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electrical_engineering_2:the_time-dependent_magnetic_field [2022/07/11 12:49]
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electrical_engineering_2:the_time-dependent_magnetic_field [2023/09/19 23:51] (aktuell)
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-====== 4. time-dependent magnetic Field ======+====== 4 Time-dependent magnetic Field ======
  
-<callout> This chapter is based on the Book 'University Physics II' ([[https://creativecommons.org/licenses/by/4.0|CC BY 4.0]], Authors: [[https://openstax.org/details/books/university-physics-volume-2|Open Stax]] ). In detail this is chapter [[https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)|11. Magnetic Forces and Fields]] </callout>+<callout> This chapter is based on the book 'University Physics II' ([[https://creativecommons.org/licenses/by/4.0|CC BY 4.0]], Authors: [[https://openstax.org/details/books/university-physics-volume-2|Open Stax]] ). In detail this is chapter [[https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_University_Physics_(OpenStax)/Book%3A_University_Physics_II_-_Thermodynamics_Electricity_and_Magnetism_(OpenStax)|11. Magnetic Forces and Fields]] </callout>
  
 <callout> <callout>
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   - use Faraday’s law to determine the magnitude of induced potential difference in a closed loop due to changing magnetic flux through the loop   - use Faraday’s law to determine the magnitude of induced potential difference in a closed loop due to changing magnetic flux through the loop
   - use Lenz’s law to determine the direction of induced potential difference whenever a magnetic flux changes   - use Lenz’s law to determine the direction of induced potential difference whenever a magnetic flux changes
-  - use Faraday’s law with Lenz’s law to determine the induced potential difference in a coil and in a solenoid+  - use Faraday’s law with Lenz’s law to determine the induced potential difference in a coil and a solenoid
  
 </callout> </callout>
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 ===== 4.1 Recap of magnetic Field ===== ===== 4.1 Recap of magnetic Field =====
  
-The first productive experiments concerning the effects of time-varying magnetic fields were performed by Michael Faraday in 1831. One of his early experiments is represented in the simulation in <imgref ImgNr02> - in the tab ''Pickup Coil''An potential difference is induced when the magnetic field in the coil is changed by pushing a bar magnet into or out of the coil. This potential difference can generate a currentwhen the circuit is closed. Potential differences of opposite signs are produced by motion in opposite directions, and the directions of potential differences are also reversed by reversing poles. The same results are produced if the coil is moved rather than the magnet — it is the relative motion that is important. The faster the motion, the greater the potential difference, and there is no potential difference when the magnet is stationary relative to the coil.+The first productive experiments concerning the effects of time-varying magnetic fields were performed by Michael Faraday in 1831. One of his early experiments is represented in the simulation in <imgref ImgNr02> - in the tab ''Pickup Coil''potential difference is induced when the magnetic field in the coil is changed by pushing a bar magnet into or out of the coil. This potential difference can generate a current when the circuit is closed. Potential differences of opposite signs are produced by motion in opposite directions, and the directions of potential differences are also reversed by reversing poles. The same results are produced if the coil is moved rather than the magnet — it is the relative motion that is important. The faster the motion, the greater the potential difference, and there is no potential difference when the magnet is stationary relative to the coil.
  
 <WRAP> <imgcaption ImgNr02 | Movement of a magnet relative to a coil produces a potential difference.> </imgcaption> <WRAP> <imgcaption ImgNr02 | Movement of a magnet relative to a coil produces a potential difference.> </imgcaption>
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 {{url>https://phet.colorado.edu/sims/cheerpj/faraday/latest/faraday.html?simulation=faraday 700,600 noborder}} </WRAP> {{url>https://phet.colorado.edu/sims/cheerpj/faraday/latest/faraday.html?simulation=faraday 700,600 noborder}} </WRAP>
  
-Faraday also discovered that a similar effect can be produced using two circuits: a changing current in one circuit induces a current in a second, nearby circuit. An example for this can be shown in the simulation in the tab ''Transformer''. When the source is changed into AC and the coils are moved nearer together, the light bulb of the second circuit momentarily lights up, indicating that a short-lived current surge has been induced in that circuit.+Faraday also discovered that a similar effect can be produced using two circuits: a changing current in one circuit induces a current in a second, nearby circuit. An example of this can be shown in the simulation in the tab ''Transformer''. When the source is changed into AC and the coils are moved nearer together, the light bulb of the second circuit momentarily lights up, indicating that a short-lived current surge has been induced in that circuit.
  
-Faraday realized that in both experiments, a current flowed in the "receiving" circuit when the magnetic field in the region occupied by that circuit was changing. As the magnet was moved, the strength of its magnetic field at the loop changed; and when the current in AC circuit changed periodically, the strength of its magnetic field at circuit 2 changed. Faraday was eventually able to interpret these and all other experiments involving magnetic fields that vary with time in terms of the following law:+Faraday realized that in both experiments, a current flowed in the "receiving" circuit when the magnetic field in the region occupied by that circuit was changing. As the magnet was moved, the strength of its magnetic field at the loop changed; and when the current in the AC circuit changed periodically, the strength of its magnetic field at circuit 2 changed. Faraday was eventually able to interpret these and all other experiments involving magnetic fields that vary with time in terms of the following law:
  
 <callout icon="fa fa-exclamation" color="red" title="Notice:"> **Faraday's Law** <callout icon="fa fa-exclamation" color="red" title="Notice:"> **Faraday's Law**
  
-Any change in the magnetic field or change in orientation of the area of a coil with respect to the magnetic field induces an electric voltage. The induced potential difference is the negative change of the so-called **magnetic flux**  $\Phi_m$ per unit time. </callout>+Any change in the magnetic field or change in the orientation of the area of a coil with respect to the magnetic field induces an electric voltage. The induced potential difference is the negative change of the so-called **magnetic flux**  $\Phi_{\rm m}$ per unit of time. </callout>
  
-The magnetic flux is a measurement of the amount of magnetic field lines through a given surface area, as seen in <imgref ImgNr03>. The magnetic flux is the amount of magnetic field lines cutting through a surface area defined by the surface vector $\vec{A}$. If the angle between the $\vec{A} = A\cdot \vec{n}$ and magnetic field vector $\vec{B}$ is parallel or antiparallel, as shown in the diagram, the abolute value of the magnetic flux is the highest possible value given the values of area and magnetic field.+The magnetic flux is a measurement of the amount of magnetic field lines through a given surface area, as seen in <imgref ImgNr03>. The magnetic flux is the amount of magnetic field lines cutting through a surface area defined by the surface vector $\vec{A}$. If the angle between the $\vec{A} = A\cdot \vec{n}$ and magnetic field vector $\vec{B}$ is parallel or antiparallel, as shown in the diagram, the absolute value of the magnetic flux is the highest possible value given the values of the area and the magnetic field.
  
 <WRAP> <imgcaption ImgNr03 | The magnetic flux is the amount of magnetic field lines cutting through a surface area A.> </imgcaption> {{drawio>MagneticFieldAndFlux.svg}} </WRAP> <WRAP> <imgcaption ImgNr03 | The magnetic flux is the amount of magnetic field lines cutting through a surface area A.> </imgcaption> {{drawio>MagneticFieldAndFlux.svg}} </WRAP>
  
-This definition lead 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 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{ U_{ind} = -{{d \Phi_m}\over{dt}} = -{{d}\over{dt}}\iint_A \vec{B} \cdot d \vec{A} } \end{align*}+\begin{align*} \boxed{ U_{\rm ind} = -{{{\rm d} \Phi_{\rm m}}\over{{\rm d}t}} = -{{\rm d}\over{{\rm d}t}}\iint_A \vec{B} \cdot {\rm d\vec{A} } \end{align*}
  
 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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 <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 alway zero: $\Phi_m = \iint_{A} \vec{B} \cdot d \vec{A} = 0$. \\ By this, it can be shown that any open surface bounded by the circuit in question can be used to evaluate $\Phi_m$ (similar to the [[:electrical_engineering_2:the_stationary_electric_flow#gauss_s_law_for_current_density|Gauss's law for current density]]). For example, $\Phi_m$ is the same for the various surfaces $S$, $S_1$, $S_2$ of the figure.+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 [[:electrical_engineering_2:the_stationary_electric_flow#gauss_s_law_for_current_density|Gauss's law for current density]]).  
 +For example, $\Phi_{\rm m}$ is the same for the various surfaces $S$, $S_1$, $S_2$ of the figure.
  
 <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_m] = [B] \cdot [A] = 1 T \cdot m^2 = 1Wb \end{align*}+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*}
  
-Occasionally, the magnetic field unit is expressed as webers per square meter ($Wb/m^2$) instead of teslas, based on this definition. 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 $\Phi_m$. Therefore, the net magnetic flux through the circuits is $N$ times the flux through one turn, and Faraday’s law is written as+Occasionally, the magnetic field unit is expressed as webers per square meter ($\rm Wb/m^2$) instead of teslas, based on this definition.  
 +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 $\Phi_{\rm m}$.  
 +Therefore, the net magnetic flux through the circuits is $N$ times the flux through one turn, and Faraday’s law is written as
  
-\begin{align*} u_{ind} = -{{d}\over{dt}}(N \cdot \Phi_m) = -N \cdot {{d \Phi_m}\over{dt}} \end{align*}+\begin{align*}  
 +u_{\rm ind} = -        { {\rm d}              \over{{\rm d}t}} (N \cdot \Phi_{\rm m} 
 +            = -N \cdot {{{\rm d\Phi_{\rm m}}\over{{\rm d}t}}  
 +\end{align*}
  
 <panel type="info" title="Exercise 4.1.1 Magnetic Field Strength around a horizontal straight Conductor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.1.1 Magnetic Field Strength around a horizontal straight Conductor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-The square coil of <imgref ImgNr06> has sides $l=0.25m$ long and is tightly wound with $N=200$ turns of wire. The resistance of the coil is $R=5.0\Omega$. The coil is placed in a spatially uniform magnetic field. The field is directed perpendicular to the face of the coil and whose magnitude is decreasing at a rate $dB/dt=−0.040T/s$.+The square coil of <imgref ImgNr06> has sides $l=0.25~\rm m$ long and is tightly wound with $N=200$ turns of wire. The resistance of the coil is $R=5.0~\Omega$. The coil is placed in a spatially uniform magnetic field. The field is directed perpendicular to the face of the coil and whose magnitude is decreasing at a rate ${\rm d}B/{\rm d}t=−0.040~ \rm T/s$.
  
