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 [2023/03/17 09:35] – mexleadmin | electrical_engineering_2:the_time-dependent_magnetic_field [2024/07/03 10:11] (aktuell) – mexleadmin | ||
|---|---|---|---|
| Zeile 1: | Zeile 1: | ||
| - | ====== 4. time-dependent magnetic Field ====== | + | ====== 4 Time-dependent magnetic Field ====== |
| < | < | ||
| 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 44: | Zeile 44: | ||
| This definition leads to a magnetic flux similar to the electric flux studied earlier: | This definition leads to a magnetic flux similar to the electric flux studied earlier: | ||
| - | \begin{align*} \Phi_m = \iint_A \vec{B} \cdot {\rm d} \vec{A} \end{align*} | + | \begin{align*} \Phi_{\rm m} = \iint_A \vec{B} \cdot {\rm d} \vec{A} \end{align*} |
| 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 54: | Zeile 54: | ||
| <imgref ImgNr05> depicts a circuit and an arbitrary surface $S$ that it bounds. Notice that $S$ is an open surface: The planar area bounded by the circuit is not part of the surface, so it is not fully enclosing a volume. | <imgref ImgNr05> depicts a circuit and an arbitrary surface $S$ that it bounds. Notice that $S$ is an open surface: The planar area bounded by the circuit is not part of the surface, so it is not fully enclosing a volume. | ||
| - | Since the magnetic field is a source-free vortex field, the flux over a closed area is always zero: $\Phi_{\rm m} = \iint_{A} \vec{B} \cdot {\rm d} \vec{A} = 0$. \\ | + | Since the magnetic field is a source-free vortex field, the flux over a closed area is always zero: $\Phi_{\rm m} = {\rlap{\Large \rlap{\int} \int} \, \LARGE \circ}_{A} \vec{B} \cdot {\rm d} \vec{A} = 0$. \\ |
| By this, it can be shown that any open surface bounded by the circuit in question can be used to evaluate $\Phi_{\rm m}$ (similar to the [[: | By this, it can be shown that any open surface bounded by the circuit in question can be used to evaluate $\Phi_{\rm m}$ (similar to the [[: | ||
| For example, $\Phi_{\rm m}$ is the same for the various surfaces $S$, $S_1$, $S_2$ of the figure. | For example, $\Phi_{\rm m}$ is the same for the various surfaces $S$, $S_1$, $S_2$ of the figure. | ||
| 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 its magnetic flux is increasing or decreasing. | + | - Determine whether |
| - | - 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*} | ||
| Furthermore, | Furthermore, | ||
| - | The situation of the single rod can be interpreted in the following way: We can calculate a motionally induced potential difference with Faraday’s law even when an actual closed circuit is not present. We simply imagine an enclosed area whose boundary includes the moving conductor, calculate $\Phi_m$, and then find the potential difference from Faraday’s law. For example, we can let the moving rod of <imgref ImgNr10> be one side of the imaginary rectangular area represented by the dashed lines. The area of the rectangle is $A = l \cdot x$, so the magnetic flux through it is $\Phi= B\cdot l \cdot x$. Differentiating this equation, we obtain | + | The situation of the single rod can be interpreted in the following way: We can calculate a motionally induced potential difference with Faraday’s law even when an actual closed circuit is not present. We simply imagine an enclosed area whose boundary includes the moving conductor, calculate $\Phi_{\rm m}$, and then find the potential difference from Faraday’s law. For example, we can let the moving rod of <imgref ImgNr10> be one side of the imaginary rectangular area represented by the dashed lines. The area of the rectangle is $A = l \cdot x$, so the magnetic flux through it is $\Phi= B\cdot l \cdot x$. Differentiating this equation, we obtain |
| \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 343: | Zeile 343: | ||
| This changes the function to time-space rather than $\varphi$. The induced potential difference, therefore, varies sinusoidally with time according to | This changes the function to time-space rather than $\varphi$. The induced potential difference, therefore, varies sinusoidally with time according to | ||
| - | \begin{align*} u_{ind} &= U_{ind,0} \cdot \sin \omega t \end{align*} | + | \begin{align*} u_{\rm ind} &= U_{\rm ind,0} \cdot \sin \omega t \end{align*} |
| where $U_{\rm ind,0} = NBA\omega$. </ | where $U_{\rm ind,0} = NBA\omega$. </ | ||
| Zeile 351: | Zeile 351: | ||
| <panel type=" | <panel type=" | ||
| - | The generator coil shown in <imgref ImgNr13> is rotated through one-fourth of a revolution (from $\phi_0=0°$ to $\phi_1=90°$) in $5.0 ~\rm ms$. | + | The generator coil shown in <imgref ImgNr13> is rotated through one-fourth of a revolution (from $\varphi_0=0°$ to $\varphi_1=90°$) in $5.0 ~\rm ms$. |
| The $200$-turn circular coil has a $5.00 ~\rm cm$ radius and is in a uniform $0.80 ~\rm T$ magnetic field. | The $200$-turn circular coil has a $5.00 ~\rm cm$ radius and is in a uniform $0.80 ~\rm T$ magnetic field. | ||
| 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 667: | Zeile 695: | ||
| So, the course of the voltage when entering or exiting is not uniquely given. | So, the course of the voltage when entering or exiting is not uniquely given. | ||
| - | < | + | < |
| 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=" | ||