1.

A current carrying conductor of certain length, kept perpendicular to the magnetic field experiences a force F. What will be the force if the current is increased four times, length is halved and magnetic field is tripled?

Answer»

F = ILB = (4I) × ( L / 2) × (3 B) = 6 F 

Therefore, the force increases six times.



Discussion

No Comment Found

Related InterviewSolutions