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A toroidal solenoid with air core has an...

A toroidal solenoid with air core has an average radius of 15cm, area of cross section `12cm^(2)` and has 2000 turns. Calculate the self- inductance of the toroid. Assume the field to be uniform across the cross-section of the toroid.

Text Solution

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Given : Radius of the toroid r = 15 cm
Area of the toroid `A=cm^(2)=12xx10^(-4)m`
No. of turns N = 2000
To find :
Self inductance of the toroid L = ?
Formula :
`L=(muN^(2)A)/(l)=(mu_(0)N^(2)A)/(2pi)`
`L=(cancel(l4pi)xx10^(-7)xx(2000)^(2)xxcancel(12)^(2)xxcancel(12)^(4)xx10^(-4))/(cancel(2pi)xxcancel(15)xx10^(-2))`
`=(8xxcancel(40)^(8)xx10^(5)xx10^(-4)xx10^(-7))/(cancel(5)xx(10^(-2)))`
`=6.4xx10^(-3)H`
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