   - What is the magnitude of the potential difference induced in the coil?   - What is the magnitude of the potential difference induced in the coil?
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 <button size="xs" type="link" collapse="Solution_4_1_1_1_Strategy">{{icon>eye}} Strategy</button><collapse id="Solution_4_1_1_1_Strategy" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_1_1_1_Strategy">{{icon>eye}} Strategy</button><collapse id="Solution_4_1_1_1_Strategy" collapsed="true">
  
-The surface $\vec{A}$ is perpendicular to area covering the loop. We will choose this to be pointing downward so that $\vec{B}$ is parallel to $\vec{A}$ and that the flux turns into multiplication of magnetic field times area. The area of the loop is not changing in time, so it can be factored out of the time derivative, leaving the magnetic field as the only quantity varying in time. Lastly, we can apply Ohm’s law once we know the induced potential difference to find the current in the loop. </collapse>+The surface $\vec{A}$ is perpendicular to the area covering the loop. We will choose this to be pointing downward so that $\vec{B}$ is parallel to $\vec{A}$ and that the flux turns into the multiplication of magnetic field times area. The area of the loop is not changing in time, so it can be factored out of the time derivative, leaving the magnetic field as the only quantity varying in time. Lastly, we can apply Ohm’s law once we know the induced potential difference to find the current in the loop. </collapse>
  
 <button size="xs" type="link" collapse="Solution_4_1_1_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_1_1_1_Solution" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_1_1_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_1_1_1_Solution" collapsed="true">
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 The flux through one turn is The flux through one turn is
  
-\begin{align*} \Phi_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 $|U_{ind}|$ from Faraday’s law:+We can calculate the magnitude of the potential difference $|U_{\rm ind}|$ from Faraday’s law:
  
-\begin{align*} |U_{ind}| &= |-{{d}\over{dt}}(N \cdot \Phi_m)| \\ &= -N \cdot {{d}\over{dt}} (B \cdot A) \\ &= -N \cdot l^2 \cdot {{dB}\over{dt}} \\ &= (200)(0.25m)^2(0.040 T/s) \\ &= 0.50 V \end{align*}+\begin{align*}  
 +|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 l^2 \cdot {{{\rm d}B}\over{{\rm d}t}}| \\  
 +              &= (200)(0.25 ~\rm m)^2(0.040 ~T/s) \\  
 +              &= 0.50 ~ 
 +\end{align*}
  
 The magnitude of the current induced in the coil is The magnitude of the current induced in the coil is
  
-\begin{align*} |I| &= {{ |U_{ind}|}\over{R}} \\ &= {{0.50V}\over{5.0\Omega}} = 0.10A \\ \end{align*} </collapse> </WRAP></WRAP></panel>+\begin{align*}  
 +|I| &= {{ |U_{\rm ind}|}\over{R}} \\  
 +    &= {{0.50 ~\rm V}\over{5.0 ~\Omega}} = 0.10 ~\rm A \\  
 +\end{align*}  
 +</collapse> </WRAP></WRAP></panel>
  
 <panel type="info" title="Exercise 4.1.2 Magnetic Field Strength around a horizontal straight Conductor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.1.2 Magnetic Field Strength around a horizontal straight Conductor"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A closely wound coil has a radius of $4.0 cm$, $50$ turns, and a total resistance of $40\Omega$.+A closely wound coil has a radius of $4.0 ~\rm cm$, $50$ turns, and a total resistance of $40~\Omega$.
  
-At what rate must a magnetic field perpendicular to the face of the coil change in order to produce Joule heating in the coil at a rate of $2.0 mW$?+At what rate must a magnetic field perpendicular to the face of the coil change in order to produce Joule heating in the coil at a rate of $2.0 ~\rm mW$?
  
 <button size="xs" type="link" collapse="Solution_4_1_2_1_Result">{{icon>eye}} Result</button><collapse id="Solution_4_1_2_1_Result" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_1_2_1_Result">{{icon>eye}} Result</button><collapse id="Solution_4_1_2_1_Result" collapsed="true">
  
-$1.1 T/s$ </collapse> </WRAP></WRAP></panel>+$1.1 ~\rm T/s$ </collapse> </WRAP></WRAP></panel>
  
 ===== 4.2 Lenz Law ===== ===== 4.2 Lenz Law =====
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 The direction of the induced potential difference drives current around a wire loop to always oppose the change in magnetic flux that causes the potential difference. </callout> The direction of the induced potential difference drives current around a wire loop to always oppose the change in magnetic flux that causes the potential difference. </callout>
  
-Lenz’s law can also be considered in terms of conservation of energy. If pushing a magnet into a coil causes current, the energy in that current must have come from somewhere. If the induced current causes a magnetic field opposing the increase in field of the magnet we pushed in, then the situation is clear. We pushed a magnet against a field and did work on the system, and that showed up as current. If it were not the case that the induced field opposes the change in the flux, the magnet would be pulled in produce a current without anything having done work. Electric potential energy would have been created, violating the conservation of energy.+Lenz’s law can also be considered in terms of the conservation of energy. If pushing a magnet into a coil causes current, the energy in that current must have come from somewhere. If the induced current causes a magnetic field opposing the increase in the field of the magnet we pushed in, then the situation is clear. We pushed a magnet against a field and did work on the system, and that showed up as current. If it were not the case that the induced field opposes the change in the flux, the magnet would be pulled in produce a current without anything having done work. Electric potential energy would have been created, violating the conservation of energy.
  
-To determine an induced potential difference $u_{ind}$, you first calculate the magnetic flux $\Phi_m$ and then obtain $d\Phi_m dt$. The magnitude of $u_{ind}$ is given by+To determine an induced potential difference $u_{\rm ind}$, you first calculate the magnetic flux $\Phi_{\rm m}$ and then obtain ${\rm d}\Phi_{\rm m} {\rm d}t$. The magnitude of $u_{\rm ind}$ is given by
  
-\begin{align*} |u_{ind}| &= \left|-{{d}\over{dt}}\Phi_m\right| \end{align*}+\begin{align*} |u_{\rm ind}| &= \left|-{{\rm d}\over{{\rm d}t}}\Phi_{\rm m}\right| \end{align*}
  
-Finally, you can apply Lenz’s law to determine the sense of $u_{ind}$. This will be developed through examples that illustrate the following problem-solving strategy.+Finally, you can apply Lenz’s law to determine the sense of $u_{\rm ind}$. This will be developed through examples that illustrate the following problem-solving strategy.
  
 <callout icon="fa fa-exclamation" color="red" title="Notice:"> **Problem-Solving Strategy: Lenz’s Law** <callout icon="fa fa-exclamation" color="red" title="Notice:"> **Problem-Solving Strategy: Lenz’s Law**
  
-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, currentsand magnetic fields:
  
   - Make a sketch of the situation for use in visualizing and recording directions.   - Make a sketch of the situation for use in visualizing and recording directions.
   - 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 the magnitude of its magnetic flux is increasing or decreasing. 
-  - Now determine the direction of the induced magnetic field $\vec{B_{ind}}$. 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 to 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 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. 
-  - Use right-hand rule to determine the direction of the induced current $i_{ind}$ that is responsible for the induced magnetic field $\vec{B}_{ind}$.+  - 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.
  
 </callout> </callout>
  
-Let’s apply Lenz’s law to the system of <imgref ImgNr06>. We designate the “front” of the closed conducting loop as the region containing the approaching bar magnet, and the “back” of the loop as the other region. The north pole of the magnet moves toward the loop. Therefore, the flux through the loop due to the field of the magnet increases because the strength of field lines directed from the front to the back of the loop is increasing. A current is consequently induced in the loop. By Lenz’s law, the direction of the induced current must be such that its own magnetic field is directed in a way to oppose the changing flux caused by the field of the approaching magnet. Hence, the induced current circulates so that its magnetic field lines through the loop are directed from the back to the front of the loop. By the rigth hand rule, place your thumb pointing against the magnetic field lines, which is toward the bar magnet. Your fingers wrap in a counterclockwise direction as viewed from the bar magnet. Alternatively, we can determine the direction of the induced current by treating the current loop as an electromagnet that opposes the approach of the north pole of the bar magnet. This occurs when the induced current flows as shown, for then the face of the loop nearer the approaching magnet is also a north pole.+Let’s apply Lenz’s law to the system of <imgref ImgNr06>. We designate the “front” of the closed conducting loop as the region containing the approaching bar magnet, and the “back” of the loop as the other region. The north pole of the magnet moves toward the loop. Therefore, the flux through the loop due to the field of the magnet increases because the strength of field lines directed from the front to the back of the loop is increasing. A current is consequently induced in the loop. By Lenz’s law, the direction of the induced current must be such that its own magnetic field is directed in a way to oppose the changing flux caused by the field of the approaching magnet. Hence, the induced current circulates so that its magnetic field lines through the loop are directed from the back to the front of the loop. By the right-hand rule, place your thumb pointing against the magnetic field lines, which is toward the bar magnet. Your fingers wrap in a counterclockwise direction as viewed from the bar magnet. Alternatively, we can determine the direction of the induced current by treating the current loop as an electromagnet that opposes the approach of the north pole of the bar magnet. This occurs when the induced current flows as shown, for then the face of the loop nearer the approaching magnet is also a north pole.
  
