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For a toroid N = 500, radius = 40 cm, an...

For a toroid N = 500, radius = 40 cm, and area of cross section `=10cm^(2)`. Find inductance

A

`125muH`

B

`250muH`

C

0.00248 H

D

zero

Text Solution

Verified by Experts

The correct Answer is:
A

If we flow a current I, then
`B=(mu_(0)Ni)/(2piR)`
`phi_("self")=(BxxA)xxN=((mu_(0)Ni)/(2piR))AxxN" "rArr" "phi_("self")=((mu_(0)N^(2)A)/(2piR))"i and "phi_("self")=Li`
`L=(mu_(0)N^(2)A)/(2piR)=(mu_(0))/(4pi)(2N^(2)A)/(R)" "rArr" "L=(10^(-7))xx(2xx(500)^(2)xx10xx(10^(-2))^(2))/(0.4)=125muH`
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