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| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_2:the_time-dependent_magnetic_field [2022/07/11 12:50] – [Bearbeiten - Panel] tfischer | electrical_engineering_2:the_time-dependent_magnetic_field [2024/07/03 10:11] (aktuell) – mexleadmin | ||
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| Zeile 1: | Zeile 1: | ||
| - | ====== 4. time-dependent magnetic Field ====== | + | ====== 4 Time-dependent magnetic Field ====== |
| - | < | + | < |
| < | < | ||
| Zeile 12: | Zeile 12: | ||
| - 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 |
| </ | </ | ||
| - | 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 ===== | ||
| - | 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 '' | + | 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 '' |
| < | < | ||
| Zeile 30: | Zeile 30: | ||
| {{url> | {{url> | ||
| - | 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 | + | 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 |
| - | Faraday realized that in both experiments, | + | Faraday realized that in both experiments, |
| <callout icon=" | <callout icon=" | ||
| - | 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** | + | 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** |
| - | 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 a measurement of the amount of magnetic field lines through a given surface area, as seen in <imgref ImgNr03> |
| < | < | ||
| - | This definition | + | This definition |
| - | \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{ | + | \begin{align*} \boxed{ |
| The negative sign describes the direction in which the induced potential difference drives current around a circuit. However, that direction is most easily determined with a rule known as Lenz’s law, which we will discuss in the next subchapter. | The negative sign describes the direction in which the induced potential difference drives current around a circuit. However, that direction is most easily determined with a rule known as Lenz’s law, which we will discuss in the next subchapter. | ||
| Zeile 54: | Zeile 54: | ||
| <imgref ImgNr05> depicts a circuit and an arbitrary surface $S$ that it bounds. Notice that $S$ is an open surface: The planar area bounded by the circuit is not part of the surface, so it is not fully enclosing a volume. | <imgref ImgNr05> depicts a circuit and an arbitrary surface $S$ that it bounds. Notice that $S$ is an open surface: The planar area bounded by the circuit is not part of the surface, so it is not fully enclosing a volume. | ||
| - | Since the magnetic field is a source free vortex field, the flux over a closed area is 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 [[: | + | Since the magnetic field is a source-free vortex field, the flux over a closed area is always |
| + | By this, it can be shown that any open surface bounded by the circuit in question can be used to evaluate $\Phi_{\rm m}$ (similar to the [[: | ||
| + | For example, $\Phi_{\rm m}$ is the same for the various surfaces $S$, $S_1$, $S_2$ of the figure. | ||
| < | < | ||
| - | 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 $\rm Weber$ (Wb), \begin{align*} [\Phi_{\rm m}] = [B] \cdot [A] = 1 ~\rm |
| - | Occasionally, the magnetic field unit is expressed as webers | + | Based on this definition, the magnetic field unit is occasionally |
| + | In many practical applications, | ||
| + | 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}) | ||
| + | | ||
| + | \end{align*} | ||
| <panel type=" | <panel type=" | ||
| - | 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? | ||
| Zeile 75: | Zeile 83: | ||
| <button size=" | <button size=" | ||
| - | 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. </ | + | 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. </ |
| <button size=" | <button size=" | ||
| Zeile 81: | Zeile 89: | ||
| 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})| \\ | ||
| + | | ||
| + | | ||
| + | | ||
| + | | ||
| + | \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*} </ | + | \begin{align*} |
| + | |I| &= {{ |u_{\rm ind}|}\over{R}} \\ | ||
| + | | ||
| + | \end{align*} | ||
| + | </ | ||
| <panel type=" | <panel type=" | ||
| - | 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=" | <button size=" | ||
| - | $1.1 T/s$ </ | + | $1.1 ~\rm T/s$ </ |
| ===== 4.2 Lenz Law ===== | ===== 4.2 Lenz Law ===== | ||
| Zeile 109: | Zeile 127: | ||
| 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. </ | 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. </ | ||
| - | 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=" | <callout icon=" | ||
| - | 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 $\vec{B_{ind}}$. The induced magnetic field tries to reinforce | + | - Now determine the direction of the induced magnetic field $\vec{B_{\rm ind}}$. |
| - | - 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. | ||
| </ | </ | ||
| - | Let’s apply Lenz’s law to the system of <imgref ImgNr06> | + | Let’s apply Lenz’s law to the system of <imgref ImgNr06> |