 <WRAP> <imgcaption ImgNr06 | The change in magnetic flux caused by the approaching magnet induces a current in the loop.> </imgcaption> {{drawio>CoilInductionExample.svg}} </WRAP> <WRAP> <imgcaption ImgNr06 | The change in magnetic flux caused by the approaching magnet induces a current in the loop.> </imgcaption> {{drawio>CoilInductionExample.svg}} </WRAP>
  
-Part (b) of the figure shows the south pole of a magnet moving toward a conducting loop. In this case, the flux through the loop due to the field of the magnet increases because the number of field lines directed from the back to the front of the loop is increasing. To oppose this change, a current is induced in the loop whose field lines through the loop are directed from the front to the back. Equivalently, we can say that the current flows in a direction so that the face of the loop nearer the approaching magnet is a south pole, which then repels the approaching south pole of the magnet. By the rigth hand rule, your thumb points away from the bar magnet. Your fingers wrap in a clockwise fashion, which is the direction of the induced current.+Part (b) of the figure shows the south pole of a magnet moving toward a conducting loop. In this case, the flux through the loop due to the field of the magnet increases because the number of field lines directed from the back to the front of the loop is increasing. To oppose this change, a current is induced in the loop whose field lines through the loop are directed from the front to the back. Equivalently, we can say that the current flows in a direction so that the face of the loop nearer the approaching magnet is a south pole, which then repels the approaching south pole of the magnet. By the right-hand rule, your thumb points away from the bar magnet. Your fingers wrap in a clockwise fashion, which is the direction of the induced current.
  
 An animation of this situation can be seen [[https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/22-Faraday-magnet|here]]. An animation of this situation can be seen [[https://www.glowscript.org/#/user/matterandinteractions/folder/matterandinteractions/program/22-Faraday-magnet|here]].
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   * the orientation of the field with the surface area.   * the orientation of the field with the surface area.
  
-If any of these quantities varies, a corresponding variation in magnetic flux occurs. So far, we’ve only considered flux changes due to a changing field. Only the causing magnet was moving. This type of induction is called **static induction**  (or stationary induction, in German: Ruheinduktion).+If any of these quantities vary, a corresponding variation in magnetic flux occurs. So far, we’ve only considered flux changes due to a changing field. Only the causing magnet was moving. This type of induction is called **static induction**  (or stationary induction, in German: //Ruheinduktion//).
  
 Now we look at another possibility: a changing area through which the field lines pass including a change in the orientation of the area. This leads us to **motional induction** Now we look at another possibility: a changing area through which the field lines pass including a change in the orientation of the area. This leads us to **motional induction**
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 <WRAP> <imgcaption ImgNr07 | Examples for motional Induction> </imgcaption> {{drawio>MotionalInductionExample}} </WRAP> <WRAP> <imgcaption ImgNr07 | Examples for motional Induction> </imgcaption> {{drawio>MotionalInductionExample}} </WRAP>
  
-It’s interesting to note that what we perceive as the cause of a particular flux change actually depends on the frame of reference we choose. For example, if you are at rest relative to the moving coils of <imgref ImgNr07> (b), you would see the flux vary because of a changing magnetic field. In part (a), the field moves from left to right in your reference frame, and in part (b), the field is rotating. It is often possible to describe a flux change through a coil that is moving in one particular reference frame in terms of a changing magnetic field in a second frame, where the coil is stationary. However, reference-frame questions related to magnetic flux are beyond this introduction. We’ll avoid such complexities by always working in a frame at rest relative to the laboratory and explain flux variations as due to either a changing field or a changing area.+It’s interesting to note that what we perceive as the cause of a particular flux change actually depends on the frame of reference we choose. For example, if you are at rest relative to the moving coils of <imgref ImgNr07> (b), you would see the flux vary because of a changing magnetic field. In part (a), the field moves from left to right in your reference frame, and in part (b), the field is rotating. It is often possible to describe a flux change through a coil that is moving in one particular reference frame in terms of a changing magnetic field in a second frame, where the coil is stationary. However, reference-frame questions related to magnetic flux are beyond this introduction. We’ll avoid such complexities by always working in a frame at rest relative to the laboratory and explaining flux variations as due to either a changing field or a changing area.
  
 ==== Single Rod ==== ==== Single Rod ====
Zeile 163: Zeile 181:
 The first step to investigate the motional induction is shown in <imgref ImgNr09>: a single conducting rod with the length $\vec{l}$ which is moving with a constant velocity $\vec{v}$ through a homogenous magnetic field $\vec{B} \perp \vec{v} \perp \vec{l}$. The first step to investigate the motional induction is shown in <imgref ImgNr09>: a single conducting rod with the length $\vec{l}$ which is moving with a constant velocity $\vec{v}$ through a homogenous magnetic field $\vec{B} \perp \vec{v} \perp \vec{l}$.
  
-  * The charges in the rod experience the Lorentz force $\vec{F}_L$.+  * The charges in the rod experience the Lorentz force $\vec{F}_{\rm L}$.
   * By this force, the positive charges move to one end of the rod and the negative to the other one.   * By this force, the positive charges move to one end of the rod and the negative to the other one.
-  * The separted charges create a patential difference and by this a Coulomb force $\vec{F}_C$ onto the charges within the rod.+  * The separated charges create a potential difference and by thisa Coulomb force $\vec{F}_{\rm C}$ onto the charges within the rod.
   * For a constant speed, the Lorentz force onto charges in the rod must have the same magnitude as the Coulomb force.   * For a constant speed, the Lorentz force onto charges in the rod must have the same magnitude as the Coulomb force.
  
Zeile 172: Zeile 190:
 This leads to: This leads to:
  
-\begin{align*} \vec{F}_C &= - \vec{F}_L \\ Q \cdot \vec{E}_{ind} &= - Q \cdot \vec{v} \times \vec{B} \\ \vec{E}_{ind} &= - \vec{v} \times \vec{B} \\ \end{align*}+\begin{align*}  
 +\vec{F}_{\rm C}   &= - \vec{F}_{\rm L} \\ Q \cdot \vec{E}_{\rm ind}  
 +                  &= - Q \cdot \vec{v} \times \vec{B} \\  
 +\vec{E}_{\rm ind} &= - \vec{v} \times \vec{B} \\  
 +\end{align*}
  
 The induced potential difference in the rod will be: The induced potential difference in the rod will be:
  
-\begin{align*} U_{ind} &= \int_l \vec{E}_{ind} \cdot d \vec{s} \\ &= - \int^0_1 \vec{v} \times \vec{B} \cdot d \vec{s} \\ \end{align*}+\begin{align*}  
 +U_{\rm ind} &  \int_l \vec{E}_{\rm ind}        \cdot {\rm d\vec{s} \\  
 +            &= - \int^0_1 \vec{v} \times \vec{B} \cdot {\rm d\vec{s} \\  
 +\end{align*}
  
-For constant $|\vec{v}|$ and $|\vec{B}|$ this leads to: \begin{align*} U_{ind} &= - v \cdot B \cdot l \\ \end{align*}+For constant $|\vec{v}|$ and $|\vec{B}|$ this leads to:  
 +\begin{align*}  
 +U_{\rm ind} &= - v \cdot B \cdot l \\  
 +\end{align*}
  
 ==== Rod in Circuit ==== ==== Rod in Circuit ====
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 <WRAP> <imgcaption ImgNr08 | Example for calculating the motional Induction> </imgcaption> {{drawio>MotionalInductionExampleCalc.svg}} </WRAP> <WRAP> <imgcaption ImgNr08 | Example for calculating the motional Induction> </imgcaption> {{drawio>MotionalInductionExampleCalc.svg}} </WRAP>
  
-The velocity of the rod is $v=dx/dt$. So the induced potential difference will get \begin{align*} U_{ind} &= - v \cdot B \cdot l \\ &= - {{dx}\over{dt}} \cdot B \cdot l \\ &= - {{B \cdot l \cdot dx}\over{dt}} \\ &= - {{B \cdot dA}\over{dt}} \\ &= - {{d\Phi_m}\over{dt}} \\ \end{align*}+The velocity of the rod is $v={\rm d}x/{\rm d}t$. So the induced potential difference will get  
 +\begin{align*}  
 +U_{\rm ind} &= - v \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 {\rm d}A}\over{{\rm d}t}} \\  
 +            &= - {{{\rm d}\Phi_{\rm m}}\over{{\rm d}t}} \\  
 +\end{align*}
  
 This is an alternative way to deduce Faraday's Law. This is an alternative way to deduce Faraday's Law.
  
-The current $I_{ind}$ induced in the given circuit is $U_{ind}$ 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*} I_{ind} = {{v \cdot B \cdot l }\over{R}} \end{align*}+\begin{align*}  
 +I_{\rm ind} = {{v \cdot B \cdot l }\over{R}}  
 +\end{align*}
  
 Furthermore, the direction of the induced potential difference satisfies Lenz’s law, as you can verify by inspection of the figure. Furthermore, the direction of the induced potential difference satisfies Lenz’s law, as you can verify by inspection of the figure.
  
-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 $A=lx$, so the magnetic flux through it is $\Phi= B\cdot lx$. 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*} U_{ind} &= - {{d}\over{dt}} \cdot \Phi_m \\ &= - B \cdot l \cdot {{dx}\over{dt}} \\ &= - B \cdot l \cdot v \\ \end{align*}+\begin{align*}  
 +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 v \\  
 +\end{align*}
  
 which is identical to the potential difference between the ends of the rod that we determined earlier. which is identical to the potential difference between the ends of the rod that we determined earlier.
Zeile 206: Zeile 247:
 <panel type="info" title="Exercise 4.3.1 Calculating the Large Motional Potential difference of an Object in Orbit"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.3.1 Calculating the Large Motional Potential difference of an Object in Orbit"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-Calculate the potential difference motional induced along a $20.0 km$ conductor moving at an orbital speed of $7.80 km/s$ perpendicular to Earth’s $5.00 \cdot 10^{-5}T$ magnetic field.+Calculate the potential difference motionally induced along a $20.0 ~\rm km$ conductor moving at an orbital speed of $7.80 ~\rm km/s$ perpendicular to Earth’s $5.00 \cdot 10^{-5} ~\rm T$ magnetic field.
  
