area of the loop depends primarily on its radius, which is half of the diameter. The area can be determined as:

{eq}\displaystyle A = \pi r^2 = \pi (\frac{d}{2})^2 {/eq}

And we substitute all given values,

{eq}\displaystyle \tau = (9)(7.2\ A)(\pi (0.055\ m)^2)(4.5\ \times\ 10^{-5}\ T) sin(51^\circ) {/eq}

we obtain:

{eq}\displaystyle \boxed{\tau = 2.15\ \times\ 10^{-5}\ TAm^2} {/eq}

(b) Since the torque will attempt to align itself with the magnetic field and it is flat on the ground, the north edge will rise up in an attempt to align itself with the magnetic field.