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The magnitude of the magnetic field at the centre of the tightly wound 150 turn coil of radius 12 cm carrying a current of 2 A isA. 18GB. 19.7GC. 15.7GD. 17.7G |
Answer» Correct Answer - C Here, `N=150, R = 12 cm = 12xx10^(-2)m, I=2A` `therefore" "B=(mu_(0)NI)/(2R)=(2pixx10^(-7)xx150xx2)/(12xx10^(-2))=1.57xx10^(-3)T` `=1.57xx10^(-3)T=15.7xx10^(-4)T=15.7G` |
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