 <button size="xs" type="link" collapse="Solution_4_3_1_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_1_1_Solution" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_3_1_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_1_1_Solution" collapsed="true">
  
-This is a great example of using the equation motional $U_{ind} = - B \cdot l \cdot v$+This is a great example of using the equation motional $U_{\rm ind} = - B \cdot l \cdot v$
  
-Entering the given values into $U_{ind} = - B \cdot l \cdot v$ gives+Entering the given values into $U_{\rm ind} = - B \cdot l \cdot v$ gives
  
-\begin{align*} U_{ind} &= - {{d \Phi_m}\over{dt}} \\ &= - B \cdot l \cdot v \\ &= - (5.00 \cdot 10^{-5}T)(20.0 \cdot 10^{3}m)(7.80\cdot 10^{3}m/s) \\ \end{align*}+\begin{align*}  
 +U_{ind} &= - {{{\rm d\Phi_{\rm m}}\over{{\rm d}t}} \\  
 +        &= - 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) \\  
 +\end{align*}
  
 </collapse> </collapse>
  
-<button size="xs" type="link" collapse="Solution_4_3_1_1_Result">{{icon>eye}} Result</button><collapse id="Solution_4_3_1_1_Result" collapsed="true"> \begin{align*} U_{ind} &= - 7.80 \cdot 10^3 V \\ \end{align*}+<button size="xs" type="link" collapse="Solution_4_3_1_1_Result">{{icon>eye}} Result</button><collapse id="Solution_4_3_1_1_Result" collapsed="true">  
 +\begin{align*} U_{\rm ind} &= - 7.80 \cdot 10^3 ~\rm V \\ \end{align*}
  
 </collapse> </collapse>
Zeile 226: Zeile 272:
 <panel type="info" title="Exercise 4.3.2 A Metal Rod Rotating in a Magnetic Field"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.3.2 A Metal Rod Rotating in a Magnetic Field"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-Part (a) of <imgref ImgNr11> shows a metal rod OS that is rotating in a horizontal plane around point O. The rod slides along a wire that forms a circular arc PST of radius $r$. The system is in a constant magnetic field $\vec{B}$ that is directed out of the page.+Part (a) of <imgref ImgNr11> shows a metal rod $\rm OSthat is rotating in a horizontal plane around point $\rm O$ 
 +The rod slides along a wire that forms a circular arc $\rm PSTof radius $r$. The system is in a constant magnetic field $\vec{B}$ that is directed out of the page.
  
-If you rotate the rod at a constant angular velocity $\omega$, what is the current $I_{ind}$ in the closed loop OPSO? Assume that the resistor $R$ furnishes all of the resistance in the closed loop.+If you rotate the rod at a constant angular velocity $\omega$, what is the current $I_{\rm ind}$ in the closed loop $\rm OPSO$ 
 +Assume that the resistor $R$ furnishes all of the resistance in the closed loop.
  
 <WRAP> <imgcaption ImgNr11 | motional Induction on a rotating single Rod> </imgcaption> {{drawio>MotionalInductionExampleRodRotating}} </WRAP> <WRAP> <imgcaption ImgNr11 | motional Induction on a rotating single Rod> </imgcaption> {{drawio>MotionalInductionExampleRodRotating}} </WRAP>
Zeile 234: Zeile 282:
 <button size="xs" type="link" collapse="Solution_4_3_2_1_Strategy">{{icon>eye}} Strategy</button><collapse id="Solution_4_3_2_1_Strategy" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_3_2_1_Strategy">{{icon>eye}} Strategy</button><collapse id="Solution_4_3_2_1_Strategy" collapsed="true">
  
-The magnetic flux is the magnetic field times the area of the quarter circle or $A = r^2 \varphi /2$. When finding the potential difference through Faraday’s law, all variables are constant in time but $\varphi$, with $\omega = d\varphi/dt$. To calculate the work per unit time, we know this is related to the torque times the angular velocity. The torque is calculated by knowing the force on a rod and integrating it over the length of the rod. </collapse>+The magnetic flux is the magnetic field times the area of the quarter circle or $A = r^2 \varphi /2$. When finding the potential difference through Faraday’s law, all variables are constant in time but $\varphi$, with $\omega = {\rm d}\varphi/{\rm d}t$. To calculate the work per unit time, we know this is related to the torque times the angular velocity. The torque is calculated by knowing the force on a rod and integrating it over the length of the rod. </collapse>
  
 <button size="xs" type="link" collapse="Solution_4_3_2_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_2_1_Solution" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_3_2_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_2_1_Solution" collapsed="true">
  
-From geometry, the area of the loop OPSO is $A=r^2\varphi /2$ . Hence, the magnetic flux through the loop is+From geometry, the area of the loop $\rm OPSOis $A=r^2\varphi /2$. Hence, the magnetic flux through the loop is
  
-\begin{align*} \Phi_m &= B\cdot A \\ &= B\cdot {{r^2\varphi}\over{2}} \\ \end{align*}+\begin{align*}  
 +\Phi_{\rm m} &= B\cdot A \\  
 +             &= B\cdot {{r^2\varphi}\over{2}} \\  
 +\end{align*}
  
 Differentiating with respect to time and using $\omega = d\varphi/dt$ , we have Differentiating with respect to time and using $\omega = d\varphi/dt$ , we have
  
-\begin{align*} U_{ind} &= |{{d}\over{dt}} \cdot \Phi_m | \\ &= B\cdot {{r^2\omega}\over{2}} \end{align*}+\begin{align*}  
 +U_{\rm ind} &= |{{\rm d}\over{{\rm d}t}} \cdot \Phi_{\rm m} | \\  
 +            &= B\cdot {{r^2\omega}\over{2}}  
 +\end{align*}
  
-When divided by the resistance $R$ of the loop, this yields for the magnitude of the induced current+When divided by the resistance $R$ of the loop, this yields the magnitude of the induced current
  
-\begin{align*} I_{ind} &= {{|U_{ind}|}\over{R}} \\ &= B\cdot {{r^2\omega}\over{2R}} \end{align*}+\begin{align*}  
 +I_{\rm ind} &= {{|U_{\rm ind}|}\over{R}} \\  
 +            &= B\cdot {{r^2\omega}\over{2R}}  
 +\end{align*}
  
 As $\varphi$ increases, so does the flux through the loop due to $\vec{B}$. To counteract this increase, the magnetic field due to the induced current must be directed into the page in the region enclosed by the loop. Therefore, as part (b) of <imgref ImgNr11> illustrates, the current circulates clockwise. </collapse> As $\varphi$ increases, so does the flux through the loop due to $\vec{B}$. To counteract this increase, the magnetic field due to the induced current must be directed into the page in the region enclosed by the loop. Therefore, as part (b) of <imgref ImgNr11> illustrates, the current circulates clockwise. </collapse>
Zeile 256: Zeile 313:
 <panel type="info" title="Exercise 4.3.3 A Rectangular Coil Rotating in a Magnetic Field"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.3.3 A Rectangular Coil Rotating in a Magnetic Field"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-The following example is the basis for an electric generator: A rectangular coil of area $A$ and $N$ turns is placed in a uniform magnetic field $\vec{B}$, as shown in <imgref ImgNr12>. The coil is rotated about the $z$-axis through its center at a constant angular velocity $\omega$.+The following example is the basis for an electric generator: A rectangular coil of area $A$ and $N$ turns is placed in a uniform magnetic field $\vec{B}$, as shown in <imgref ImgNr12> 
 +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 $U_{ind}$ in the coil.+Obtain an expression for the induced potential difference $U_{\rm ind}$ in the coil.
  
 <WRAP> <imgcaption ImgNr12 | A rectangular coil rotating in a uniform magnetic field> </imgcaption> {{drawio>MotionalInductionExampleCoilRotating2}} </WRAP> <WRAP> <imgcaption ImgNr12 | A rectangular coil rotating in a uniform magnetic field> </imgcaption> {{drawio>MotionalInductionExampleCoilRotating2}} </WRAP>
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 <button size="xs" type="link" collapse="Solution_4_3_3_1_Strategy">{{icon>eye}} Strategy</button><collapse id="Solution_4_3_3_1_Strategy" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_3_3_1_Strategy">{{icon>eye}} Strategy</button><collapse id="Solution_4_3_3_1_Strategy" collapsed="true">
  
-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 simplify quickly. The induced potential difference is written out using Faraday’s law. </collapse>+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 simplify quick. The induced potential difference is written out using Faraday’s law. </collapse>
  
 <button size="xs" type="link" collapse="Solution_4_3_3_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_3_1_Solution" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_3_3_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_3_3_1_Solution" collapsed="true">
Zeile 270: Zeile 328:
 When the coil is in a position such that its surface vector $\vec{A}$ makes an angle $\varphi$ with the magnetic field $\vec{B}$ the magnetic flux through a single turn of the coil is When the coil is in a position such that its surface vector $\vec{A}$ makes an angle $\varphi$ with the magnetic field $\vec{B}$ the magnetic flux through a single turn of the coil is
  
-\begin{align*} \Phi_m &= \iint \vec{B} d\vec{A} \\ &= BA\cdot cos \varphi \\ \end{align*}+\begin{align*}  
 +\Phi_{\rm m} &= \iint \vec{B} {\rm d}\vec{A} \\  
 +             &= BA\cdot \cos \varphi \\  
 +\end{align*}
  
 From Faraday’s law, the induced potential difference in the coil is From Faraday’s law, the induced potential difference in the coil is
  
