1.

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`


Discussion

No Comment Found

Related InterviewSolutions