Question

A wire with current I = 15 A flows at an angle of 30 degrees to the direction of a uniform magnetic field of 0.9 T. What is the magnitude of the force exerted on a 10 m length of wire? (in units of N)

Answer 1

Answer:

67.5

Explanation:

$F= BILsin\theta$ where F is the magnitude of the force, B is the magnetic field, I is current, $\theta$ is the angle of inclination, L is the length of the wire.

Substituting 0.9 T for magnetic field B, 10 m for length, 15 A for current I and 30 degrees for $\theta$ then

$F=0.9\times 15\times 10\times sin 30^{\circ}=67.5$