Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
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electrical_engineering_2:the_time-dependent_magnetic_field [2024/04/29 20:45] mexleadmin [4 Time-dependent magnetic Field] |
electrical_engineering_2:the_time-dependent_magnetic_field [2024/05/07 03:52] (aktuell) mexleadmin [Rod in Circuit] |
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| Zeile 16: | Zeile 16: | ||
| </ | </ | ||
| - | We have been considering electric fields created by fixed charge distributions and magnetic fields produced by constant currents, but electromagnetic phenomena are not restricted to these stationary situations. Most of the interesting applications of electromagnetism are, in fact, time-dependent. To investigate some of these applications, | + | We have been considering electric fields created by fixed charge distributions and magnetic fields produced by constant currents, but electromagnetic phenomena are not restricted to these stationary situations. Most of the interesting applications of electromagnetism are, in fact, time-dependent. To investigate some of these applications, |
| Lastly, we describe applications of these principles, such as the card reader shown in <imgref ImgNr01> | Lastly, we describe applications of these principles, such as the card reader shown in <imgref ImgNr01> | ||
| - | < | + | < |
| ===== 4.1 Recap of magnetic Field ===== | ===== 4.1 Recap of magnetic Field ===== | ||
| Zeile 60: | Zeile 60: | ||
| < | < | ||
| - | The SI unit for magnetic flux is the Weber (Wb), \begin{align*} [\Phi_{\rm m}] = [B] \cdot [A] = 1 ~\rm T \cdot m^2 = 1 ~ Wb \end{align*} | + | The SI unit for magnetic flux is the $\rm Weber$ (Wb), \begin{align*} [\Phi_{\rm m}] = [B] \cdot [A] = 1 ~\rm T \cdot m^2 = 1 ~ Wb \end{align*} |
| - | Occasionally, the magnetic field unit is expressed as webers | + | Based on this definition, the magnetic field unit is occasionally |
| In many practical applications, | In many practical applications, | ||
| Each turn experiences the same magnetic flux $\Phi_{\rm m}$. | Each turn experiences the same magnetic flux $\Phi_{\rm m}$. | ||
| Zeile 139: | Zeile 139: | ||
| To use Lenz’s law to determine the directions of induced potential difference, currents, and magnetic fields: | To use Lenz’s law to determine the directions of induced potential difference, currents, and magnetic fields: | ||
| - | - Make a sketch of the situation | + | - Make a sketch of the situation |
| - Determine the direction of the applied magnetic field $\vec{B}$. | - Determine the direction of the applied magnetic field $\vec{B}$. | ||
| - Determine whether the magnitude of its magnetic flux is increasing or decreasing. | - Determine whether the magnitude of its magnetic flux is increasing or decreasing. | ||
| - | - Now determine the direction of the induced magnetic field $\vec{B_{\rm ind}}$. The induced magnetic field tries to reinforce | + | - Now determine the direction of the induced magnetic field $\vec{B_{\rm ind}}$. |
| - Use the right-hand rule to determine the direction of the induced current $i_{\rm ind}$ that is responsible for the induced magnetic field $\vec{B}_{\rm ind}$. | - Use the right-hand rule to determine the direction of the induced current $i_{\rm ind}$ that is responsible for the induced magnetic field $\vec{B}_{\rm ind}$. | ||
| - The direction (or polarity) of the induced potential difference can now drive a conventional current in this direction. | - The direction (or polarity) of the induced potential difference can now drive a conventional current in this direction. | ||
| Zeile 371: | Zeile 371: | ||
| \begin{align*} | \begin{align*} | ||
| - | u_{\rm ind} &= N B \cdot \sin \varphi \cdot {{{\rm d} \varphi}\over{{\rm d}t}} | + | u_{\rm ind} &= N B A \cdot \sin \varphi \cdot {{{\rm d} \varphi}\over{{\rm d}t}} |
| \end{align*} | \end{align*} | ||
| </ | </ | ||
| Zeile 377: | Zeile 377: | ||
| <button size=" | <button size=" | ||
| - | We are given that $N=200$, $B=0.80~\rm T$ , $\varphi = 90°$ , $d\varphi=90°=\pi/ | + | We are given that $N=200$, $B=0.80~\rm T$ , $\varphi = 90°$ , $\Delta\varphi=90°=\pi/ |
