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Block 20 — Electromagnetic Induction and Energy

Learning objectives

Preparation at Home

Well, again

• read through the present chapter and write down anything you did not understand.
• Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting).

For checking your understanding please do the following exercises:

• …

90-minute plan

1. Warm-up (x min):
   1. ….
2. Core concepts & derivations (x min):
   1. …
3. Practice (x min): …
4. Wrap-up (x min): Summary box; common pitfalls checklist.

Conceptual overview

Core content

Self-Induction

Up to now, we investigated the induction of electric voltages and currents based on the change of an external flux ${\rm d}\Psi / {\rm d}t$. For the induced current $i_{\rm ind}$, we found that it counteracts the change of the external flux (Lenz law).

But what happens, when there is no external field - only a coil which creates the flux change itself (see Abbildung 1)?

To understand this, we will investigate the situation for a long coil (Abbildung 2).

The created field density of the coil can be derived from Ampere's Circuital Law

\begin{align*} \theta(t) &= \int & \vec{H}(t) \cdot {\rm d}\vec{s} \\ &= \int & \vec{H}_{\rm inner}(t) \cdot {\rm d}\vec{s} & + & \int \vec{H}_{\rm outer}(t) \cdot {\rm d} \vec{s} \\ &= \int & \vec{H}(t) \cdot {\rm d}\vec{s} & + & 0 \\ &= & {H}(t) \cdot l \\ \end{align*}

With magnetic voltage $\theta(t) = N \cdot i$ this lead to the magnetic flux density $B(t)$

\begin{align*} N \cdot i &= {H}(t) \cdot l \\ {H}(t) &= {{N \cdot i }\over {l}} \\ {B}(t) &= \mu_0 \mu_{\rm r} \cdot {{N \cdot i }\over {l}} \\ \end{align*}

Based on the magnetic flux density $B(t)$ it is possible to calculate the flux $\Phi(t)$:

\begin{align*} \Phi(t) &= \iint_A \vec{B}(t) \cdot {\rm d}\vec{A} \\ &= \iint_A \mu_0 \mu_{\rm r} \cdot {{N \cdot i }\over {l}} \cdot {\rm d}A \\ &= \mu_0 \mu_{\rm r} \cdot {{N \cdot i }\over {l}} \cdot A \\ \end{align*}

The changing flux $\Phi$ is now creating an induced electric voltage and current, which counteracts the initial change of the current. This effect is called Self Induction. The induced electric voltage $u_{\rm ind}$ is given by:

\begin{align*} u_{\rm ind} &= - N \cdot {{{\rm d} \Phi(t)}\over{{\rm d}t}} \\ &= - N \cdot {{{\rm d} (\mu_0 \mu_{\rm r} \cdot {{N \cdot i }\over {l}} \cdot A)}\over{{\rm d}t}} \\ &= - N \cdot \mu_0 \mu_{\rm r} \cdot {{N \cdot A }\over {l}} \cdot {{{\rm d}i}\over{{\rm d}t}} \\ \end{align*}

\begin{align*} \boxed{ u_{\rm ind} = - \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A }\over {l}} \cdot {{{\rm d}i}\over{{\rm d}t}} \\ } \\ \text{for a long coil} \end{align*}

The result means that the induced electric voltage $u_{\rm ind}$ is proportional to the change of the current ${{\rm d}\over{{\rm d}t}}i$. The proportionality factor is also called Self-inductance $L$ (or often simply called inductance).

4.5 Inductance

The inductance is another passive basic component of the electric circuit. Besides the ohmic resistor $R$ and the capacitor $C$, the inductor $L$ is the lump component entailing the inductance.

Generally, the inductance is defined by: \begin{align*} \boxed{ L = \left|{{u_{\rm ind}}\over{{\rm d}i / {\rm d}t}}\right| \\ } \end{align*}

The inductance $L$ can also be described differently based on Lenz law $u_{\rm ind} = - {{\rm d}\over{{\rm d}t}}\Psi(t)$ :

\begin{align*} L &= \left|{{u_{\rm ind}}\over{{\rm d}i / {\rm d}t}}\right| \\ &= {{{d \Psi(t)}/{dt}}\over{{\rm d}i / {\rm d}t}} \\ \end{align*}

\begin{align*} \boxed{ L = {{ \Psi(t)}\over{i}} } \end{align*}

One can also consider an inductor a „conservative person“: it does not like to see abrupt changes in the passing current. It reacts to any change in the current with a counteracting voltage since the current change leads to a changing flux and - therefore - an induced voltage. The Abbildung 3 shows an inductor in series with a resistor and a switch (any real switch also behaves as a capacitor, when open). Once the simulation is started, the inductor directly counteracts the current, which is why the current only slowly increases.