-\begin{align*} u_{ind} &= - N {{d}\over{dt}} \Phi_m \\ &= NBA \cdot sin \varphi \cdot {{d\varphi}\over{dt}} \end{align*}+\begin{align*}  
 +u_{\rm ind} &= - N {{\rm d}\over{{\rm d}t}} \Phi_{\rm m} \\  
 +            &= NBA \cdot \sin \varphi \cdot {{{\rm d}\varphi}\over{{\rm d}t}}  
 +\end{align*}
  
-The constant angular velocity is $\omega = d\varphi/dt$. The angle $\varphi$ represents the time evolution of the angular velocity or $\omega t$. This is changes the function to time space rather than $\varphi$. The induced potential difference therefore varies sinusoidally with time according to+The constant angular velocity is $\omega = {\rm d}\varphi/{\rm d}t$. The angle $\varphi$ represents the time evolution of the angular velocity or $\omega t$.  
 +This changes the function to time-space rather than $\varphi$. The induced potential differencethereforevaries sinusoidally with time according to
  
-\begin{align*} u_{ind} &= U_{ind,0} \cdot sin \omega t \end{align*}+\begin{align*} u_{ind} &= U_{ind,0} \cdot \sin \omega t \end{align*}
  
-where $U_{ind,0} = NBA\omega$. </collapse>+where $U_{\rm ind,0} = NBA\omega$. </collapse>
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
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 <panel type="info" title="Exercise 4.3.4 Calculating the Potential Difference Induced in a Generator Coil"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.3.4 Calculating the Potential Difference Induced in a Generator Coil"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-The generator coil shown in <imgref ImgNr13> is rotated through one-fourth of a revolution (from $\phi_0=0°$ to $\phi_1=90°$) in $5.0 ms$. The $200$-turn circular coil has a $5.00 cm$ radius and is in a uniform $0.80 T$ magnetic field.+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.
  
 What is the value of the induced potential difference? What is the value of the induced potential difference?
  
-<WRAP> <imgcaption ImgNr13 | When this generator coil is rotated through one-fourth of a revolution, the magnetic flux changes from its maximum to zero, inducing an potential difference.> </imgcaption> {{drawio>MotionalInductionExampleCoilRotating.svg}} </WRAP>+<WRAP> <imgcaption ImgNr13 | When this generator coil is rotated through one-fourth of a revolution, the magnetic flux changes from its maximum to zero, inducing potential difference.> </imgcaption> {{drawio>MotionalInductionExampleCoilRotating.svg}} </WRAP>
  
 <button size="xs" type="link" collapse="Solution_4_3_4_1_Strategy">{{icon>eye}} Strategy</button><collapse id="Solution_4_3_4_1_Strategy" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_3_4_1_Strategy">{{icon>eye}} Strategy</button><collapse id="Solution_4_3_4_1_Strategy" collapsed="true">
Zeile 296: Zeile 362:
 Faraday’s law of induction is used to find the potential difference induced: Faraday’s law of induction is used to find the potential difference induced:
  
-\begin{align*} u_{ind} &= - N {{d \Phi_m}\over{dt}} \end{align*}+\begin{align*}  
 +u_{\rm ind} &= - N {{{\rm d} \Phi_{\rm m}}\over{{\rm d}t}}  
 +\end{align*}
  
-We recognize this situation as the same one in the exercise before. According to the diagram, the projection of the surface vector $\vec{A}$ to the magnetic field is initially ${A}\cdot cos \varphi$ and this is inserted by the definition of the dot product. The magnitude of the magnetic field and area of the loop are fixed over time, which makes the integration simplify quickly. The induced potential difference is written out using Faraday’s law:+We recognize this situation as the same one in the exercise before.  
 +According to the diagram, the projection of the surface vector $\vec{A}$ to the magnetic field is initially ${A}\cdot \cos \varphi$and this is inserted by the definition of the dot product.  
 +The magnitude of the magnetic field and the area of the loop are fixed over time, which makes the integration simplify quick. The induced potential difference is written out using Faraday’s law:
  
-\begin{align*} u_{ind} &= N B \cdot sin \varphi \cdot {{d \varphi}\over{dt}} \end{align*} </collapse>+\begin{align*}  
 +u_{\rm ind} &= N B \cdot \sin \varphi \cdot {{{\rm d\varphi}\over{{\rm d}t}}  
 +\end{align*}  
 +</collapse>
  
 <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 $N=200$, $B=0.80T$ , $\varphi = 90°$ , $d\varphi=90°=\pi/2$ , and $dt=5.0ms$ .+We are given that $N=200$, $B=0.80~\rm T$ , $\varphi = 90°$ , $d\varphi=90°=\pi/2$ , and $\Delta t=5.0~\rm ms$ .
  
 The area of the loop is The area of the loop is
  
-\begin{align*} A = \pi r^2 = 3.14 \cdot (0.0500m)^2 = 7.85 \cdot 10^{-3} m^2 \end{align*}+\begin{align*}  
 +A = \pi r^2 = 3.14 \cdot (0.0500~\rm m)^2 = 7.85 \cdot 10^{-3} ~\rm m^2  
 +\end{align*}
  
 </collapse> <button size="xs" type="link" collapse="Solution_4_3_4_1_Result">{{icon>eye}} Result</button><collapse id="Solution_4_3_4_1_Result" collapsed="true"> </collapse> <button size="xs" type="link" collapse="Solution_4_3_4_1_Result">{{icon>eye}} Result</button><collapse id="Solution_4_3_4_1_Result" collapsed="true">
Zeile 314: Zeile 389:
 Entering this value gives Entering this value gives
  
-\begin{align*} U_{ind} &= 200 \cdot 0.80 T \cdot (7.85 \cdot 10^{-3} m^2) \cdot sin 90° \cdot {{\pi /2}\over{5.0 \cdot 10^{-3} s}} = 395 V \end{align*}+\begin{align*}  
 +U_{\rm ind} &= 200 \cdot 0.80 ~\rm T \cdot (7.85 \cdot 10^{-3} ~\rm m^2) \cdot \sin 90° \cdot {{\pi /2}\over{5.0 \cdot 10^{-3} ~\rm s}} = 395 ~\rm  
 +\end{align*}
  
 </collapse> </WRAP></WRAP></panel> </collapse> </WRAP></WRAP></panel>
Zeile 322: Zeile 399:
 ==== Linked Flux ==== ==== Linked Flux ====
  
-When looking onto the magnetic field in a coil multiple windings capture the passing flux, see <imgref ImgNr14> (a). Each winding will create a potential difference. It can also be interpreted in such a way that the flux is going through the closed surface of the circuit multiple times (in the picture (b)).+When looking at the magnetic field in a coil multiple windings capture the passing flux, see <imgref ImgNr14> (a).  
 +Each winding will create a potential difference.  
 +It can also be interpreted in such a way that the flux is going through the closed surface of the circuit multiple times (in picture (b)).
  
 <WRAP> <imgcaption ImgNr14 | Example for a linked Flux> </imgcaption> {{drawio>LinkedFlux.svg}} </WRAP> <WRAP> <imgcaption ImgNr14 | Example for a linked Flux> </imgcaption> {{drawio>LinkedFlux.svg}} </WRAP>
  
-The resulting electric voltage in such a situation given by the sum of the induced potential differences in each winding.+The resulting electric voltage in such a situation is given by the sum of the induced potential differences in each winding.
  
-\begin{align*} u_{ind} &= - {{d \Phi_{sum}}\over{dt}} \\ &= - \sum_{i=1}^n {{d \Phi_{i}}\over{dt}} \\ \end{align*}+\begin{align*}  
 +u_{\rm ind} &= - {{{\rm d\Phi_{\rm sum}}\over{{\rm d}t}} \\  
 +            &= - \sum_{i=1}^n {{{\rm d\Phi_{i}}\over{{\rm d}t}} \\  
 +\end{align*}
  
-The **linked flux**  $\Psi$ is defined as the resulting flux given by the sum of the partial fluxes of the closed ciruit.+The **linked flux** $\Psi$ is defined as the resulting flux given by the sum of the partial fluxes of the closed circuit.
  
-\begin{align*} \boxed{ \Psi = \sum_{i=1}^n \Phi_{i} } \end{align*}+\begin{align*}  
 +\boxed{ \Psi = \sum_{i=1}^n \Phi_{i} }  
 +\end{align*}
  
 The linked flux simplifies the induced electric voltage of a coil to: The linked flux simplifies the induced electric voltage of a coil to:
  
-\begin{align*} u_{ind} &= - N \cdot {{d \Phi}\over{dt}} \\ &= - {{d}\over{dt}} \Psi \\ \end{align*}+\begin{align*}  
 +u_{\rm ind} &= - N \cdot {{{\rm d\Phi}\over{{\rm d}t}} \\  
 +            &= -         {{\rm d}\over{{\rm d}t}} \Psi \\  
 +\end{align*}
  
 ==== Self-Induction ==== ==== Self-Induction ====
  
-Up to now, we investigated the induction of electric voltages and currents based on the change of an external flux $d\Psi / dt$. For the induced current $i_{ind}$, we found that it counteracts the change of the external flux (Lenz law).+Up to now, we investigated the induction of electric voltages and currents based on the change of an external flux ${\rm d}\Psi / {\rm d}t$.  
 +For the induced current $i_{\rm ind}$, we found that it counteracts the change of the external flux (Lenz law).
  
 But what happens, when there is no external field - only a coil which creates the flux change itself (see <imgref ImgNr46>)? But what happens, when there is no external field - only a coil which creates the flux change itself (see <imgref ImgNr46>)?
Zeile 346: Zeile 434:
 <WRAP> <imgcaption ImgNr46 | Induction Phenomenons> </imgcaption> {{drawio>InductionPhenomenons.svg}} </WRAP> <WRAP> <imgcaption ImgNr46 | Induction Phenomenons> </imgcaption> {{drawio>InductionPhenomenons.svg}} </WRAP>
  
-In order to undestand this, we will investigate the situation for a long coil (<imgref ImgNr15>).+To understand this, we will investigate the situation for a long coil (<imgref ImgNr15>).
  
