Question

A wire in a uniform magnetic field of 0.350 T carries a current of 3.50 A. If the magnitude of the magnetic force per unit length on the wire is 0.278 N/m, what is the angle between the magnetic field and the current?

Answer 1

Answer:

13.12°

Explanation:

Parameters given:

Magnetic field strength, B = 0.350 T

Current in the wire, I = 3.5 A

Force per unit length, F/L = 0.278 N/m

The force on a wire due to the presence of a magnetic field is given as:

F = I*L*B*sinФ

where Ф = angle between the magnetic field and current

The force per unit length, F/L, will be:

F/L = I*B*sinФ

0.278 = 3.5 * 0.35 * sinФ

=> sinФ = $\frac{0.278}{3.5 * 0.35}$

sinФ = 0.227

Ф = $sin^{-1}(0.227)$

Ф = 13.12°

The angle between the magnetic field and the current is 13.12°.