First slide

Inductance

Question

A circular coil of radius 5 cm has 500 turns of a wire. The approximate value of the coefficient of self-induction of the coil will be

Moderate

A

25 mH
B

$25 \times 10^{- 3} mH$
C

$50 \times 10^{- 3} mH$
D

$50 \times 10^{- 3} H$

Solution

$ϕ = Li \Rightarrow NBA = Li$
Since magnetic field at the centre of circular coil carrying current is given by $B = \frac{μ_{0}}{4 π} \cdot \frac{2 πNi}{r}$
$∴ N \cdot \frac{μ_{0}}{4 π} \cdot \frac{2 πNi}{r} \cdot {πr}^{2} = Li \Rightarrow L = \frac{μ_{0} N^{2} πr}{2}$
Hence, self-inductance of a coil
$= \frac{4 π \times 10^{- 7} \times 500 \times 500 \times π \times 0.05}{2} = 25 mH$

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