Question

A single-turn current loop, carrying a current of 4.75 A, is in the shape of a right triangle with sides 50.0, 120, and 130 cm. The loop is in a uniform magnetic field of magnitude 71.5 mT whose direction is parallel to the current in the 130 cm side of the loop. What are the magnitude of the magnetic forces on each of the three sides?
(a) the 130 cm side

(b) the 50.0 cm side

(c) the 120 cm side

(d) What is the magnitude of the net force on the loop?

Answer 1

Given,

I = 4.75 A ; ; B = 71.5 mT ;

for this to be a right angled triangle,

P = 50 ; B = 120 ; H = 130

H = 130 = sqrt (50^2 + 120^2)

so using trig, we can find the angle theta

theta = tan^-1(50/120) = 22.62 deg

Bx = 71.5 mT x cos22.62 = 66 mT

By = 71.5 mT x sin22.62 = 27.5 mT

a)For the side 130

F = 0 N

b)for 50 cm

F = i L Bx

F = 4.75 x 0.5 x 66 x 10^-3 = 0.157 N

Hence, F(50) 0.157 N

c)F(120) = i L By

F(120) = 4.75 x 1.2 x 27.5 x 10^-3 = 0.157 N

Hence, F(120) = 0.157 N

d)Fnet = 0 N