 <WRAP> <imgcaption ImgNr15 | Self-Induction of a Coil> </imgcaption> {{drawio>SelfInductionCoil.svg}} </WRAP> <WRAP> <imgcaption ImgNr15 | Self-Induction of a Coil> </imgcaption> {{drawio>SelfInductionCoil.svg}} </WRAP>
Zeile 352: Zeile 440:
 The created field density of the coil can be derived from Ampere's Circuital Law The created field density of the coil can be derived from Ampere's Circuital Law
  
-\begin{align*} \theta(t) &= \int & \vec{H}(t) \cdot d\vec{s} \\ &= \int & \vec{H}_{inner}(t) \cdot d\vec{s} &+ & \int \vec{H}_{outer}(t) \cdot d\vec{s} \\ &= \int & \vec{H}(t) \cdot d\vec{s} &+ & 0 \\ &= & {H}(t) \cdot l \\ \end{align*}+\begin{align*}  
 +\theta(t) &= \int & \vec{H}(t) \cdot {\rm d}\vec{s} \\  
 +          &= \int & \vec{H}_{\rm inner}(t) \cdot {\rm d}\vec{s} & + & \int \vec{H}_{\rm outer}(t) \cdot {\rm d\vec{s} \\  
 +          &= \int & \vec{H}(t) \cdot {\rm d}\vec{s}             & + &   0 \\  
 +          &     & {H}(t) \cdot l \\  
 +\end{align*}
  
 With magnetic voltage $\theta(t) = N \cdot i$ this lead to the magnetic flux density $B(t)$ With magnetic voltage $\theta(t) = N \cdot i$ this lead to the magnetic flux density $B(t)$
  
-\begin{align*} N \cdot i &= {H}(t) \cdot l \\ {H}(t) &= {{N \cdot i }\over {l}} \\ {B}(t) &= \mu_0 \mu_r \cdot {{N \cdot i }\over {l}} \\ \end{align*}+\begin{align*}  
 +N \cdot i &= {H}(t) \cdot l \\  
 +   {H}(t) &                        {{N \cdot i }\over {l}} \\  
 +   {B}(t) &= \mu_0 \mu_{\rm r} \cdot {{N \cdot i }\over {l}} \\  
 +\end{align*}
  
 Based on the magnetic flux density $B(t)$ it is possible to calculate the flux $\Phi(t)$: Based on the magnetic flux density $B(t)$ it is possible to calculate the flux $\Phi(t)$:
  
-\begin{align*} \Phi(t) &= \iint_A \vec{B}(t) \cdot d\vec{A} \\ &= \iint_A \mu_0 \mu_r \cdot {{N \cdot i }\over {l}} \cdot dA \\ &= \mu_0 \mu_r \cdot {{N \cdot i }\over {l}} \cdot A \\ \end{align*}+\begin{align*}  
 +\Phi(t) &= \iint_A \vec{B}(t)                                      \cdot {\rm d}\vec{A} \\  
 +        &= \iint_A \mu_0 \mu_{\rm r} \cdot {{N \cdot i }\over {l}} \cdot {\rm d}A \\  
 +        &        \mu_0 \mu_{\rm r} \cdot {{N \cdot i }\over {l}} \cdot A \\  
 +\end{align*}
  
-The changing flux $\Phi$ is now creating an induced electric voltage and current, which counteracts the initial change of the current. This effect is called **Self Induction**. The induced electric voltage $u_{ind}$ is given by:+The changing flux $\Phi$ is now creating an induced electric voltage and current, which counteracts the initial change of the current.  
 +This effect is called **Self Induction**. The induced electric voltage $u_{\rm ind}$ is given by:
  
-\begin{align*} u_{ind} &= - N \cdot {{d \Phi(t)}\over{dt}} \\ &= - N \cdot {{d (\mu_0 \mu_r \cdot {{N \cdot i }\over {l}} \cdot A)}\over{dt}} \\ &= - N \cdot \mu_0 \mu_r \cdot {{N \cdot A }\over {l}} \cdot {{di}\over{dt}} \\ \end{align*}+\begin{align*}  
 +u_{\rm ind} &= - N \cdot {{{\rm d}                                                   \Phi(t)}\over{{\rm d}t}} \\  
 +            &= - N \cdot {{{\rm d(\mu_0 \mu_{\rm r} \cdot {{N \cdot i }\over {l}} \cdot A)}\over{{\rm d}t}} \\  
 +            &= - N \cdot            \mu_0 \mu_{\rm r} \cdot {{N \cdot A }\over {l}} \cdot {{{\rm d}i}\over{{\rm d}t}} \\  
 +\end{align*}
  
-\begin{align*} \boxed{ u_{ind} = - \mu_0 \mu_r \cdot N^2 \cdot {{A }\over {l}} \cdot {{di}\over{dt}} \\ } \\ \text{for a long coil} \end{align*}+\begin{align*}  
 +\boxed{ u_{\rm ind} = - \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A }\over {l}} \cdot {{{\rm d}i}\over{{\rm d}t}} \\ } \\  
 +\text{for a long coil}  
 +\end{align*}
  
-The result means that the induced electric voltage $u_{ind}$ is proportional to the change of the current ${{d}\over{dt}}i$. The proportionality factor is also called **Self-inductance**  $L$ (or often simply called inductance).+The result means that the induced electric voltage $u_{\rm ind}$ is proportional to the change of the current ${{\rm d}\over{{\rm d}t}}i$.  
 +The proportionality factor is also called **Self-inductance**  $L$ (or often simply called inductance).
  
 ===== 4.5 Inductance ===== ===== 4.5 Inductance =====
  
-The inductance is another passive basic component of the electric circuit. Beside the ohmic resistor $R$ and the capacitor $C$, the inductor $L$ is the lump component entailing the inductance.+The inductance is another passive basic component of the electric circuit.  
 +Besides the ohmic resistor $R$ and the capacitor $C$, the inductor $L$ is the lump component entailing the inductance.
  
-Generally the inductance is defined by: \begin{align*} \boxed{ L = \left|{{u_{ind}}\over{di dt}}\right| \\ } \end{align*}+Generallythe inductance is defined by:  
 +\begin{align*}  
 +\boxed{ L = \left|{{u_{\rm ind}}\over{{\rm d}i {\rm d}t}}\right| \\ }  
 +\end{align*}
  
-The inductance $L$ can also be described differently based on Lenz law $u_{ind} = - {{d}\over{dt}}\Psi(t)$ :+The inductance $L$ can also be described differently based on Lenz law $u_{\rm ind} = - {{\rm d}\over{{\rm d}t}}\Psi(t)$ :
  
-\begin{align*} L &= \left|{{u_{ind}}\over{di dt}}\right| \\ &= {{{d \Psi(t)}/{dt}}\over{di dt}} \\ \end{align*}+\begin{align*}  
 +L &= \left|{{u_{\rm ind}}\over{{\rm d}i {\rm d}t}}\right| \\  
 +  & {{{d \Psi(t)}/{dt}}\over{{\rm d}i {\rm d}t}} \\  
 +\end{align*}
  
 \begin{align*} \boxed{ L = {{ \Psi(t)}\over{i}} } \end{align*} \begin{align*} \boxed{ L = {{ \Psi(t)}\over{i}} } \end{align*}
  
-One can also consider an inductor a conservative: it does not like to see abrupt changes of the passing current. It reacts to any change in the current with a counteracting voltagesince the current change lead to a changing flux and - therefore - an induced voltage. 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.+One can also consider an inductor a "conservative person": it does not like to see abrupt changes in the passing current.  
 +It reacts to any change in the current with a counteracting voltage since the current change leads to a changing flux and - therefore - an induced voltage.  
 +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.
  
 <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>
Zeile 388: Zeile 508:
 Mathematically the voltages can be described in the following way: Mathematically the voltages can be described in the following way:
  
-\begin{align*} u_0 &= u_R &+ &u_L \\ &= i \cdot R &+ &{{d\Psi}\over{dt}} \\ &= i \cdot R &+ &L \cdot {{di}\over{dt}} \\ \end{align*}+\begin{align*}  
 +u_0 &= u_R &+ &u_L \\  
 +    &= i \cdot R & + &     {{{\rm d}\Psi}\over{{\rm d}t}} \\  
 +    &= i \cdot R & + &L \cdot {{{\rm d}i}\over{{\rm d}t}} \\  
 +\end{align*}
  
 ==== Inductance of different Components ==== ==== Inductance of different Components ====
Zeile 394: Zeile 518:
 === Long Coil === === Long Coil ===
  
-In the last sub-chapter, the formula of a long coil was already investigated. By these, the inductance of a long coil is+In the last sub-chapter, the formula of a long coil was already investigated.  
 +By these, the inductance of a long coil is
  
-\begin{align*} \boxed{L_{long \; coil} = \mu_0 \mu_r \cdot N^2 \cdot {{A }\over {l}}} \end{align*}+\begin{align*}  
 +\boxed{L_{\rm long \; coil} = \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A }\over {l}}}  
 +\end{align*}
  
 === Toroidal Coil === === Toroidal Coil ===
  
-The toroidal coil was analysed in the last chapter(see [[:electrical_engineering_2:the_magnetostatic_field#magnetic_field_strength_part_1toroidal_coil|magnetic Field Strength Part 1: Toroidal Coil]]). Here, a rectangular intersection a assumed (see <imgref ImgNr16>).+The toroidal coil was analyzed in the last chapter(see [[:electrical_engineering_2:the_magnetostatic_field#magnetic_field_strength_part_1toroidal_coil|magnetic Field Strength Part 1: Toroidal Coil]]).  
 +Here, a rectangular intersection a assumed (see <imgref ImgNr16>).
  
