A straight wire of mass 200g and length 1.5m carries a current of 2A. It is suspended in mid air by a uniform horizontal magnetic field B.
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Answer:
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Explanation:
Given, mass of the wire =200g=0.2kg
Length of the wire =15m
Current i=2A
Magnetic field B=?
The force acting on the current carrying wire in uniform magnetic field
F=Bilsinθ
F=Bil (∵θ=90 o )
Weight of the wire w=mg=0.2×9.8N
In the position of suspension
Bil=mg
B= ilmg = 2×150.2×9.8=0.65T