Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_2:the_time-dependent_magnetic_field [2024/04/29 21:03] – [4 Time-dependent magnetic Field] mexleadmin | electrical_engineering_2:the_time-dependent_magnetic_field [2024/07/03 10:11] (aktuell) – mexleadmin | ||
|---|---|---|---|
| Zeile 48: | Zeile 48: | ||
| Therefore, the induced potential difference generated by a conductor or coil moving in a magnetic field is | Therefore, the induced potential difference generated by a conductor or coil moving in a magnetic field is | ||
| - | \begin{align*} \boxed{ | + | \begin{align*} \boxed{ |
| The negative sign describes the direction in which the induced potential difference drives current around a circuit. However, that direction is most easily determined with a rule known as Lenz’s law, which we will discuss in the next subchapter. | The negative sign describes the direction in which the induced potential difference drives current around a circuit. However, that direction is most easily determined with a rule known as Lenz’s law, which we will discuss in the next subchapter. | ||
| 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 91: | Zeile 91: | ||
| \begin{align*} \Phi_{\rm m} = B \cdot A \end{align*} | \begin{align*} \Phi_{\rm m} = B \cdot A \end{align*} | ||
| - | We can calculate the magnitude of the potential difference $|U_{\rm ind}|$ from Faraday’s law: | + | We can calculate the magnitude of the potential difference $|u_{\rm ind}|$ from Faraday’s law: |
| \begin{align*} | \begin{align*} | ||
| - | |U_{\rm ind}| &= |- {{\rm d}\over{{\rm d}t}}(N \cdot \Phi_{\rm m})| \\ | + | |u_{\rm ind}| &= |- {{\rm d}\over{{\rm d}t}}(N \cdot \Phi_{\rm m})| \\ |
| &= |-N \cdot {{\rm d}\over{{\rm d}t}}(B \cdot A) | \\ | &= |-N \cdot {{\rm d}\over{{\rm d}t}}(B \cdot A) | \\ | ||
| &= |-N \cdot l^2 \cdot {{{\rm d}B}\over{{\rm d}t}}| \\ | &= |-N \cdot l^2 \cdot {{{\rm d}B}\over{{\rm d}t}}| \\ | ||
| Zeile 104: | Zeile 104: | ||
| \begin{align*} | \begin{align*} | ||
| - | |I| &= {{ |U_{\rm ind}|}\over{R}} \\ | + | |I| &= {{ |u_{\rm ind}|}\over{R}} \\ |
| &= {{0.50 ~\rm V}\over{5.0 ~\Omega}} = 0.10 ~\rm A \\ | &= {{0.50 ~\rm V}\over{5.0 ~\Omega}} = 0.10 ~\rm A \\ | ||
| \end{align*} | \end{align*} | ||
| 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 199: | Zeile 199: | ||
| \begin{align*} | \begin{align*} | ||
| - | U_{\rm ind} & | + | u_{\rm ind} & |
| &= - \int^0_1 \vec{v} \times \vec{B} \cdot {\rm d} \vec{s} \\ | &= - \int^0_1 \vec{v} \times \vec{B} \cdot {\rm d} \vec{s} \\ | ||
| \end{align*} | \end{align*} | ||
| Zeile 205: | Zeile 205: | ||
| For constant $|\vec{v}|$ and $|\vec{B}|$ this leads to: | For constant $|\vec{v}|$ and $|\vec{B}|$ this leads to: | ||
| \begin{align*} | \begin{align*} | ||
| - | U_{\rm ind} &= - v \cdot B \cdot l \\ | + | u_{\rm ind} &= - v \cdot B \cdot l \\ |
| \end{align*} | \end{align*} | ||
| Zeile 216: | Zeile 216: | ||
| The velocity of the rod is $v={\rm d}x/{\rm d}t$. So the induced potential difference will get | The velocity of the rod is $v={\rm d}x/{\rm d}t$. So the induced potential difference will get | ||
| \begin{align*} | \begin{align*} | ||
| - | U_{\rm ind} &= - v \cdot B \cdot l \\ | + | u_{\rm ind} &= - v \cdot B \cdot l \\ |
| &= - {{{\rm d}x}\over{{\rm d}t}} \cdot B \cdot l \\ | &= - {{{\rm d}x}\over{{\rm d}t}} \cdot B \cdot l \\ | ||
| &= - {{B \cdot l \cdot {\rm d}x}\over{{\rm d}t}} \\ | &= - {{B \cdot l \cdot {\rm d}x}\over{{\rm d}t}} \\ | ||
| Zeile 225: | Zeile 225: | ||
| This is an alternative way to deduce Faraday' | This is an alternative way to deduce Faraday' | ||
| - | The current $I_{\rm ind}$ induced in the given circuit is $U_{\rm ind}$ divided by the resistance $R$ | + | The current $i_{\rm ind}$ induced in the given circuit is $u_{\rm ind}$ divided by the resistance $R$ |
