Consider a 150 turn square loop of wire 17.5 cm on a side that carries a 42 a current in a 1.7 t. a) what is the maximum torque on the loop of wire in n. m? b) what is the magnitude of the torque in n·m when the angle between the field and the normal to the plane of the loop θ is 14°?

(a) 328 Nm

(b) 79.35 Nm

Explanation:

N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T

A = side^2 = 0.175^2 = 0.030625 m^2

(a) Torque = N x i x A x B x Sinθ

For maximum torque, θ = 90 degree

Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90

Torque = 328 Nm

(b) θ = 14 degree

Torque =  150 x 42 x 0.030625 x 1.7 x Sin 14

Torque = 79.35 Nm

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6