A straight conductor 25m long carrying a current of 5A is kept in a uniform magnetic field of 0.05 T. Find the force acting on the wire when it is (1) at right angles to the field (2) at 30° to the field.
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Answer:
Given : B=0.5T L=0.5m I=2.5A
As the wire is suspended in air, weight of the wire is balanced by the force acting on the wire in upward direction.
∴ mg=BIL
Or m×10=0.5×2.5×0.5
⟹ m=0.0625kg=62.5 gm
Explanation:
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