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
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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:
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