Question

A current of 20.0 mA is maintained in a single circular loop with a circumference of 2.15 m. A magnetic field of 0.735 T is directed parallel to the plane of the loop. What is the magnitude of the torque exerted by the magnetic field on the loop

Answer 1

`5.39\cdot 10^(-3) N\cdot m`

Step-by-step explanation:

The torque exerted on a current single loop of wires is given by:

`\tau = \mu B sin \theta`

where

`\mu` is the magnetic moment of the loop

B is the strength of the magnetic field

`\theta` is the angle between the direction of the field and the normal to the plane of the loop

The magnetic moment of a loop is given by

`\mu =IA`

where

I is the current in the loop

A is the area enclosed by the coil

Substituting into the original equation,

`\tau=IABsin \theta`

in this problem, we have:

I = 20.0 mA = 0.020 A is the current in the loop

`c=2.15 m` is the circumference of the loop, so we can find its radius:

`c=2\pi r\\r=(c)/(2\pi)=(2.15)/(2\pi)=0.342 m`

So the area of the coil is

`A=\pi r^2=\pi(0.342)^2=0.367 m^2`

Also we have

B = 0.735 T (strength of the magnetic field)

`\theta=90^(\circ)` (because the field is parallel to the plane of the coil)

So, the magnitude of the torque is:

`\tau=(0.020)(0.367)(0.735)(sin 90^(\circ))=5.39\cdot 10^(-3) N\cdot m`