A circular coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is in a region where the magnetic field is 0.56 T. (a) What orientation of the coil gives the maximum torque on the coil, and what is this maximum torque? (b) For what orientation of the coil is the magnitude of the torque 71% of the maximum found in part (a)?

(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰

(a)(ii) The maximum torque is 0.132 Nm

(b) The orientation of the coil is 45⁰

Explanation:

Given;

diameter of the circular wire, d = 8.6 cm = 0.086 m

radius of the wire, r = d /2 = 0.043 m

number of turns, N = 15 turns

magnetic field, B = 0.56 T

The torque on the wire is given by;

τ = NIABsinθ

where;

θ is the orientation of the wire

(a) maximum torque occurs when the orientation of the wire is at 90⁰

The maximum torque is given by;

τ = NIABsin(90⁰)

τ = NIAB

τ = (15)(2.7)(π x 0.043²)(0.56)

τ = 0.132 Nm

(b)

71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm

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