A current is 1.5 ampere in a 800 turns coil produce 1.5 x 18–⁵ Weber magnetised flux in each turn. Find out the self induction of coil
Gives
Current=1.5A
Turn=500
Magnetic flux= "1.5\\times10^{-5}Wb"
"\\phi=nLi"
"L=\\frac{\\phi}{ni}"
Put value
"L=\\frac{1.5\\times10^{-5}}{500\\times1.5}=0.02\\times10^{-6} H"
"L=0.02\\mu H"
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