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Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
electrical_engineering_2:the_time-dependent_magnetic_field [2024/04/29 21:13] mexleadmin [Bearbeiten - Panel] |
electrical_engineering_2:the_time-dependent_magnetic_field [2024/05/07 03:52] (aktuell) mexleadmin [Rod in Circuit] |
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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 604: | 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 680: | 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 689: | 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=" |