Question

A 0.5 m long straight wire in which a current of 1.2 a is flowing is kept at right angle to a uniform magnetic field of 2.0 t. The force acting on wire will be

Answer 1

Let B be uniform magnetic field ,current in the wire of length L be I, then force due to magnetic field is given by

F=ILxB.

L=0.5m

I=1.2 A

B=2T

Therefore,

|F|=(1.2)(0.5)(2)(sin pi/2)=1.2N. In the direction perpendicular to the plane formed by LandB according to right hand screw law for cross product of two vectors.