| \begin{align*} | \begin{align*} | ||
| - | I_{\rm ind} = {{v \cdot B \cdot l }\over{R}} | + | i_{\rm ind} = {{v \cdot B \cdot l }\over{R}} |
| \end{align*} | \end{align*} | ||
| Zeile 236: | Zeile 236: | ||
| \begin{align*} | \begin{align*} | ||
| - | U_{\rm ind} &= - {{\rm d}\over{{\rm dt}}} \cdot \Phi_{\rm m} \\ | + | u_{\rm ind} &= - {{\rm d}\over{{\rm dt}}} \cdot \Phi_{\rm m} \\ |
| &= - B \cdot l \cdot {{{\rm d}x}\over{{\rm d}t}} \\ | &= - B \cdot l \cdot {{{\rm d}x}\over{{\rm d}t}} \\ | ||
| &= - B \cdot l \cdot v \\ | &= - B \cdot l \cdot v \\ | ||
| Zeile 251: | Zeile 251: | ||
| <button size=" | <button size=" | ||
| - | This is a great example of using the equation motional $U_{\rm ind} = - B \cdot l \cdot v$ | + | This is a great example of using the equation motional $u_{\rm ind} = - B \cdot l \cdot v$ |
| - | Entering the given values into $U_{\rm ind} = - B \cdot l \cdot v$ gives | + | Entering the given values into $u_{\rm ind} = - B \cdot l \cdot v$ gives |
| \begin{align*} | \begin{align*} | ||
| - | U_{ind} &= - {{{\rm d} \Phi_{\rm m}}\over{{\rm d}t}} \\ | + | u_{ind} &= - {{{\rm d} \Phi_{\rm m}}\over{{\rm d}t}} \\ |
| &= - B \cdot l \cdot v \\ | &= - B \cdot l \cdot v \\ | ||
| &= - (5.00 \cdot 10^{-5}~\rm T)(20.0 \cdot 10^{3} ~\rm m)(7.80\cdot 10^{3} ~\rm m/s) \\ | &= - (5.00 \cdot 10^{-5}~\rm T)(20.0 \cdot 10^{3} ~\rm m)(7.80\cdot 10^{3} ~\rm m/s) \\ | ||
| Zeile 264: | Zeile 264: | ||
| <button size=" | <button size=" | ||
| - | \begin{align*} | + | \begin{align*} |
| </ | </ | ||
| Zeile 296: | Zeile 296: | ||
| \begin{align*} | \begin{align*} | ||
| - | U_{\rm ind} &= |{{\rm d}\over{{\rm d}t}} \cdot \Phi_{\rm m} | \\ | + | u_{\rm ind} &= |{{\rm d}\over{{\rm d}t}} \cdot \Phi_{\rm m} | \\ |
| &= B\cdot {{r^2\omega}\over{2}} | &= B\cdot {{r^2\omega}\over{2}} | ||
| \end{align*} | \end{align*} | ||
| Zeile 303: | Zeile 303: | ||
| \begin{align*} | \begin{align*} | ||
| - | I_{\rm ind} &= {{|U_{\rm ind}|}\over{R}} \\ | + | i_{\rm ind} &= {{|u_{\rm ind}|}\over{R}} \\ |
| &= B\cdot {{r^2\omega}\over{2R}} | &= B\cdot {{r^2\omega}\over{2R}} | ||
| \end{align*} | \end{align*} | ||
| Zeile 316: | Zeile 316: | ||
| The coil is rotated about the $z$-axis through its center at a constant angular velocity $\omega$. | The coil is rotated about the $z$-axis through its center at a constant angular velocity $\omega$. | ||
| - | Obtain an expression for the induced potential difference $U_{\rm ind}$ in the coil. | + | Obtain an expression for the induced potential difference $u_{\rm ind}$ in the coil. |
| < | < | ||
| Zeile 322: | Zeile 322: | ||
| <button size=" | <button size=" | ||
| - | According to the diagram, the angle between the surface vector $\vec{A}$ and the magnetic field $\vec{B}$ is $\varphi$. The dot product of $\vec{A} \cdot \vec{B}$ simplifies to only the $\cos \varphi$ component of the magnetic field times the area, namely where the magnetic field projects onto the unit surface vector $\vec{A}$. The magnitude of the magnetic field and the area of the loop are fixed over time, which makes the integration | + | According to the diagram, the angle between the surface vector $\vec{A}$ and the magnetic field $\vec{B}$ is $\varphi$. The dot product of $\vec{A} \cdot \vec{B}$ simplifies to only the $\cos \varphi$ component of the magnetic field times the area, namely where the magnetic field projects onto the unit surface vector $\vec{A}$. The magnitude of the magnetic field and the area of the loop are fixed over time, which makes the integration |
| <button size=" | <button size=" | ||
| 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=" | ||