A) 2
B) 1.5
C) 0.55
D) 0.65
Correct Answer: D
Solution :
Magnetic force on straight wire \[F=Bil\text{ }sin\text{ }\theta =\text{ }Bil\text{ }sin\text{ }{{90}^{o}}=\text{ }Bil\] For equilibrium of wire in mid-air,\[F=mg\] \[Bil=mg\] \[\therefore \] \[B=\frac{mg}{il}=\frac{200\times {{10}^{-3}}\times 9.8}{2\times 1.5}=0.65\,T\]You need to login to perform this action.
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