Question

A horizontal wire 0.10 m long carries a current of 5A find the magnitude and direction of the field that can support the wire assuming it's mass to be 3.0

Answer 1

Answer:

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.

Explanation: