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electrical_engineering_2:the_time-dependent_magnetic_field [2024/04/29 23:31]
mexleadmin [4.5 Inductance] 		electrical_engineering_2:the_time-dependent_magnetic_field [2024/05/07 03:52] (aktuell)
mexleadmin [Rod in Circuit]
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 <WRAP> <imgcaption ImgNr05 | A circuit bounding an arbitrary open surface S. > </imgcaption> {{drawio>FluxThroughSurfaces.svg}} </WRAP> <WRAP> <imgcaption ImgNr05 | A circuit bounding an arbitrary open surface S. > </imgcaption> {{drawio>FluxThroughSurfaces.svg}} </WRAP>
  
-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 per square meter ($\rm Wb/m^2$) instead of teslas, based on this definition. +Based on this definition, the magnetic field unit is occasionally expressed as Weber per square meter ($\rm Wb/m^2$) instead of teslas. 
 In many practical applications, the circuit of interest consists of a number $N$ of tightly wound turns (similar to <imgref ImgNr06>).  In many practical applications, the circuit of interest consists of a number $N$ of tightly wound turns (similar to <imgref ImgNr06>). 
 Each turn experiences the same magnetic flux $\Phi_{\rm m}$.  Each turn experiences the same magnetic flux $\Phi_{\rm m}$. 
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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 for use in visualizing and recording directions.+  - Make a sketch of the situation to visualize and record the directions of fields, movements etc. .
   - 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 a magnetic flux that is decreasing or opposes a magnetic flux that is increasing. Therefore, the induced magnetic field adds or subtracts from the applied magnetic field, depending on the change in magnetic flux.+  - Now determine the direction of the induced magnetic field $\vec{B_{\rm ind}}$. \\ The induced magnetic field attempts to withstand its cause: It tries to strengthen a decreasing magnetic flux (or to counteract an increasing magnetic flux). Therefore, the induced magnetic field adds or subtracts from the applied magnetic field, depending on the change in magnetic flux.
   - 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} &= 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*} 
 </collapse> </collapse>
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 <button size="xs" type="link" collapse="Solution_4_3_4_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_4_1_Solution" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_3_4_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_4_1_Solution" collapsed="true">
  
-We are given that $N=200$, $B=0.80~\rm T$ , $\varphi = 90°$ , $d\varphi=90°=\pi/2$ , and $\Delta t=5.0~\rm ms$ .+We are given that $N=200$, $B=0.80~\rm T$ , $\varphi = 90°$ , $\Delta\varphi=90°=\pi/2$ , and $\Delta t=5.0~\rm ms$ .
  
 The area of the loop is The area of the loop is
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 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 = {{Vs}\over{A}} = {{Wb}\over{A}} $+The unit of the inductance is $\rm 1 ~Henry = 1 ~H = 1 {{Vs}\over{A}} = 1{{Wb}\over{A}} $
  
 <WRAP> <imgcaption BildNr5 | Example of a Circuit with an Inductor> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCBMCmC0CcIAsA6ADAdgBwFYMYGYcd4CyCsQc0rlIq4wwAoANyiRoLSQ5qfo0anWkJBoUOZgBsQZAGzh5XAosiReQ9GghDmAJzmqoGo4u6bwaNMwIYVigWaX9x45gHc+cns4vuADyUCKDRKMAw1JmQOEABXZiCIykgwRywUsEReMEobIKRUqAxeJDJijBjckAAlROQ0NXlFJCyoeWzYhIL1UJCy+khGqsoAGXrOSqycOn54Sl5TbuQwGbAy2fAkFtiAS09vJwVwMEF6oZpUlKGTCFKaQGTCers1NBmCeFeFhpAnoLtFh05PBFhh6PdfswAPbgEDySwBLBgHTQEBYLS6KBYsSXdwwkLwtyI5FgAAmonQayxgjktIkaDOQA noborder}} </WRAP> <WRAP> <imgcaption BildNr5 | Example of a Circuit with an Inductor> </imgcaption> \\ {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=CQAgjCBMCmC0CcIAsA6ADAdgBwFYMYGYcd4CyCsQc0rlIq4wwAoANyiRoLSQ5qfo0anWkJBoUOZgBsQZAGzh5XAosiReQ9GghDmAJzmqoGo4u6bwaNMwIYVigWaX9x45gHc+cns4vuADyUCKDRKMAw1JmQOEABXZiCIykgwRywUsEReMEobIKRUqAxeJDJijBjckAAlROQ0NXlFJCyoeWzYhIL1UJCy+khGqsoAGXrOSqycOn54Sl5TbuQwGbAy2fAkFtiAS09vJwVwMEF6oZpUlKGTCFKaQGTCers1NBmCeFeFhpAnoLtFh05PBFhh6PdfswAPbgEDySwBLBgHTQEBYLS6KBYsSXdwwkLwtyI5FgAAmonQayxgjktIkaDOQA noborder}} </WRAP>