====== Block 20 — Electromagnetic Induction and Energy ====== ===== Learning objectives ===== After this 90-minute block, you can * ... ===== 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 ===== - Warm-up (x min): - .... - Core concepts & derivations (x min): - ... - Practice (x min): ... - 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 )? {{drawio>InductionPhenomenons.svg}} To understand this, we will investigate the situation for a long coil (). {{drawio>SelfInductionCoil.svg}} 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 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}} $ \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCBMCmC0CcIAsA6ADAdgBwFYMYGYcd4CyCsQc0rlIq4wwAoANyiRoLSQ5qfo0anWkJBoUOZgBsQZAGzh5XAosiReQ9GghDmAJzmqoGo4u6bwaNMwIYVigWaX9x45gHc+cns4vuADyUCKDRKMAw1JmQOEABXZiCIykgwRywUsEReMEobIKRUqAxeJDJijBjckAAlROQ0NXlFJCyoeWzYhIL1UJCy+khGqsoAGXrOSqycOn54Sl5TbuQwGbAy2fAkFtiAS09vJwVwMEF6oZpUlKGTCFKaQGTCers1NBmCeFeFhpAnoLtFh05PBFhh6PdfswAPbgEDySwBLBgHTQEBYLS6KBYsSXdwwkLwtyI5FgAAmonQayxgjktIkaDOQA noborder}} 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 [[:electrical_engineering_2:the_magnetostatic_field#magnetic_field_strength_part_1toroidal_coil|magnetic Field Strength Part 1: Toroidal Coil]]). Here, a rectangular intersection a assumed (see ). {{drawio>SelfInductionToroCoil.svg}} 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*} The inductor behaves in the parallel and series circuit similar to the resistor. ===== Common pitfalls ===== * ... ===== Exercises ===== {{page>electrical_engineering_and_electronics:task_ljxf80q7vxywehqf_with_calculation&nofooter}} {{page>electrical_engineering_and_electronics:task_unkkahm3u0v9azny_with_calculation&nofooter}} {{fa>pencil?32}} Calculate the inductance for the following settings 1. Cylindrical long air coil with $N=390$, winding diameter $d=3.0 ~\rm cm$ and length $l=18 ~\rm cm$ #@HiddenBegin_HTML~4511S,Solution~@# \begin{align*} L_1 &= \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A }\over {l}} \\ &= 4\pi \cdot 10^{-7} {\rm {{H}\over{m}}} \cdot 1 \cdot (390)^2 \cdot {{\pi \cdot (0.03~\rm m)^2 }\over {0.18 ~\rm m}} \end{align*} #@HiddenEnd_HTML~4511S,Solution ~@# #@HiddenBegin_HTML~4511R,Result~@# \begin{align*} L_1 &= 3.0 ~\rm mH \end{align*} #@HiddenEnd_HTML~4511R,Result~@# 2. Similar coil geometry as explained in 1. , but with double the number of windings #@HiddenBegin_HTML~4512S,Solution~@# \begin{align*} L_2 &= \mu_0 \mu_{\rm r} \cdot N_2^2 \cdot {{A }\over {l}} \\ &= \mu_0 \mu_{\rm r} \cdot (2\cdot N)^2 \cdot {{A }\over {l}} \\ &= \mu_0 \mu_{\rm r} \cdot 4\cdot N^2 \cdot {{A }\over {l}} \\ &= 4\cdot L_1 \\ \end{align*} #@HiddenEnd_HTML~4512S,Solution ~@# #@HiddenBegin_HTML~4512R,Result~@# \begin{align*} L_1 &= 12 ~\rm mH \end{align*} #@HiddenEnd_HTML~4512R,Result~@# 3. Two coils as explained in 1. in series #@HiddenBegin_HTML~4513S,Solution~@# multiple inductances in series just add up. One can think of adding more windings to the first coil in the formula.. #@HiddenEnd_HTML~4513S,Solution ~@# #@HiddenBegin_HTML~4513R,Result~@# \begin{align*} L_1 &= 6.0 ~\rm mH \end{align*} #@HiddenEnd_HTML~4513R,Result~@# 4. Similar coil geometry and number of windings as explained in 1. , but with an iron core ($\mu_{\rm r}=1000$) #@HiddenBegin_HTML~4514S,Solution~@# \begin{align*} L_4 &= \mu_0 \mu_{\rm r,4} \cdot N^2 \cdot {{A }\over {l}} \\ &= \mu_0 \cdot 1000 \cdot N^2 \cdot {{A }\over {l}} \\ &= 1000 \cdot L_4 \\ \end{align*} #@HiddenEnd_HTML~4514S,Solution ~@# #@HiddenBegin_HTML~4514R,Result~@# \begin{align*} L_4 &= 3.0 ~\rm H \end{align*} #@HiddenEnd_HTML~4514R,Result~@# {{fa>pencil?32}} A cylindrical air coil (length $l=40 ~\rm cm$, diameter $d=5.0 ~\rm cm$, and a number of windings $N=300$) passes a current of $30 ~\rm A$. The current shall be reduced linearly in $2.0 ~\rm ms$ down to $0.0 ~\rm A$. What is the amount of the induced voltage $u_{\rm ind}$? #@HiddenBegin_HTML~4521S,Solution~@# The requested induced voltage can be derived by: \begin{align*} L &= \left|{{u_{\rm ind}}\over{{\rm d}i / {\rm d}t}}\right| \\ \rightarrow \left|u_{\rm ind}\right| &= L \cdot \left|{{{\rm d}i}\over{{\rm d}t}}\right| \\ &= L \cdot \left|{{\Delta i}\over{\Delta t}}\right| \\ \end{align*} Therefore, we just need the inductance $L$, since ${{\Delta i}\over{\Delta t}}$ is defined as $30 ~\rm A$ per $2 ~\rm ms$: \begin{align*} L &= \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A }\over {l}} \\ \end{align*} So, the result can be derived as: \begin{align*} \left|u_{\rm ind}\right| &= \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A }\over {l}} \cdot \left|{{\Delta i}\over{\Delta t}}\right| \\ &= 4\pi \cdot 10^{-7} {\rm {{H}\over{m}}} \cdot 1 \cdot (300)^2 \cdot {{\pi \cdot (0.05~\rm m)^2 }\over {0.40 ~\rm m}} \cdot {{30 ~\rm A}\over{2 ~\rm ms}} \end{align*} #@HiddenEnd_HTML~4521S,Solution ~@# #@HiddenBegin_HTML~1,Result~@# \begin{align*} \left|u_{\rm ind}\right| &= 33 ~\rm V\end{align*} #@HiddenEnd_HTML~1,Result~@# {{fa>pencil?32}} A coil with the inductance $L=20 ~\rm µH$ passes a current of $40 ~\rm A$. The current shall be reduced linearly in $5 ~\rm µs$ down to $0 ~\rm A$ (see ). * What is the amount of the induced voltage $u_{\rm ind}$? * Sketch the course of $u_{\rm ind}(t)$! {{drawio>CircuitAndTiming.svg}} \\ ===== Embedded resources ===== Explanation (video): ... ~~PAGEBREAK~~ ~~CLEARFIX~~