Question

What is the angle between a wire carrying an 8.40 A current and the 1.20 T field it is in, if 50.0 cm of the wire experiences a magnetic force of 2.55 N? ° (b) What is the force (in N) on the wire if it is rotated to make an angle of 90° with the field? N

Answer 1

A. 30.38°

B 5.04N

Step-by-step explanation:

Using

F= ILBsin theta

2 .55N= 8.4Ax 0.5mx 1.2T x sintheta

Theta = 30.38°

B. If theta is 90°

Then

F= 8.4Ax 0.5mx 1.2x sin 90°

F= 5.04N