The unit of the inductance is $\rm 1 ~Henry = 1 ~H = 1 {{Vs}\over{A}} = 1{{Wb}\over{A}} $

Mathematically the voltages can be described in the following way:

\begin{align*} u_0 &= u_R &+ &u_L \\ &= i \cdot R & + & {{{\rm d}\Psi}\over{{\rm d}t}} \\ &= i \cdot R & + &L \cdot {{{\rm d}i}\over{{\rm d}t}} \\ \end{align*}

Inductance of different Components

Long Coil

In the last sub-chapter, the formula of a long coil was already investigated. By these, the inductance of a long coil is

\begin{align*} \boxed{L_{\rm long \; coil} = \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A }\over {l}}} \end{align*}

Toroidal Coil

The toroidal coil was analyzed in the last chapter(see magnetic Field Strength Part 1: Toroidal Coil). Here, a rectangular intersection a assumed (see Abbildung 4).

This leads to

\begin{align*} H(t) = {{N \cdot i}\over {l}} \end{align*}

with the mean magnetic path length (= length of the average field line) $l = \pi(r_{\rm o} + r_{\rm i})$:

\begin{align*} H(t) = {{N \cdot i}\over { \pi(r_{\rm o} + r_{\rm i})}} \end{align*}

The inductance $L$ can be calculated by

\begin{align*} L_{\rm toroidal \; coil} &= {{ \Psi(t)}\over{i}} \\ &= {{ N \cdot \Phi(t)}\over{i}} \\ \end{align*}

With the magnetic flux density $B(t) = \mu_0 \mu_{\rm r} H(t) = \mu_0 \mu_{\rm r} {{i \cdot N }\over {l}}$ and the cross section $A = h (r_{\rm o} - r_{\rm i})$, we get:

\begin{align*} \quad \quad L_{\rm toroidal \; coil} &= {{ N \cdot \mu_0 \mu_{\rm r} {{i \cdot N } \over { \pi(r_{\rm o} + r_{\rm i})}} \cdot h(r_{\rm o} - r_{\rm i})}\over{i}} \\ &= {{ N^2 \cdot \mu_0 \mu_{\rm r} \cdot h(r_{\rm o} - r_{\rm i})}\over{ \pi(r_{\rm o} + r_{\rm i})}} \\ \end{align*}

\begin{align*} \boxed{ L_{\rm toroidal \; coil} = \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{ h(r_{\rm o} - r_{\rm i})}\over{ \pi(r_{\rm o} + r_{\rm i})}} } \end{align*}

6 Inductances in Circuits

Focus here: uncoupled inductors!

Series Circuits

Based on $L = {{ \Psi(t)}\over{i}}$ and Kirchhoff's mesh law ($i=\rm const$) the series circuit of inductions can be interpreted as a single current $i$ which generates multiple linked fluxes $\Psi$. Since the current must stay constant in the series circuit, the following applies for the equivalent inductor of a series connection of single ones:

\begin{align*} L_{\rm eq} &= {{\sum_i \Psi_i}\over{I}} = \sum_i L_i \end{align*}

A similar result can be derived from the induced voltage $u_{ind}= L {{{\rm d}i}\over{{\rm d}t}}$, when taking the situation of a series circuit (i.e. $i_1 = i_2 = i_1 = ... = i_{\rm eq}$ and $u_{\rm eq}= u_1 + u_2 + ...$):

\begin{align*} & u_{\rm eq} & = &u_1 & + &u_2 &+ ... \\ & L_{\rm eq} {{{\rm d}i_{\rm eq} }\over{{\rm d}t}} & = &L_{1} {{{\rm d}i_{1} }\over{{\rm d}t}} & + &L_{2} {{di_{2} }\over{dt}} &+ ... \\ & L_{\rm eq} {{{\rm d}i }\over{{\rm d}t}} & = &L_{1} {{{\rm d}i }\over{{\rm d}t}} & + &L_{2} {{di }\over{dt}} &+ ... \\ & L_{\rm eq} & = &L_{1} & + &L_{2} &+ ... \\ \end{align*}

Parallel Circuits

For parallel circuits, one can also start with the principles based on Kirchhoff's mesh law:

\begin{align*} u_{\rm eq}= u_1 = u_2 = ... \\ \end{align*}

and Kirchhoff's nodal law:

\begin{align*} i_{\rm eq}= i_1 + i_2 + ... \\ \end{align*}

Here, the formula for the induced voltage has to be rearranged:

\begin{align*} u_{\rm ind} &= L {{{\rm d}i}\over{{\rm d}t}} \quad \quad \quad \quad \bigg| \int(){\rm d}t \\ \int u_{\rm ind} {\rm d}t &= L \cdot i \\ i &= {{1}\over{L}} \cdot \int u_{\rm ind} {\rm d}t \\ \end{align*}