| < | < | ||
| - | 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, | + | 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, |
| An animation of this situation can be seen [[https:// | An animation of this situation can be seen [[https:// | ||
| Zeile 146: | Zeile 164: | ||
| * the orientation of the field with the surface area. | * the orientation of the field with the surface area. | ||
| - | If any of these quantities | + | If any of these quantities |
| Now we look at another possibility: | Now we look at another possibility: | ||
| Zeile 157: | Zeile 175: | ||
| < | < | ||
| - | 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 | + | 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 |
| ==== Single Rod ==== | ==== Single Rod ==== | ||
| Zeile 163: | Zeile 181: | ||
| The first step to investigate the motional induction is shown in <imgref ImgNr09>: | The first step to investigate the motional induction is shown in <imgref ImgNr09>: | ||
| - | * 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 | + | * The separated |
| * 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} | ||
| + | | ||
| + | \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*} | + | \begin{align*} |
| + | u_{\rm ind} & | ||
| + | | ||
| + | \end{align*} | ||
| - | For constant $|\vec{v}|$ and $|\vec{B}|$ this leads to: \begin{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 ==== | ||
| Zeile 186: | Zeile 214: | ||
| < | < | ||
| - | The velocity of the rod is $v=dx/dt$. So the induced potential difference will get \begin{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 \\ | ||
| + | | ||
| + | | ||
| + | | ||
| + | | ||
| + | \end{align*} | ||
| This is an alternative way to deduce Faraday' | This is an alternative way to deduce Faraday' | ||
| - | 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*} | + | \begin{align*} |
| + | i_{\rm ind} = {{v \cdot B \cdot l }\over{R}} | ||
| + | \end{align*} | ||
| Furthermore, | Furthermore, | ||
| - | The situation of the single rod can be interpreted in the following way: We can calculate a motionally induced potential difference with Faraday’s law even when an actual closed circuit is not present. We simply imagine an enclosed area whose boundary includes the moving conductor, calculate $\Phi_m$ , and then find the potential difference from Faraday’s law. For example, we can let the moving rod of <imgref ImgNr10> be one side of the imaginary rectangular area represented by the dashed lines. The area of the rectangle is $A=lx$, so the magnetic flux through it is $\Phi= B\cdot | + | 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 |
| - | \begin{align*} | + | \begin{align*} |
| + | u_{\rm ind} &= - {{\rm d}\over{{\rm dt}}} \cdot \Phi_{\rm m} \\ | ||
| + | | ||
| + | | ||
| + | \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=" | <panel type=" | ||
| - | Calculate the potential difference | + | Calculate the potential difference |
| <button size=" | <button size=" | ||
| - | 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*} | + | \begin{align*} |
| + | u_{ind} &= - {{{\rm d} \Phi_{\rm m}}\over{{\rm d}t}} \\ | ||
| + | | ||
| + | | ||
| + | \end{align*} | ||
| </ | </ | ||
| - | <button size=" | + | <button size=" |
| + | \begin{align*} | ||
| </ | </ | ||
| Zeile 226: | Zeile 272: | ||
| <panel type=" | <panel type=" | ||
| - | 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 $\rm OS$ that is rotating in a horizontal plane around point $\rm O$. |
| + | The rod slides along a wire that forms a circular arc $\rm PST$ of 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. | ||
| < | < | ||
| Zeile 234: | Zeile 282: | ||
| <button size=" | <button size=" | ||
| - | 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. </ | + | 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. </ |
| <button size=" | <button size=" | ||
| - | 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 $\rm OPSO$ is $A=r^2\varphi /2$. 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/ | Differentiating with respect to time and using $\omega = d\varphi/ | ||
| - | \begin{align*} | + | \begin{align*} |
| + | u_{\rm ind} &= |{{\rm d}\over{{\rm d}t}} \cdot \Phi_{\rm m} | \\ | ||
| + | | ||
| + | \end{align*} | ||
| - | When divided by the resistance $R$ of the loop, this yields | + | When divided by the resistance $R$ of the loop, this yields the magnitude of the induced current |
| - | \begin{align*} | + | \begin{align*} |
| + | i_{\rm ind} &= {{|u_{\rm ind}|}\over{R}} \\ | ||
| + | | ||
| + | \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, | 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, | ||
| Zeile 256: | Zeile 313: | ||
| <panel type=" | <panel type=" | ||
| - | 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 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. |
| < | < | ||
| Zeile 264: | Zeile 322: | ||
| <button size=" | <button size=" | ||
