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

Explanation:

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

where

is the magnetic moment of the loop

B is the strength of the magnetic field

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

where

I is the current in the loop

A is the area enclosed by the coil

Substituting into the original equation,

in this problem, we have:

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

is the circumference of the loop, so we can find its radius:

So the area of the coil is

Also we have

B = 0.735 T (strength of the magnetic field)

(because the field is parallel to the plane of the coil)

So, the magnitude of the torque is:

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form