By this, we get:

\begin{align*} i_{\rm eq} &=& i_1 &+& i_2 &+& ... \\ {{1}\over{L_{\rm eq}}} \cdot \int u_{\rm eq} {\rm d}t &=& {{1}\over{L_1}} \cdot \int u_{1} {\rm d}t &+& {{1}\over{L_2}} \cdot \int u_{2} {\rm d}t &+& ... \\ {{1}\over{L_{\rm eq}}} \cdot \int u {\rm d}t &=& {{1}\over{L_1}} \cdot \int u {\rm d}t &+& {{1}\over{L_2}} \cdot \int u {\rm d}t &+& ... \\ {{1}\over{L_{\rm eq}}} &=& {{1}\over{L_1}} &+& {{1}\over{L_2}} &+& ... \\ \end{align*}

Common pitfalls

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Exercises

Exercise 4.1.5 Effects of induction II

A changing of magnetic flux is passing a coil with a single winding and induces the voltage $u_{\rm ind}(t)$. The following pictures Abbildung ## show different voltage-time diagrams as examples.

• Create for each $u_{\rm ind}(t)$-diagram the corresponding $\Phi(t)$-diagram!
• Write down each maximum value of $\Phi(t)$

Note the given start value $\Phi_0$ for each flux.

Abb. ##: Voltage-Time-Diagrams

Solution for (a)

For partwise linear $u_{\rm ind}$ one can derive: \begin{align*} u_{\rm ind} &= -{{{\rm d}\Phi}\over{{\rm d}t}} \\ \rightarrow \Phi &= -\int_0^t{ u_{\rm ind} \;{\rm d}t} \\ \Phi &= \Phi_0 -\sum_k {u_{{\rm ind},~k} \; \Delta t} \\ \end{align*}

For diagram (a):

• $t= 0.00 ... 0.04 ~\rm s\quad$: $\quad \Phi = \Phi_0 - {0 \cdot \; \Delta t} \quad\quad\quad\quad\quad\quad\quad= 0 ~\rm Wb$
• $t= 0.04 ... 0.10 ~\rm s\quad$: $\quad \Phi = 0 {~\rm Wb} - {{30 ~\rm mV} \cdot \; (t - 0.04 ~\rm s)} = \quad {1.2 ~\rm mWb} - t \cdot 30 ~\rm mV$
• $t= 0.10 ... 0.14 ~\rm s\quad$: $\quad \Phi = {1.2 ~\rm mWb} - {0.10 ~\rm s} \cdot 30 ~\rm mV \quad = - {1.8 ~\rm mWb}$

Result for (a)

Exercise E2 Self Induction
(written test, approx. 8 % of a 120-minute written test, SS2022)

Accidentally, a motor with an inductance of $L=50 ~\rm \mu H$ is connected to a DC voltage source, which is fused with a circuit breaker.
The circuit breaker opens the circuit At $t_0=0$ with a current of $I=63 ~\rm A$. The current is reduced linearly down to $0 ~\rm A$ within $1 ~\rm \mu s$.
(The inner resistance of the motor shall be neglected.)

1. Draw the circuit (the circuit breaker can be drawn as a switch), with all voltage and current arrows.

Result

2. What is the maximum magnitude of the voltage, which the circuit breaker has to withstand?
Sketch the diagrams $i(t)$ and $u_{\rm ind} (t)$ with the calculated value of the induced voltage.

Path

For the maximum voltage on the circuit breaker one has to consider the following:

• external voltage of the voltage source $U_\rm s$
• voltage $u_{\rm ind} (t)$ induced by the change of the current

The first one is not given in the exercise, and therefore not considered here.

The induced voltage can be calculated by linearizing the following: \begin{align*} u_{\rm ind} (t) &= - L {{ {\rm d} i}\over{ {\rm d} t }}\\ \rightarrow u_{\rm ind} (t) &= - L {{ \Delta i}\over{ \Delta t }}\\ \end{align*}

With the given details: \begin{align*} u_{\rm ind} (t) &= - L {{ 0 - I}\over{ t_1 - t_0 }}\\ &= 50 \cdot 10^{-6} {~\rm H} {{ 63 ~\rm A}\over{ 1 \cdot 10^{-6} ~\rm s}}\\ &= 3150 {{ ~\rm Vs}\over{ ~\rm A}} \; {{ ~\rm A}\over{ ~\rm s}}\\ \end{align*}

Result

$u_{\rm ind} (t) = 3150 ~\rm V$

Embedded resources