A toroidal solenoid with an air core has an average radius of 15 cm , area of cross-section `12 "cm"^(2)` and 1200 turns . Ignoring the field variation across the cross-section of the toroid the self-inductance of the toroid is

Options

A. 4.6mH
B. 6.9mH
C. 2.3mH
D. 9.2mH

Answer

C

Explanation

For a solenoid `B=mu_(0)nI`
where `n=N/(2pir)`
`impliesB=(mu_(0)NI)/(2pir)`
Flux linked with the solenoid is
`phi=NBA`
`impliesphi=(mu_(0)N^(2)IA)/(2pir)`
`:.L=(phi)/I=(mu_(0)N^(2)A)/(2pir)`
`=(4pixx10^(-7)xx(1200)^(2)xx12xx10^(-4))/(2pixx15xx10^(-2))`
`L=2.3xx10^(-3)H`
`=2.3mH`