Guided Derivation — Inductance $L$

Derive $L$ (self-inductance), $M$ (mutual inductance), and $U = \frac{1}{2} L I^2$ from first principles.

Try each step yourself before expanding the answer.

1. Self-Inductance of a Solenoid

1.1 Magnetic field inside a solenoid

A solenoid has $N$ turns, length $\ell$, and carries current $I$. What is $B$ inside?

[!danger]- Derive it $$ B = \mu_0 n I = \mu_0 \frac{N}{\ell} I $$ Where it comes from: Ampère's law $\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{\text{enc}}$. For a solenoid, the field is uniform inside and negligible outside, so $B \ell = \mu_0 N I$.

1.2 Magnetic flux through one turn

Each turn has area $A$. What is $\Phi$ through a single turn?

[!danger]- Derive it $$ \Phi = BA = \mu_0 \frac{N}{\ell} I A $$

1.3 Flux linkage

The total flux linking all $N$ turns is $N\Phi$. Write it out.

[!danger]- Derive it $$ N\Phi = N \cdot \left( \mu_0 \frac{N}{\ell} I A \right) = \frac{\mu_0 N^2 A}{\ell} I $$

1.4 Where does $L = N\Phi / I$ come from?

From the Biot–Savart law, $B \propto I$ for a fixed geometry in a linear medium. So $\Phi \propto I$, and the flux linkage $N\Phi \propto I$. Write the proportionality constant as $L$:

$$ N\Phi = L I \quad\Longrightarrow\quad L = \frac{N\Phi}{I} $$

$L$ is called the self-inductance. It depends only on geometry (turns, area, length, core material), not on current.

Analogy: $Q = CV ;\Longleftrightarrow; N\Phi = LI$

Now substitute your expression for $N\Phi$ from §1.3 and obtain $L$ for a solenoid.

[!danger]- Derive it $$ L = \frac{N\Phi}{I} = \frac{ \frac{\mu_0 N^2 A}{\ell} I }{I} = \frac{\mu_0 N^2 A}{\ell} $$ Result: $\displaystyle L = \frac{\mu_0 N^2 A}{\ell}$

1.5 Alternative route — back EMF definition

Inductance is also defined via Faraday's Law:

$$ \mathcal{E} = -L\frac{dI}{dt} $$

Starting from Faraday's Law $\mathcal{E} = -N \frac{d\Phi}{dt}$, substitute $\Phi$ from step 1.2 and recover the same $L$.

[!danger]- Derive it $$ \mathcal{E} = -N \frac{d}{dt}!\left( \mu_0 \frac{N}{\ell} I A \right) = -\frac{\mu_0 N^2 A}{\ell} \frac{dI}{dt} $$ Comparing with $\mathcal{E} = -L\frac{dI}{dt}$ gives $L = \dfrac{\mu_0 N^2 A}{\ell}$.

2. Mutual Inductance of Coaxial Solenoids

Two coaxial solenoids share the same length $\ell$ and area $A$. Primary has $N_p$ turns, secondary $N_s$ turns.

2.1 Field from the primary

The primary carries current $I_p$. What $B$-field does it produce inside?

[!danger]- Derive it $$ B_p = \mu_0 \frac{N_p}{\ell} I_p $$

2.2 Flux through the secondary

The coils are coaxial — the $B$-field of the primary is parallel to the axis, and the normal to each secondary turn is also axial, so $\theta = 0$ and $\cos\theta = 1$.

Find $\Phi_s$ (flux per turn) and $N_s\Phi_s$ (flux linkage).

[!danger]- Derive it $$ \Phi_s = B_p A \cos 0 = B_p A = \mu_0 \frac{N_p}{\ell} I_p A $$ $$ N_s \Phi_s = \frac{\mu_0 N_p N_s A}{\ell} I_p $$

[!warning]- What if the coils aren't perfectly coaxial? If the secondary's area normal makes an angle $\theta$ to the primary's field: $$ \Phi_s = B_p A \cos\theta \quad\Rightarrow\quad M = \frac{\mu_0 N_p N_s A}{\ell} \cos\theta $$ Mutual inductance is maximised when $\theta = 0$ and zero when $\theta = 90^\circ$.

2.3 Definition of mutual inductance

Mutual inductance $M$ is defined as the flux linkage in the secondary per unit current in the primary:

$$ M = \frac{N_s \Phi_s}{I_p} $$

Substitute and obtain $M$.

[!danger]- Derive it $$ M = \frac{N_s \Phi_s}{I_p} = \frac{ \frac{\mu_0 N_p N_s A}{\ell} I_p }{I_p} = \frac{\mu_0 N_p N_s A}{\ell} $$ Result: $\displaystyle M = \frac{\mu_0 N_p N_s A}{\ell}$

2.4 Reciprocity

Swap roles — if $I_s$ drives the secondary, the flux linkage in the primary is $N_p \Phi_p = \frac{\mu_0 N_p N_s A}{\ell} I_s$. Show $M$ is the same.

