Consider two wire loops close to each other

• An increasing current $I_1$ flows in loop $1$
• This creates an increasing $B$-field $\vec B_1$
• $B$-field passes through loop 2, flux $\Phi_2$
• $\Phi_2$ linked by loop 2 increases
• $\mathcal E_2$ induced in loop 2
• Current $I_2$ flows in loop 2

⭐ This is mutual inductance

🗒️ only works for changing currents

$\Phi_{21}$ is flux through loop 2 due to loop 1, and $M_{21}$ is the mutual inductance

$$ \Phi_{21}=M_{21}I_1 $$

• For a changing current $I_1$

$$ \mathcal E_2=-\frac{\text d \Phi_{21}}{\text d t}=-M_{21}\frac{\text d I_1}{\text d t} $$

• For a changing current $I_2$

$$ \mathcal E_1=-\frac{\text d \Phi_{12}}{\text d t}=-M_{12}\frac{\text d I_2}{\text d t} $$

⭐ $M_{12}=M_{21}=M$ where $M$ is measured in Henrys $1$ H = $1$ Wb$/$A

Transformers

Consider 2 coils wrapped around an iron core with $N_1$ and $N_2$ turns

• Flux for each turn is the same
• Hence total flux for each proportional to $N$

$$ \begin{aligned} \Phi_1\propto B N_1 \;\; &\;\, ; \quad \Phi_2\propto B N_2 \\ \Rightarrow \quad \frac{\Phi_1}{\Phi_2}&=\frac{N_1}{N_2} \end{aligned} $$

• With changing current then:

$$ \frac{\text d\Phi_1}{\text dt}=\frac{N_1}{N_2}\frac{\text d \Phi_2}{\text dt} $$

• Faraday’s Law gives: $\mathcal E_1=\frac{N_1}{N_2}\mathcal E_2$
• Power conservation so: $\mathcal E_1 I_1=\mathcal E_2 I_2$

⭐ A transformer allows AC voltage changes

Calculating mutual inductance

Consider solenoid 1. previous result

$$ B_1=\mu_0 n_1 I_1 =\frac{\mu_0 N_1 I_1}{l} $$

• For each turn turn $\Phi=B_1 A$ for either coil so:

$$ \begin{aligned} M&=M_{21}=\frac{\Phi_2}{I_1}=\frac{N_2B_1A}{I_1} \\ \Rightarrow M&=M_{12}=\frac{\mu_0 N_1 N_2 I_1 A}{l I_1}= \frac{\mu_0 N_1 N_2 A}{l} \end{aligned} $$