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| electrical_engineering_2:the_time-dependent_magnetic_field [2023/04/23 08:56] – ott | electrical_engineering_2:the_time-dependent_magnetic_field [2025/05/06 11:22] (current) – [Bearbeiten - Panel] mexleadmin | ||
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| - | ====== 4. Time-dependent magnetic Field ====== | + | ====== 4 Time-dependent magnetic Field ====== |
| < | < | ||
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| </ | </ | ||
| - | 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 ===== | ||
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| 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. | ||
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| < | < | ||
| - | 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 $\Phi_{\rm m}$. | ||
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| \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}}| \\ | ||
| Line 104: | Line 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*} | ||
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| 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. | ||
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| \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*} | ||
| Line 205: | Line 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*} | ||
| Line 216: | Line 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}} \\ | ||
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| 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*} | ||
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| \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 \\ | ||
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| <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) \\ | ||
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| <button size=" | <button size=" | ||
| - | \begin{align*} | + | \begin{align*} |
| </ | </ | ||
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| \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*} | ||
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| \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*} | ||
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| 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. |
| < | < | ||
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| <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=" | ||
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| 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 $U_{\rm ind,0} = NBA\omega$. </ | ||
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| <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. | ||
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| \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*} | ||
| </ | </ | ||
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| <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 | ||
| Line 503: | Line 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}} $ | ||
| < | < | ||
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| < | < | ||
| - | <button size=" | + | <button size=" |
| For partwise linear $u_{\rm ind}$ one can derive: | For partwise linear $u_{\rm ind}$ one can derive: | ||
| Line 585: | Line 587: | ||
| </ | </ | ||
| - | <button size=" | + | <button size=" |
| + | {{icon> | ||
| + | < | ||
| + | </ | ||
| + | </ | ||
| </ | </ | ||
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| < | < | ||
| + | |||
| + | # | ||
| + | |||
| + | 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> | ||
| + | # | ||
| + | |||
| </ | </ | ||
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| 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. | ||
| - | < | + | < |
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| 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 \cdot 1000 \cdot N^2 \cdot {{A }\over {l}} \\ | ||
| + | &= 1000 \cdot L_4 \\ | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_4 &= 3.0 ~\rm H | ||
| + | \end{align*} | ||
| + | # | ||
| </ | </ | ||
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| <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=" | ||