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Number of turns in a coil are 100. When ...

Number of turns in a coil are 100. When a current of 5A is flowing through the coil, the magnetic flux is `10^(-6)Wb.` Find the self induction.

Text Solution

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Self inductance, `L= (nphi)/(i)`
number of turns, n=100, magnetic flux, `phi =10^(-6)" "Wb, Current, i=5A"`
`L=(100 xx 10^(-6))/(5)=20xx10-6 =20muH therefore "Self inductance ,L"=20muH`
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