(i) Define the coefficient of self-induction and obtain an expression for the self-inductance of a long solenoid of length $\ell$, area of cross-section $A$, having $N$ turns. (ii) Calculate the self-inductance of a coil using the data obtained when an AC source of frequency $\left(\dfrac{200}{\pi}\right)\,\mathrm{Hz}$ and a DC source are applied across the coil: AC source — $(V,I)$: $(3.0\,\mathrm{V}, 0.5\,\mathrm{A})$, $(6.0\,\mathrm{V}, 1.0\,\mathrm{A})$, $(9.0\,\mathrm{V}, 1.5\,\mathrm{A})$; DC source — $(V,I)$: $(4.0\,\mathrm{V}, 1.0\,\mathrm{A})$, $(6.0\,\mathrm{V}, 1.5\,\mathrm{A})$, $(8.0\,\mathrm{V}, 2.0\,\mathrm{A})$.

Self-inductance = flux linkage per unit current; $L = \dfrac{\mu_{0}N^{2}A}{\ell} = \mu_{0}n^{2}A\ell$. From the data, $L = 5\sqrt{5}\,\mathrm{mH} \approx 11.2\,\mathrm{mH}$.

CBSE Class 12 · 2023 · Electromagnetic InductionLong Answer

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(i) Define the coefficient of self-induction and obtain an expression for the self-inductance of a long solenoid of length

ℓ

, area of cross-section

A

, having

N

turns.
(ii) Calculate the self-inductance of a coil using the data obtained when an AC source of frequency

(\frac{200}{π}) Hz

and a DC source are applied across the coil: AC source —

(V, I)

(3.0 V, 0.5 A)

(6.0 V, 1.0 A)

(9.0 V, 1.5 A)

; DC source —

(V, I)

(4.0 V, 1.0 A)

(6.0 V, 1.5 A)

(8.0 V, 2.0 A)

Solution

(i) The coefficient of self-induction

L

is the magnetic flux linked with a coil per unit current through it,

N ϕ = L I

.
For a long solenoid the interior field is

B = μ_{0} n I

with

n = N / ℓ

. Flux through one turn is

B A

, and the total flux linkage is

N ϕ = (n ℓ) (μ_{0} n I) (A) = μ_{0} n^{2} A ℓ I

Hence

L = μ_{0} n^{2} A ℓ = \frac{μ _{0} N ^{2} A}{ℓ}

(ii) The DC readings give the ohmic resistance:

R = \frac{V}{I} = \frac{4}{1} = 4 Ω

(consistent for all rows).
The AC readings give the impedance:

Z = \frac{V}{I} = \frac{3}{0.5} = 6 Ω

.
For a coil,

Z^{2} = R^{2} + X_{L}^{2}

, so

X_{L} = Z^{2} - R^{2} = 36 - 16 = 20 Ω

With

X_{L} = 2 π f L

and

2 π f = 2 π (\frac{200}{π}) = 400 s^{- 1}

L = \frac{X _{L}}{400} = \frac{20}{400}

L = 55 mH \approx 11.2 mH

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CBSE Class 12 · 2023 · Electromagnetic InductionLong Answer

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