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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 22:56] 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 768: | 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=" |