 <WRAP> <imgcaption ImgNr16 | Self-Induction of a toroidal Coil> </imgcaption> {{drawio>SelfInductionToroCoil.svg}} </WRAP> <WRAP> <imgcaption ImgNr16 | Self-Induction of a toroidal Coil> </imgcaption> {{drawio>SelfInductionToroCoil.svg}} </WRAP>
Zeile 408: Zeile 536:
 \begin{align*} H(t) = {{N \cdot i}\over {l}} \end{align*} \begin{align*} H(t) = {{N \cdot i}\over {l}} \end{align*}
  
-with the mean magnetic path length (= length of the average field line) $l = \pi(r_o+r_i)$:+with the mean magnetic path length (= length of the average field line) $l = \pi(r_{\rm o} r_{\rm i})$:
  
-\begin{align*} H(t) = {{N \cdot i}\over { \pi(r_o+r_i)}} \end{align*}+\begin{align*} H(t) = {{N \cdot i}\over { \pi(r_{\rm o} r_{\rm i})}} \end{align*}
  
 The inductance $L$ can be calculated by The inductance $L$ can be calculated by
  
-\begin{align*} L_{toroidal \; coil} &= {{ \Psi(t)}\over{i}} \\ &= {{ N \cdot \Phi(t)}\over{i}} \\ \end{align*}+\begin{align*}  
 +L_{\rm toroidal \; coil} &        {{ \Psi(t)}\over{i}} \\  
 +                         &= {{ N \cdot \Phi(t)}\over{i}} \\  
 +\end{align*}
  
-With the magnetic flux density $B(t) = \mu_0 \mu_r H(t) = \mu_0 \mu_r {{i \cdot N }\over {l}}$ and the cross section $A = h(r_o-r_i)$, we get:+With the magnetic flux density $B(t) = \mu_0 \mu_{\rm r} H(t) = \mu_0 \mu_{\rm r} {{i \cdot N }\over {l}}$ and the cross section $A = h (r_{\rm o} r_{\rm i})$, we get:
  
-\begin{align*} \quad \quad L_{toroidal \; coil} &= {{ N \cdot \mu_0 \mu_r {{i \cdot N }\over { \pi(r_o+r_i)}} \cdot h(r_o-r_i)}\over{i}} \\ &= {{ N^2 \cdot \mu_0 \mu_r \cdot h(r_o-r_i)}\over{ \pi(r_o+r_i)}} \\ \end{align*}+\begin{align*}  
 +\quad \quad L_{\rm toroidal \; coil} &= {{   N \cdot \mu_0 \mu_{\rm r} {{i \cdot N } \over { \pi(r_{\rm o} r_{\rm i})}} \cdot h(r_{\rm o} r_{\rm i})}\over{i}} \\  
 +                                     &= {{ N^2 \cdot \mu_0 \mu_{\rm r}                                                    \cdot h(r_{\rm o} r_{\rm i})}\over{ \pi(r_{\rm o} r_{\rm i})}} \\  
 +\end{align*}
  
-\begin{align*} \boxed{ L_{toroidal \; coil} = \mu_0 \mu_r \cdot N^2 \cdot {{ h(r_o-r_i)}\over{ \pi(r_o+r_i)}} } \end{align*}+\begin{align*}  
 +\boxed{ L_{\rm toroidal \; coil} = \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{ h(r_{\rm o} r_{\rm i})}\over{ \pi(r_{\rm o} r_{\rm i})}} }  
 +\end{align*}
  
 === Exercises === === Exercises ===
Zeile 426: Zeile 562:
 <panel type="info" title="Exercise 4.1.4 Effects of induction I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.1.4 Effects of induction I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-changing of magnetic flux is passing a coil with a single winding. The following pictures <imgref ImgNrEx01> show different flux-time-diagrams as examples.+change of magnetic flux is passing a coil with a single winding. The following pictures <imgref ImgNrEx01> show different flux-time-diagrams as examples.
  
-  * Create for each $\Phi(t)$-diagram the corresponding $u_{ind}(t)$-diagram! +  * Create for each $\Phi(t)$-diagram the corresponding $u_{\rm ind}(t)$-diagram! 
-  * Write down each maximum value of $u_{ind}(t)$+  * Write down each maximum value of $u_{\rm ind}(t)$
  
 <WRAP> <imgcaption ImgNrEx01| Flux-Time-Diagrams> </imgcaption> <WRAP> {{drawio>FluxTimeDia1.svg}} \\ </WRAP></WRAP> <WRAP> <imgcaption ImgNrEx01| Flux-Time-Diagrams> </imgcaption> <WRAP> {{drawio>FluxTimeDia1.svg}} \\ </WRAP></WRAP>
Zeile 435: Zeile 571:
 <button size="xs" type="link" collapse="Solution_4_1_4_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_1_4_1_Solution" collapsed="true"> <button size="xs" type="link" collapse="Solution_4_1_4_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_1_4_1_Solution" collapsed="true">
  
-For partwise linear $u_{ind}$ one can derive: \begin{align*} u_{ind} &= -{{d\Phi}\over{dt}} \\ &= -{{\Delta \Phi}\over{\Delta t}} \end{align*}+For partwise linear $u_{\rm ind}$ one can derive:  
 +\begin{align*}  
 +u_{\rm ind} &= -{{{\rm d}\Phi}\over{{\rm d}t}} \\  
 +            &= -{{\Delta \Phi}\over{\Delta t}}  
 +\end{align*}
  
 For diagram (a): For diagram (a):
  
-  * $t= 0.0 ... 0.6s$: $u_{ind} = -{{0Vs}\over{0.6s}}= 0$ +  * $t= 0.0 ... 0.6 ~\rm s$: $u_{\rm ind} = -{{0 ~\rm Vs}\over{0.6 ~\rm s}}= 0$ 
-  * $t= 0.6 ... 1.5s$: $u_{ind} = -{{-3.75\cdot 10^{-3}Vs}\over{0.9s}}= +4.17 mV$ +  * $t= 0.6 ... 1.5 ~\rm s$: $u_{\rm ind} = -{{-3.75\cdot 10^{-3} ~\rm Vs}\over{0.9 ~\rm s}}= +4.17 ~\rm mV$ 
-  * $t= 1.5 ... 2.1s$: $u_{ind} = -{{0Vs}\over{0.6s}}= 0$+  * $t= 1.5 ... 2.1 ~\rm s$: $u_{\rm ind} = -{{0 ~\rm Vs}\over{0.6 ~\rm s}}= 0$
  
 </collapse> </collapse>
Zeile 451: Zeile 591:
 <panel type="info" title="Exercise 4.1.5 Effects of induction II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.1.5 Effects of induction II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A changing of magnetic flux is passing a coil with a single winding and induces the voltage $u_{ind}(t)$. The following pictures <imgref ImgNrEx02> show different voltage-time-diagrams as examples.+A changing of magnetic flux is passing a coil with a single winding and induces the voltage $u_{\rm ind}(t)$.  
 +The following pictures <imgref ImgNrEx02> show different voltage-time diagrams as examples.
  
-  * Create for each $u_{ind}(t)$-diagram the corresponding $\Phi(t)$-diagram!+  * Create for each $u_{\rm ind}(t)$-diagram the corresponding $\Phi(t)$-diagram!
   * Write down each maximum value of $\Phi(t)$   * Write down each maximum value of $\Phi(t)$
  
Zeile 464: Zeile 605:
 <panel type="info" title="Exercise 4.1.6 Coil in magnetic Field I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.1.6 Coil in magnetic Field I"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A single winding ist located in a homogenius magnetic field ($B = 0.5T$) between the pole pieces. The winding has a length of $150mm$ and a distance between the conductors of $50mm$ (see <imgref ImgNrEx03>).+A single winding is located in a homogenous magnetic field ($B = 0.5 ~\rm T$) between the pole pieces.  
 +The winding has a length of $150 ~\rm mm$ and a distance between the conductors of $50 ~\rm mm$ (see <imgref ImgNrEx03>).
  
-  * Determine the function $u_{ind}(t)$, when the coil is rotating with $3000 min^{-1}$. +  * Determine the function $u_{\rm ind}(t)$, when the coil is rotating with $3000 ~\rm min^{-1}$. 
-  * Given a current of $20A$ through the winding: What is the torque $M(\varphi)$ depending on the angel between the surface vector of the winding an the magnetic field?+  * Given a current of $20 ~\rm A$ through the winding: What is the torque $M(\varphi)$ depending on the angle between the surface vector of the winding and the magnetic field?
  
 <WRAP> <imgcaption ImgNrEx03| Winding between Pole Pieces> </imgcaption> <WRAP> {{drawio>WindingPolePieces.svg}} \\ </WRAP></WRAP> <WRAP> <imgcaption ImgNrEx03| Winding between Pole Pieces> </imgcaption> <WRAP> {{drawio>WindingPolePieces.svg}} \\ </WRAP></WRAP>
Zeile 473: Zeile 615:
 <button size="xs" type="link" collapse="Solution_4_1_6_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_1_6_1_Solution" collapsed="true">  <button size="xs" type="link" collapse="Solution_4_1_6_1_Solution">{{icon>eye}} Solution</button><collapse id="Solution_4_1_6_1_Solution" collapsed="true"> 
 \begin{align*}  \begin{align*} 
-u_{ind} &= -{{d\Phi}\over{dt}} \\  +u_{\rm ind} &= -   {{{\rm d}\Phi}\over{{\rm d}t}} \\  
-        &= -{{d}\over{dt}} B \cdot A \\ +            &= -         {{\rm d}\over{{\rm d}t}} B \cdot A \\ 
-        &= - B \cdot {{d}\over{dt}} l \cdot d \cdot cos(\omega t)\\ +            &= - B \cdot {{\rm d}\over{{\rm d}t}} A\\ 
-        &= + B \cdot l \cdot d \cdot \omega \cdot sin(\omega t)\\ +            &= - B \cdot {{\rm d}\over{{\rm d}t}} \left(l \cdot d \cdot \cos(\omega t) \right)\\ 
 +            &= + B \cdot l \cdot d \cdot \omega \cdot \sin(\omega t)\\ 
 \end{align*} \end{align*}
  
Zeile 483: Zeile 626:
 <panel type="info" title="Exercise 4.1.7 Coil in magnetic Field II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.1.7 Coil in magnetic Field II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A rectangular coil is given by the sizes $a=10cm$, $b=4cm$ and the number of windings $N=200$. This coil moves with a constant speed of $v=2 m/s$ perpendicular to a homogenious magnetic field ($B=1.3T$ on a length of $l=5cm$). The area of the coil is tilted with regard to the field in $\alpha=60°$ and enters the field from the left side (see <imgref ImgNrEx04>).+A rectangular coil is given by the sizes $a=10 ~\rm cm$, $b=4 ~\rm cm$and the number of windings $N=200$.  
 +This coil moves with a constant speed of $v=2 ~\rm m/s$ perpendicular to a homogeneous magnetic field ($B=1.3 ~\rm T$ on a length of $l=5 ~\rm cm$).  
 +The area of the coil is tilted with regard to the field in $\alpha=60°$ and enters the field from the left side (see <imgref ImgNrEx04>).
  
