3B · Inductance Numerics + Proofs (Full Solutions)

Solenoid L and energy numerics, parallel-coil proof, series-opposing numeric, U = ½Li² proof. Step-by-step, SI units.

This page covers every inductance numeric in the GE-2 papers (2022, 2019, 2023) plus the two proofs (parallel coils 2022, energy ×2 in 2018 and 2023). Four master formulas solve everything.

Formula 1. Solenoid inductance: $L = \dfrac{\mu_0 N^2 A}{l}$, where $A = \pi r^2$ is the cross-section and $\mu_0 = 4\pi\times 10^{-7}$ H/m.

Formula 2. Similar-coil scaling: $L \propto N^2$ (same shape, area and length, only turns change).

Formula 3. Series combination: $L = L_1 + L_2 \pm 2M$ (upper sign aiding, lower sign opposing).

Formula 4. Magnetic energy stored in an inductor: $U = \dfrac{1}{2}LI^2$ (work done against the induced e.m.f. while building the current).

Solenoid with inside field B, length l, area A, N turns

Fig 1. Solenoid: inside field $B = \mu_0 NI/l$, flux linkage $N\Phi = NBA$, hence $L = \mu_0 N^2 A/l$.

Solved PYQs

Q12022

A coil of 800 turns has self-inductance 400 mH. What will be the self-inductance of a similar coil with 500 turns?

Recall

Same shape and size means only $N$ changes. Then $L \propto N^2$, so $L_2/L_1 = (N_2/N_1)^2$.

Steps

Write the ratio: $L_2/L_1 = (N_2/N_1)^2$ (Eq. 2 — "similar coil" means only $N$ changes).
Substitute $N_2/N_1 = 500/800 = 0.625$.
Square: $0.625^2 = 0.390625$.
Multiply: $L_2 = 400 \times 0.390625 = 156.25$ mH.

Answer

$\boxed{\,L_2 \approx 156\,\text{mH}\,}$.

Exam tip

Write "$L \propto N^2$, not $N$" first — that single line is the mark.

Q22019

Solenoid 1 m long, 10 cm diameter, 5000 turns: (i) inductance, (ii) energy stored with 2 A current.

Recall

$L = \mu_0 N^2 A/l$ with $A = \pi r^2$; energy $U = \tfrac{1}{2}LI^2$. Diameter 10 cm → $r = 0.05$ m.

Steps

Radius and area: $r = 0.05$ m, $A = \pi(0.05)^2 = 7.854\times 10^{-3}$ m² (diameter must be halved first).
Inductance: $L = (4\pi\times 10^{-7})(5000^2)(7.854\times 10^{-3})/1.0$.
Compute numerator: $4\pi\times 10^{-7}\times 25\times 10^6 \times 7.854\times 10^{-3} = 0.2467$ H.
Energy with $I = 2$ A: $U = \tfrac{1}{2}(0.2467)(2^2) = 0.5\times 0.2467 \times 4 = 0.4934$ J.

Answer

(i) $\boxed{\,L \approx 0.247\,\text{H}\,}$. (ii) $\boxed{\,U \approx 0.49\,\text{J}\,}$.

Exam tip

Show $A = \pi(0.05)^2$ explicitly — diameter-vs-radius is the classic trap.

Q32023

Find the self-inductance of a solenoid 40 cm long and radius 4 cm having 200 turns ($\mu_r = 1$).

Recall

Same Eq. (1); air core ($\mu_r = 1$) so $\mu = \mu_0$.

Steps

Convert: $l = 0.40$ m, $r = 0.04$ m, $N = 200$. $A = \pi(0.04)^2 = 5.0265\times 10^{-3}$ m².
Substitute: $L = (4\pi\times 10^{-7})(200^2)(5.0265\times 10^{-3})/0.40$.
Compute $200^2 = 40\,000$ and the numerator: $4\pi\times 10^{-7}\times 40\,000\times 5.0265\times 10^{-3} = 2.5266\times 10^{-4}$.
Divide by $0.40$: $L = 6.3166\times 10^{-4}$ H.