| The area of the loop is | The area of the loop is | ||
| Zeile 503: | Zeile 503: | ||
| The <imgref BildNr5> shows an inductor in series with a resistor and a switch (any real switch also behaves as a capacitor, when open). | The <imgref BildNr5> 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. | 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}} $ | ||
| < | < | ||
| Zeile 569: | Zeile 571: | ||
| < | < | ||
| - | <button size=" | + | <button size=" |
| For partwise linear $u_{\rm ind}$ one can derive: | For partwise linear $u_{\rm ind}$ one can derive: | ||
| Zeile 585: | Zeile 587: | ||
| </ | </ | ||
| - | <button size=" | + | <button size=" |
| + | {{icon> | ||
| + | < | ||
| + | </ | ||
| + | </ | ||
| </ | </ | ||
| Zeile 600: | Zeile 606: | ||
| < | < | ||
| + | |||
| + | # | ||
| + | |||
| + | For partwise linear $u_{\rm ind}$ one can derive: | ||
| + | \begin{align*} | ||
| + | u_{\rm ind} &= -{{{\rm d}\Phi}\over{{\rm d}t}} \\ | ||
| + | \rightarrow | ||
| + | \Phi & | ||
| + | \end{align*} | ||
| + | |||
| + | For diagram (a): | ||
| + | |||
| + | * $t= 0.00 ... 0.04 ~\rm s\quad$: $\quad \Phi = \Phi_0 - {0 \cdot \; \Delta t} \quad\quad\quad\quad\quad\quad\quad= 0 ~\rm Wb$ | ||
| + | * $t= 0.04 ... 0.10 ~\rm s\quad$: $\quad \Phi = 0 {~\rm Wb} - {{30 ~\rm mV} \cdot \; (t - 0.04 ~\rm s)} = \quad {1.2 ~\rm mWb} - t \cdot 30 ~\rm mV$ | ||
| + | * $t= 0.10 ... 0.14 ~\rm s\quad$: $\quad \Phi = {1.2 ~\rm mWb} - {0.10 ~\rm s} \cdot 30 ~\rm mV \quad = - {1.8 ~\rm mWb}$ | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | {{drawio> | ||
| + | # | ||
| + | |||
| </ | </ | ||
| Zeile 676: | Zeile 704: | ||
| Calculate the inductance for the following settings | Calculate the inductance for the following settings | ||
| - | - cylindrical | + | 1. Cylindrical long air coil with $N=390$, winding diameter $d=3.0 ~\rm cm$ and length $l=18 ~\rm cm$ |
| - | - similar | + | # |
| - | - two coils as explained in 1. in series | + | |
| - | - similar | + | \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*} | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_1 &= 3.0 ~\rm mH | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | 2. Similar | ||
| + | |||
| + | # | ||
| + | \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*} | ||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_1 &= 12 ~\rm mH | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | 3. Two coils as explained in 1. in series | ||
| + | |||
| + | # | ||
| + | multiple inductances in series just add up. One can think of adding more windings to the first coil in the formula.. | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_1 &= 6.0 ~\rm mH | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | 4. Similar | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_4 &= \mu_0 \mu_{\rm r,4} \cdot N^2 \cdot {{A }\over {l}} \\ | ||
| + | &= \mu_0 \codt 1000 \cdot N^2 \cdot {{A }\over {l}} \\ | ||
| + | &= 1000 \cdot L_4 \\ | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_4 &= 3.0 ~\rm H | ||
| + | \end{align*} | ||
| + | # | ||
| </ | </ | ||
| Zeile 685: | Zeile 770: | ||
| <panel type=" | <panel type=" | ||
| - | A cylindrical air coil (length $l=40 ~\rm cm$, diameter $d=5 ~\rm cm$, and a number of windings $N=300$) passes a current of $30 ~\rm A$. The current shall be reduced linearly in $2 ~\rm ms$ down to $0 ~\rm A$. | + | 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 $u_{\rm ind}$? </ | + | What is the amount of the induced voltage $u_{\rm ind}$? |
| + | |||
| + | # | ||
| + | |||
| + | The requested induced voltage can be derived by: | ||
| + | |||
| + | \begin{align*} | ||
| + | L &= \left|{{u_{\rm ind}}\over{{\rm d}i / {\rm d}t}}\right| \\ | ||
| + | \rightarrow | ||
| + | & | ||
| + | \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| \\ | ||
| + | & | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | \left|u_{\rm ind}\right| &= 33 ~\rm V\end{align*} | ||
| + | # | ||
| + | |||
| + | |||
| + | </ | ||
| <panel type=" | <panel type=" | ||