-  * Determine the function $u_{ind}(t)$ on the coil along the given path. Draw a sketch of the $u_{ind}(t)$ diagram. +  * Determine the function $u_{\rm ind}(t)$ on the coil along the given path. Sketch of the $u_{\rm ind}(t)$ diagram. 
-  * What is the maximum induced voltage $u_{ind,Max}$?+  * What is the maximum induced voltage $u_{\rm ind,Max}$?
  
 <WRAP> <imgcaption ImgNrEx04| Winding between Pole Pieces> </imgcaption> <WRAP> {{drawio>WindingPolePieces2.svg}} \\ </WRAP></WRAP> <WRAP> <imgcaption ImgNrEx04| Winding between Pole Pieces> </imgcaption> <WRAP> {{drawio>WindingPolePieces2.svg}} \\ </WRAP></WRAP>
Zeile 497: Zeile 642:
 **Step 1**: Calculate the effective area, perpendicular to the $\vec{B}$-field (independent from whether the area is in the $\vec{B}$-field or not). **Step 1**: Calculate the effective area, perpendicular to the $\vec{B}$-field (independent from whether the area is in the $\vec{B}$-field or not).
  
-For this $b$ has to be projected onto the plane perpendicular to the $\vec{B}$-field: $b_{eff}= b \cdot cos \alpha$+For this $b$ has to be projected onto the plane perpendicular to the $\vec{B}$-field: $b_{\rm eff}= b \cdot \cos \alpha$
 \begin{align*}  \begin{align*} 
-A_{eff} &= a \cdot b \cdot cos \alpha+A_{\rm eff} &= a \cdot b \cdot \cos \alpha
 \end{align*} \end{align*}
  
 **Step 2**: Focus on entering and exiting the $\vec{B}$-field. \\ **Step 2**: Focus on entering and exiting the $\vec{B}$-field. \\
-Induction only occurs for ${{d}\over{dt}}(A\cdot B)\neq 0$, so here: when the area $A_{eff}$ enters and leave the constant $\vec{B}$-field. +Induction only occurs for ${{\rm d}\over{{\rm d}t}}(A\cdot B)\neq 0$, so here: when the area $A_{\rm eff}$ enters and leave the constant $\vec{B}$-field. 
  
 When entering the $\vec{B}$-field the area $A$ with $0<A<A_{eff}$ is in the field.  When entering the $\vec{B}$-field the area $A$ with $0<A<A_{eff}$ is in the field. 
-The area moves with $v$. Therefore, after $\Delta t = b_{eff} \cdot v$ the full $\vec{B}$-field is provided onto the area $A_{eff}$: +The area moves with $v$. Therefore, after $\Delta t = b_{\rm eff} \cdot v$ the full $\vec{B}$-field is provided onto the area $A_{\rm eff}$: 
 \begin{align*}  \begin{align*} 
-u_{ind} &= -{{d\Psi}\over{dt}} \\  +u_{\rm ind} &= -        {{{\rm d}\Psi}\over{{\rm d}t}} \\  
-        &= -N \cdot {{d}\over{dt}} B \cdot A \\  +            &= -N \cdot       {{\rm d}\over{{\rm d}t}} B \cdot A \\  
-        &= -N \cdot {{1}\over{\Delta t}} B \cdot A_{eff} \\  +            &= -N \cdot           {{1}\over{\Delta t}} B \cdot A_{\rm eff} \\  
-        &= -N \cdot {{1}\over{b \cdot cos \alpha \cdot v}} B \cdot a \cdot b \cdot cos \alpha \\  +            &= -N \cdot {{1}\over{b \cdot \cos \alpha \cdot v}} B \cdot a \cdot b \cdot \cos \alpha \\  
-        &= -N \cdot B \cdot {{a}\over{v}}\\ +            &= -N \cdot B \cdot {{a}\over{v}}\\ 
 \end{align*} \end{align*}
  
 The following diagram shows ... The following diagram shows ...
-  * ... how one can derive the effective width $b_{eff}$, which is projected onto the plane perpendicular to the $\vec{B}$-field: $b_{eff}= b \cdot cos \alpha$ +  * ... how one can derive the effective width $b_{\rm eff}$, which is projected onto the plane perpendicular to the $\vec{B}$-field: $b_{\rm eff}= b \cdot \cos \alpha$ 
-  * ... what happens on the effective area $A_{eff}$: there is a change of the field lines in the area only for entering and leaving the $\vec{B}$-field.  +  * ... what happens on the effective area $A_{\rm eff}$: there is a change of the field lines in the area only for entering and leaving the $\vec{B}$-field.  
-  * ... how the $u_{ind}(t)$ looks as a graph: the part of $A_{eff}$, where the $\vec{B}$-field passes through increase linearly due to the constant speed $v$ +  * ... how the $u_{\rm ind}(t)$ looks as a graph: the part of $A_{\rm eff}$, where the $\vec{B}$-field passes through increase linearly due to the constant speed $v$ 
-Be aware, that the task did not provide a clue for the direction of windings and therefore no clued for the polarisation of the induced voltage. \\ Therefore, the course of the voltage when entering or exiting is not uniquely given.+Be aware, that the task did not provide a clue for the direction of windings and therefore it provides no clue for the polarization of the induced voltage. \\  
 +So, the course of the voltage when entering or exiting is not uniquely given.
  
-<WRAP> <imgcaption ImgNrEx04s| Solution> </imgcaption> <WRAP> {{drawio>WindingPolePieces2solution}}  \\ </WRAP></WRAP>+<WRAP> <imgcaption ImgNrEx04s| Solution> </imgcaption> <WRAP>{{drawio>WindingPolePieces2solution.svg}}  \\ </WRAP></WRAP>
  
  
Zeile 530: Zeile 676:
 Calculate the inductance for the following settings Calculate the inductance for the following settings
  
-  - cylindrical air coil with $N=400$, winding diameter $d=3 cm$ and length $l=18cm$+  - cylindrical air coil with $N=400$, winding diameter $d=3 ~\rm cm$ and length $l=18 ~\rm cm$
   - similar coil geometry as explained in 1. , but with double the number of windings   - similar coil geometry as explained in 1. , but with double the number of windings
-  - two coils as explained in 1. in seriea +  - two coils as explained in 1. in series 
-  - similar coil geometry and number of windings as explained in 1. , but with an iron core ($\mu_r=1000$)+  - similar coil geometry and number of windings as explained in 1. , but with an iron core ($\mu_{\rm r}=1000$)
  
 </WRAP></WRAP></panel> </WRAP></WRAP></panel>
Zeile 539: Zeile 685:
 <panel type="info" title="Exercise 4.5.2 Self Induction II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.5.2 Self Induction II"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A cylindrical air coil (length $l=40cm$, diameter $d=5cm$ and number of windings $N=300$) passes a current of $30A$. The current shall be reduced linearly in $2ms$ down to $0A$.+A cylindrical air coil (length $l=40 ~\rm cm$, diameter $d=5 ~\rm cm$and number of windings $N=300$) passes a current of $30 ~\rm A$. The current shall be reduced linearly in $2 ~\rm ms$ down to $0 ~\rm A$.
  
-What is the amount of the induced voltage $u_{ind}$? </WRAP></WRAP></panel>+What is the amount of the induced voltage $u_{\rm ind}$? </WRAP></WRAP></panel>
  
 <panel type="info" title="Exercise 4.5.3 Self Induction III"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> <panel type="info" title="Exercise 4.5.3 Self Induction III"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%>
  
-A coil with the inductance $L_=20\mu H$ passes a current of $40A$. The current shall be reduced linearly in $5\mu s$ down to $0A$ (see <imgref ImgNrEx05>).+A coil with the inductance $L=20 ~\rm µH$ passes a current of $40 ~\rm A$. The current shall be reduced linearly in $5 ~\rm µs$ down to $0 ~\rm A$ (see <imgref ImgNrEx05>).
  
-  * ListenpunktWhat is the amount of the induced voltage $u_{ind}$? +  * What is the amount of the induced voltage $u_{\rm ind}$? 
-  * Sketch the course of $u_{ind}(t)$!+  * Sketch the course of $u_{\rm ind}(t)$!
  
 <WRAP> <imgcaption ImgNrEx05| Circuit and timing Diagram> </imgcaption> <WRAP> {{drawio>CircuitAndTiming.svg}} \\ </WRAP></WRAP> <WRAP> <imgcaption ImgNrEx05| Circuit and timing Diagram> </imgcaption> <WRAP> {{drawio>CircuitAndTiming.svg}} \\ </WRAP></WRAP>