====== X. Inductance ======
===== - Mutual inductance =====
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Mutual inductance of coil 2 caused by coil 1 is

$$ M_{21} = \frac{N_2 \Phi_{21}}{I_1} $$

where $\Phi_{21}$ is the flux through the coil 2 caused by the magnetic field created by the current in coil 1 $I_1$. A similar equation of mutual inductance can be written for coil 1:

$$ M_{12} = \frac{N_1 \Phi_{12}}{I_2}. $$

It can be proven that $M_{21} = M_{12}$. So dropping the subscripts we get, **mutual inductance** of two coils

$$ M = \frac{N_2 \Phi_{21}}{I_1} = \frac{N_1 \Phi_{12}}{I_2}. $$

The unit is **henry** (H) named after American scientist Joseph Henry who discovered induced emf independently of Faraday.

Prove that 1 H = 1 V s / A.

And prove that the induced emf in the two coils

$$ \varepsilon_1 = -M\frac{dI_2}{dt} \text{ and } \varepsilon_2 = -M\frac{dI_1}{dt}. $$

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In this setup, $N_1=500$, $N_2=10$, $R_1=3.10$ cm, $l_1=75.0$ cm. In the solenoid $dI_1/dt=200$ A/s. What is emf induced in the outer coil?

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A transformer should have high $M$, but a clothes dryer shown above should not have a large mutual inductance. How can you reduce $M$ in a dryer?
===== - Self-inductance and inductors =====
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If the current varies, the self-induced emf in the circuit

$$ \varepsilon = - \frac{d\Phi_m}{dt} $$

where $\Phi_m \propto I$ or $\Phi_m=LI$. Here, $L$ is called the **self-inductance** of the wire loop. If the loop has $N$ turns, then $ N\Phi_m = LI$ and, hence, the induced emf

$$ \varepsilon = -L \frac{dI}{dt} $$

and so the self-inductance $L=|\varepsilon| / |dI/dt|$. This is an **inductor** used in a circuit.

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The induced emf always opposes the original current in the circuit according to Lenz's law.

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Inductors are used in street intersections in the developed world, and in metal detectors everywhere.

==== - Solenoid ====
For a cylindrical solenoid the interior magnetic field $B=\mu_0 NI$ and per unit length it becomes $B=\mu_0 NI / l$.

So the magnetic flux $\Phi_m = BA = \mu_0 NAI / l$.

And the self-inductance of the solenoid $L=N\Phi_m / I = \mu_0 N^2 A / l$.

If the turns per unit length $n=N/l$, then

$$ L = \mu_0 \left(\frac{N}{l}\right)^2 Al = \mu_0 n^2 Al = \mu_0 n^2 V. $$

==== - Toroid ====

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For the rectangular toroid above apply Ampere's law to find $B=\mu_0 NI / (2\pi r)$ and so

$$ \Phi_m = \int B da = \int_{R_1}^{R_2} \frac{\mu_0 NI}{2\pi r} hdr = \frac{\mu_0 NhI}{2\pi} \ln\frac{R_2}{R_1}. $$

So the inductance of such an inductor will be

$$ L = \frac{N\Phi_m}{I} = \frac{\mu_0 N^2h}{2\pi} \ln\frac{R_2}{R_1} $$

which is only dependent on the geometry of the toroid.
===== - Energy in a magnetic field =====
The density of electric energy stored in a capacitor was $u_E=\epsilon_0 E^2 / 2$. And the density of magnetic energy stored in an inductor

$$ u_B = \frac{B^2}{2\mu_0}. $$

Find the total energy by multiplying this with the total volume inside an inductor $V$. Total energy

$$ U = u_B \times V = \frac{B^2}{2\mu_0} Al = \frac{(\mu_0 nI)^2}{2\mu_0}Al = \frac{1}{2} (\mu_0 n^2 Al) I^2. $$

But $L=\mu_0 n^2 Al$ is the inductance of an inductor made of a cylindrical solenoid. So

$$ U = \frac{1}{2}LI^2. $$

This equation is valid for any type of inductors.

Can you compare this with the equation of the potential energy in a spring?
===== - RL circuits =====
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Apply Kirchhoff's loop rule to get

$$ \varepsilon - L\frac{dI}{dt} - IR = 0 $$

wich gives the current through the inductor as a function of time

$$ I(t) = \frac{\varepsilon}{R} \left(1-e^{-Rt/L}\right) = \frac{\varepsilon}{R} \left(1-e^{-t/\tau_L}\right) $$

$$ V_L = -L\frac{dI}{dt} = -\varepsilon e^{-t/\tau_L}. $$

Thus the magnetic field energy stored in the inductor

$$ U_L = \frac{1}{2}LI^2 $$

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Think about the significance of $\tau_L$ and the energy stored.

Now if there is no source emf in the RL circuit as shown in panel (c) of the first figure, the inductor will act as a source emf and according to the loop rule,

$$ IR + L\frac{dI}{dt} = 0. $$

Rearranging and evaluating the integral give

$$ I(t) = \frac{\varepsilon}{R} e^{-t/\tau_L} $$

$$ V_L(t) = -L\frac{dI}{dt} = \varepsilon e^{-t/\tau_L}. $$

===== - Oscillations in LC circuits =====
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$$ U = \frac{1}{2}\frac{q^2}{C} + \frac{1}{2} Li^2 = \frac{1}{2}\frac{q_0^2}{C} = \frac{1}{2}LI_0^2. $$

Compare this with the conservation of energy in a mass-spring system:

$$ E = \frac{1}{2}mv^2+\frac{1}{2}kx^2. $$

If the equation of motion of mass is $x(t)=A\cos(\omega t+\phi)$ then for the charge we get

$$ q(t) = q_0\cos(\omega t+\phi) $$

and the angular frequency of the osillation

$$ \omega = \sqrt{\frac{1}{LC}}. $$

And the current as a function of time

$$ i(t) = \frac{dq(t)}{dt} = -\omega q_0 \sin(\omega t+\phi). $$
===== - RLC series circuits =====
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The electromagnetic energy is dissipated by the resistor as

$$ \frac{dU}{dt} = \frac{q}{C}\frac{dq}{dt} + Li\frac{di}{dt} = -i^2R $$

$$ \Rightarrow L\frac{d^2q}{dt^2} + R\frac{dq}{dt} + \frac{1}{C}q=0 $$

which is analogous to


$$ m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx=0 $$

which has the solution

$$ x(t) = A_0 e^{-\frac{b}{2m}t} \cos(\omega t+\phi) $$

and we can write an analogous equation for the charge

$$ q(t) = q_0 e^{-Rt/(2L)} \cos(\omega' t+\phi). $$

where the angular frequency

$$ \omega' = \sqrt{\frac{1}{LC}-\left(\frac{R}{2L}\right)^2}. $$