A) 2
B) 1.5
C) 0.55
D) 0.65
Correct Answer: D
Solution :
Megnetic force on straight wire\[F=Bil\text{ }sin\theta =Bil\text{ }sin\,{{90}^{o}}=Bil\] 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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