Calculate the self inductance (L) of toroid for major radius (R) = 15 cm, cross-section of toroid having radius (r) = 2.0 cm and the number of turns (n) = 1200.
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Given: radius = 0.15m
number of turns in a coil = 1200
cross section = 0.02m
To find : self inductance (L)
Solution : Self inductance is defined as the property of coil due to which it opposes any change of current flowing through the circuit.
Magnetic field in a coil will be , B = μ₀n₁i / l
= μ₀ni / 2πr
Total magnetic flux Ф = Li
But, Ф = N₁BA = μ₀N₁²iA\ 2πr
Therefore, L = μ₀N₁²A / 2πr
L= (4π×10^-7×1200×1200×0.02×0.02)/ 2π×0.15 H
L= 7680 ×10^-7 H
L= 0.7680 mH
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