A conducting rod of length 20 cm undergoes pure translation at a uniform velocity of 10 cm s ⁻¹ perpendicular to its length. A uniform magnetic field of 0.10 T is present, directed perpendicular to the plane of motion. Determine the average magnetic force exerted on the free electrons of the rod.

Step-by-step solution

The magnetic force acting on a moving charge in a magnetic field is given by F = qvB . For a free electron in the conducting rod, the charge is q = e = 1.6 10⁻¹⁹ C . The velocity of the rod is v = 10 cm s ⁻¹ = 0.1 m s ⁻¹ . The magnetic field is B = 0.10 T . Since the magnetic field is perpendicular to the plane of motion, the angle between the velocity and the magnetic field is = 90^ , giving 90^ = 1 . The average magnetic force on each free electron is: F = evB F = (1.6 10⁻¹⁹) (0.1) (0.10) F = 1.6 10⁻²¹ N