A current-carrying wire 0.50 m long is positioned perpendicular to a uniform magnetic field. If the current is 10.0 A and there is a resultant force of 3.0 N on the wire due to the interaction of the current and field, what is the magnetic field strength?

Answer

0.6 teslas

Explanation

When a conductor is inside a magnetic field it experiences a force given by;

Force = ILBsinθ

Where I⇒ current

L ⇒length of the conductor

B ⇒ magnetic field strength

θ ⇒ Angle between the conductor and magnetic field.

F = ILBsinθ

When θ = 90°, Then sin 90 = 1 and the formula becomes;

F = ILB

3 = 10 * 0.5 * B

3 = 5B

B = 3/5

= 0.6

magnetic field strength = 0.6 teslas