Question

A straight current-carrying conductor 30 cm long carries a current of 5 A. It is placed in a uniform magnetic field of induction 0.2 T, with its length making an angle of 60° with the direction of the field. Find the force acting on the conductor.

Answer 1

Data : l = 30 cm = 0.3 m, 

I = 5 A, B = 0.2 T, θ = 60°

The magnitude of the force on the conductor,

F = I\(\vec l\) x\(\vec B\)| = IlB sin θ 

= (5 A) (0.3 m) (0.2 T) sin 60° 

= 0.3 × 0.866 = 0.2598 N 

The direction of the force is given by the cross product rule.