| - | According to the diagram, the angle between the surface vector $\vec{A}$ and the magnetic field $\vec{B}$ is $\varphi$. The dot product of $\vec{A} \cdot \vec{B}$ simplifies to only the $cos \varphi$ component of the magnetic field times the area, namely where the magnetic field projects onto the unit surface vector $\vec{A}$. The magnitude of the magnetic field and the area of the loop are fixed over time, which makes the integration | + | According to the diagram, the angle between the surface vector $\vec{A}$ and the magnetic field $\vec{B}$ is $\varphi$. The dot product of $\vec{A} \cdot \vec{B}$ simplifies to only the $\cos \varphi$ component of the magnetic field times the area, namely where the magnetic field projects onto the unit surface vector $\vec{A}$. The magnitude of the magnetic field and the area of the loop are fixed over time, which makes the integration |
| <button size=" | <button size=" | ||
| Zeile 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} | ||
| + | &= BA\cdot | ||
| + | \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} \\ | ||
| + | | ||
| + | \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 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 $U_{ind,0} = NBA\omega$. </ | + | where $U_{\rm ind,0} = NBA\omega$. </ |
| </ | </ | ||
| Zeile 286: | Zeile 351: | ||
| <panel type=" | <panel type=" | ||
| - | The generator coil shown in <imgref ImgNr13> is rotated through one-fourth of a revolution (from $\phi_0=0°$ to $\phi_1=90°$) in $5.0 ms$. 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? | ||
| - | < | + | < |
| <button size=" | <button size=" | ||
| 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 | + | 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 | ||
| + | The magnitude of the magnetic field and the area of the loop are fixed over time, which makes the integration simplify | ||
| - | \begin{align*} u_{ind} &= N B \cdot sin \varphi \cdot {{d \varphi}\over{dt}} \end{align*} </ | + | \begin{align*} |
| + | u_{\rm ind} &= N B A \cdot \sin \varphi \cdot {{{\rm d} \varphi}\over{{\rm d}t}} | ||
| + | \end{align*} | ||
| + | </ | ||
| <button size=" | <button size=" | ||
| - | We are given that $N=200$, $B=0.80T$ , $\varphi = 90°$ , $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 | ||
| - | \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} | ||
| + | \end{align*} | ||
| </ | </ | ||
| 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 / | + | \begin{align*} |
| + | U_{\rm ind} &= 200 \cdot 0.80 ~\rm T \cdot (7.85 \cdot 10^{-3} | ||
| + | \end{align*} | ||
| </ | </ | ||
| Zeile 322: | Zeile 399: | ||
| ==== Linked Flux ==== | ==== Linked Flux ==== | ||
| - | When looking | + | When looking |
| + | 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)). | ||
| < | < | ||
| - | 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 |
| - | \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}} \\ | ||
| + | | ||
| + | \end{align*} | ||
| - | The **linked flux** | + | The **linked flux** $\Psi$ is defined as the resulting flux given by the sum of the partial fluxes of the closed |
| - | \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}} \\ | ||
| + | | ||
| + | \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: | ||
| < | < | ||
| - | In order to undestand | + | To understand |
| < | < | ||
| Zeile 352: | Zeile 440: | ||
| The created field density of the coil can be derived from Ampere' | The created field density of the coil can be derived from Ampere' | ||
| - | \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} \\ | ||
| + | | ||
| + | | ||
| + | | ||
| + | \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) & | ||
| + | {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) | ||
| + | | ||
| + | | ||
| + | \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}} \\ | ||
| + | | ||
| + | | ||
| + | \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** | + | 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** | ||
| ===== 4.5 Inductance ===== | ===== 4.5 Inductance ===== | ||
| - | The inductance is another passive basic component of the electric circuit. | + | The inductance is another passive basic component of the electric circuit. |
| + | Besides | ||
| - | Generally the inductance is defined by: \begin{align*} \boxed{ L = \left|{{u_{ind}}\over{di / dt}}\right| \\ } \end{align*} | + | Generally, the 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)}/ | + | \begin{align*} |
| + | L &= \left|{{u_{\rm ind}}\over{{\rm d}i / {\rm d}t}}\right| \\ | ||
| + | | ||
| + | \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: | + | One can also consider an inductor a "conservative |
| + | It reacts to any change in the current with a counteracting voltage since the current change | ||
| + | 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. | ||
| + | |||
| + | The unit of the inductance is $\rm 1 ~Henry = 1 ~H = 1 {{Vs}\over{A}} = 1{{Wb}\over{A}} $ | ||
| < | < | ||
| Zeile 388: | Zeile 510: | ||
| 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 &+ & | + | \begin{align*} |
