A current of 5.0 A flows in a semicircular wire having a radius of 5.0 cm . A magnetic field B with a magnitude of 0.50 T is present perpendicular to the plane of the wire. Determine the magnitude of the magnetic force exerted on this wire.

Step-by-step solution

The magnetic force on a current-carrying wire in a uniform magnetic field is given by F = I( L B ) , where L is the displacement vector from the initial point to the final point of the wire. For a semicircular wire of radius R , the magnitude of the displacement vector is the diameter of the semicircle, so L = 2R . Since the magnetic field is perpendicular to the plane of the wire, the angle between L and B is 90^ . The magnitude of the magnetic force is F = I L B (90^ ) = I(2R)B . Given I = 5.0 A , R = 5.0 cm = 0.