Answer

$\boxed{\,L \approx 6.3\times 10^{-4}\,\text{H} = 0.63\,\text{mH}\,}$.

Exam tip

Write $\mu_r = 1$ explicitly so the examiner sees the air-core assumption.

Q42022

Show that the equivalent inductance of two coils $L_1$, $L_2$ in parallel is $(L_1 L_2 − M^2)/(L_1 + L_2 − 2M)$.

Recall

Parallel means the same voltage across both coils; total current splits $i = i_1 + i_2$. Sign of $M$ follows aiding vs opposing connection.

Steps

Write voltage equations: $V = L_1\,di_1/dt \pm M\,di_2/dt$ and $V = L_2\,di_2/dt \pm M\,di_1/dt$ (each coil sees its own plus mutual e.m.f.).
Solve the two-equation system for the slopes $di_1/dt$, $di_2/dt$. The determinant is $L_1 L_2 - M^2$.
Apply Cramer's rule: $di_1/dt = V(L_2 \mp M)/(L_1 L_2 - M^2)$ and $di_2/dt = V(L_1 \mp M)/(L_1 L_2 - M^2)$.
Add for total: $di/dt = di_1/dt + di_2/dt = V(L_1 + L_2 \mp 2M)/(L_1 L_2 - M^2)$.
Define $V = L_{eq}\,di/dt$ and read off $L_{eq}$.
Sanity check: set $M = 0$ and recover $L_{eq} = L_1 L_2/(L_1 + L_2)$ — the resistor-parallel form.

Answer

$\boxed{\,L_{eq} = \dfrac{L_1 L_2 - M^2}{L_1 + L_2 \mp 2M}\,}$ — upper sign ($-2M$) is the opposing case asked in the question; the lower sign is aiding.

Exam tip

End with the $M = 0$ sanity check — it proves your algebra and earns the last mark.

Q52023

Coils of 50 mH and 100 mH in series give effective inductance 75 mH. Determine the coefficient of mutual inductance.

Recall

Series combination: $L = L_1 + L_2 \pm 2M$ — upper sign aiding, lower sign opposing.

Steps

Compare $L_1 + L_2 = 150$ mH with measured $75$ mH. The measured value is less than the sum, so the minus sign applies (opposing).
Write the equation: $75 = 150 - 2M$.
Solve: $2M = 150 - 75 = 75$, so $M = 37.5$ mH.
Sanity: $M \le \sqrt{L_1 L_2} = \sqrt{50\times 100} = \sqrt{5000} \approx 70.7$ mH, and $37.5 < 70.7$ — physically allowed.

Answer

$\boxed{\,M = 37.5\,\text{mH, opposing connection}\,}$.

Exam tip

First line must be "$75 < 150$, hence opposing" — that justifies the minus sign.

Q620182023×2

Find the magnetic energy stored in an inductor of self-inductance $L$ carrying current $i$.

Recall

Work done against the induced e.m.f. $e = -L\,di/dt$ while building the current from 0 to $i$ is stored as field energy.

Steps

Power needed to push current against $e$: $P = e_{applied}\times i = L\,i\,(di/dt)$ (the applied voltage must overcome $L\,di/dt$).
Small work in time $dt$: $dW = P\,dt = L\,i\,di$.
Integrate from $0$ to $i$: $W = \int_0^i L\,i\,di = \tfrac{1}{2}Li^2$.
No heat loss in an ideal coil — all the work sits in the magnetic field.

Answer

$\boxed{\,U = \dfrac{1}{2}Li^2\,}$, in joules.

Exam tip

Mention "no heat loss, ideal coil — energy stored in the field" to nail the energy-conservation line.

Reference. Brij Lal & Subrahmanyam Ch-10; H.C. Verma Vol-1 Ch-38; S.L. Arora Vol-1 Ch-6.