A straight wire of mass 250 g and length 2.5 m carries a current of 4 A . It can be suspended in mid air by a uniform horizontal magnetic field of magnitude:

Step-by-step solution

The wire is suspended in mid-air, which implies that the magnetic force acting on the wire must balance the gravitational force acting on it. The gravitational force F_g acting on the wire is given by F_g = mg , where m = 250 g = 0.25 kg and g = 9.8 m/s ^2 . F_g = 0.25 9.8 = 2.45 N . The magnetic force F_m on a current-carrying wire in a uniform magnetic field is given by F_m = IlB ( ) . Since the magnetic field is horizontal and the wire is suspended, the force is perpendicular to the wire, so = 90^ and (90^ ) = 1