Question

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15T?

Answer 1

I = 8A, B = 0.15 T and θ = 30°
F = BI1 sin θ
Therefore, force per unit length on the wire,
F = 0.15 × 8 x 1 x sin30°
= 0.15 x 8 x 1 x 0.5
= 0.6 Nm^-1