[!danger]- Derive it $$ M = \frac{N_p \Phi_p}{I_s} = \frac{ \frac{\mu_0 N_p N_s A}{\ell} I_s }{I_s} = \frac{\mu_0 N_p N_s A}{\ell} $$ So $M_{12} = M_{21} = M$ — mutual inductance depends only on geometry.

2.5 Mutually induced EMF

From Faraday's Law, the EMF induced in the secondary is $\mathcal{E}_s = -N_s \frac{d\Phi_s}{dt}$. Substitute $\Phi_s$ from 2.2 and express in terms of $M$.

[!danger]- Derive it $$ \mathcal{E}_s = -N_s \frac{d}{dt}!\left( \mu_0 \frac{N_p}{\ell} I_p A \right) = -\frac{\mu_0 N_p N_s A}{\ell} \frac{dI_p}{dt} = -M \frac{dI_p}{dt} $$

2.6 Coupling coefficient $k$

The coaxial derivation assumed all flux from the primary links the secondary. In reality, some flux leaks out. The coupling coefficient $k$ measures what fraction actually links:

$$ k = \frac{M}{\sqrt{L_p L_s}}, \qquad 0 \le k \le 1 $$

Derive $k$ for the ideal coaxial case using $L_p = \frac{\mu_0 N_p^2 A}{\ell}$, $L_s = \frac{\mu_0 N_s^2 A}{\ell}$, and $M = \frac{\mu_0 N_p N_s A}{\ell}$.

[!danger]- Derive it $$ k = \frac{M}{\sqrt{L_p L_s}} = \frac{ \frac{\mu_0 N_p N_s A}{\ell} } { \sqrt{ \frac{\mu_0 N_p^2 A}{\ell} \cdot \frac{\mu_0 N_s^2 A}{\ell} } } = \frac{ \cancel{\mu_0 N_p N_s A / \ell} } { \cancel{\mu_0 N_p N_s A / \ell} } = 1 $$ Coaxial solenoids with a shared core and no leakage have $k = 1$ (perfect coupling). Real transformers have $k < 1$ due to flux leakage.

[!tip]- What $k$ means physically

  • $k = 1$ → all flux from one coil links the other (ideal transformer)
  • $k \approx 0.5$ — loosely coupled (e.g. air-core coils, wireless charging with large gap)
  • $k \approx 0.99$ — tightly coupled (iron-core transformer)
  • $k$ depends on geometry and core material, not on current

3. Energy Stored in an Inductor

3.1 Power delivered to an inductor

The voltage across an inductor is $V_L = L \frac{dI}{dt}$. Power is $P = V_L I$. Write $P$ in terms of $L$ and $I$.

[!danger]- Derive it $$ P = V_L I = L \frac{dI}{dt} \cdot I $$

3.2 Energy from power

Energy is the time integral of power: $U = \int P , dt$. Substitute $P$ and change variable to $I$.

[!danger]- Derive it $$ U = \int P , dt = \int L \frac{dI}{dt} \cdot I , dt = \int_0^I L I , dI $$

3.3 Evaluate the integral

$L$ is constant (geometry). Complete the integral.

[!danger]- Derive it $$ U = L \int_0^I I , dI = L \cdot \frac{1}{2} I^2 $$ Result: $\displaystyle U = \frac{1}{2} L I^2$

3.4 Energy density (bonus)

For a solenoid, substitute $L = \mu_0 n^2 A \ell$ and $B = \mu_0 n I$ to express $U$ in terms of $B$ and volume $A\ell$. What is the magnetic energy density $u = U / (A\ell)$?

[!danger]- Derive it $$ U = \frac{1}{2} L I^2 = \frac{1}{2} (\mu_0 n^2 A \ell) I^2 = \frac{1}{2} \mu_0 n^2 I^2 A \ell $$ Using $B = \mu_0 n I$ → $n I = B / \mu_0$ → $n^2 I^2 = B^2 / \mu_0^2$: $$ U = \frac{1}{2} \mu_0 \cdot \frac{B^2}{\mu_0^2} \cdot A \ell = \frac{B^2}{2\mu_0} (A\ell) $$ $$ u = \frac{U}{A\ell} = \frac{B^2}{2\mu_0} $$

Summary of Results

Quantity Formula Derived in
Self-inductance (solenoid) $\displaystyle L = \frac{\mu_0 N^2 A}{\ell}$ §1.4
Back EMF $\displaystyle \mathcal{E} = -L\frac{dI}{dt}$ §1.5
Mutual inductance (coaxial) $\displaystyle M = \frac{\mu_0 N_p N_s A}{\ell}$ §2.3
Mutually induced EMF $\displaystyle \mathcal{E}_s = -M\frac{dI_p}{dt}$ §2.5
Stored energy $\displaystyle U = \frac{1}{2} L I^2$ §3.3
Magnetic energy density $\displaystyle u = \frac{B^2}{2\mu_0}$ §3.4

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