| + | u_0 &= u_R &+ &u_L \\ | ||
| + | | ||
| + | | ||
| + | \end{align*} | ||
| ==== Inductance of different Components ==== | ==== Inductance of different Components ==== | ||
| Zeile 394: | Zeile 520: | ||
| === Long Coil === | === Long Coil === | ||
| - | In the last sub-chapter, | + | In the last sub-chapter, |
| + | 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 | + | The toroidal coil was analyzed |
| + | Here, a rectangular intersection a assumed (see <imgref ImgNr16> | ||
| < | < | ||
| Zeile 408: | Zeile 538: | ||
| \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} & | ||
| + | &= {{ 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^2 \cdot \mu_0 \mu_{\rm r} | ||
| + | \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 564: | ||
| <panel type=" | <panel type=" | ||
| - | A changing | + | A change |
| - | * 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 $u_{\rm ind}(t)$ |
| < | < | ||
| - | <button size=" | + | <button size=" |
| - | 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}} \\ | ||
| + | | ||
| + | \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} |
| - | * $t= 1.5 ... 2.1s$: $u_{ind} = -{{0Vs}\over{0.6s}}= 0$ | + | * $t= 1.5 ... 2.1 ~\rm s$: $u_{\rm ind} = -{{0 ~\rm Vs}\over{0.6 ~\rm s}}= 0$ |
| </ | </ | ||
| - | <button size=" | + | <button size=" |
| + | {{icon> | ||
| + | < | ||
| + | </ | ||
| + | </ | ||
| </ | </ | ||
| Zeile 451: | Zeile 597: | ||
| <panel type=" | <panel type=" | ||
| - | 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> | + | 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> | ||
| - | * 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 459: | Zeile 606: | ||
| < | < | ||
| + | |||
| + | # | ||
| + | |||
| + | For partwise linear $u_{\rm ind}$ one can derive: | ||
| + | \begin{align*} | ||
| + | u_{\rm ind} &= -{{{\rm d}\Phi}\over{{\rm d}t}} \\ | ||
| + | \rightarrow | ||
| + | \Phi & | ||
| + | \end{align*} | ||
| + | |||
| + | For diagram (a): | ||
| + | |||
| + | * $t= 0.00 ... 0.04 ~\rm s\quad$: $\quad \Phi = \Phi_0 - {0 \cdot \; \Delta t} \quad\quad\quad\quad\quad\quad\quad= 0 ~\rm Wb$ | ||
| + | * $t= 0.04 ... 0.10 ~\rm s\quad$: $\quad \Phi = 0 {~\rm Wb} - {{30 ~\rm mV} \cdot \; (t - 0.04 ~\rm s)} = \quad {1.2 ~\rm mWb} - t \cdot 30 ~\rm mV$ | ||
| + | * $t= 0.10 ... 0.14 ~\rm s\quad$: $\quad \Phi = {1.2 ~\rm mWb} - {0.10 ~\rm s} \cdot 30 ~\rm mV \quad = - {1.8 ~\rm mWb}$ | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | {{drawio> | ||
| + | # | ||
| + | |||
| </ | </ | ||
| Zeile 464: | Zeile 633: | ||
| <panel type=" | <panel type=" | ||
| - | A single winding | + | A single winding |
| + | 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 $u_{ind}(t)$, | + | * 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 | + | * 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 |
| < | < | ||
| Zeile 473: | Zeile 643: | ||
| <button size=" | <button size=" | ||
| \begin{align*} | \begin{align*} | ||
| - | u_{ind} &= -{{d\Phi}\over{dt}} \\ | + | u_{\rm ind} &= - |
| - | &= -{{d}\over{dt}} B \cdot A \\ | + | &= - |
| - | &= - B \cdot {{d}\over{dt}} A\\ | + | &= - B \cdot {{\rm d}\over{{\rm d}t}} A\\ |
| - | &= - B \cdot {{d}\over{dt}} \left(l \cdot d \cdot cos(\omega t) \right)\\ | + | &= - 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)\\ | + | &= + B \cdot l \cdot d \cdot \omega \cdot \sin(\omega t)\\ |
| \end{align*} | \end{align*} | ||
| Zeile 484: | Zeile 654: | ||
| <panel type=" | <panel type=" | ||
| - | 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 | + | 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 | ||
| + | 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 | + | * Determine the function $u_{\rm ind}(t)$ on the coil along the given path. Sketch |
| - | * What is the maximum induced voltage $u_{ind, | + | * What is the maximum induced voltage $u_{\rm ind,Max}$? |
| < | < | ||
| Zeile 498: | Zeile 670: | ||
| **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: | + | For this $b$ has to be projected onto the plane perpendicular to the $\vec{B}$-field: |
| \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< | When entering the $\vec{B}$-field the area $A$ with $0< | ||
| - | 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 |
| - | &= -N \cdot {{1}\over{\Delta t}} B \cdot A_{eff} \\ | + | &= -N \cdot |
| - | &= -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: | + | * ... how one can derive the effective width $b_{\rm eff}$, which is projected onto the plane perpendicular to the $\vec{B}$-field: |
| - | * ... 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 | + | Be aware, that the task did not provide a clue for the direction of windings and therefore |
| + | So, the course of the voltage when entering or exiting is not uniquely given. | ||
| - | < | + | < |
| Zeile 531: | Zeile 704: | ||
| Calculate the inductance for the following settings | Calculate the inductance for the following settings | ||
| - | - cylindrical | + | 1. Cylindrical long air coil with $N=390$, winding diameter $d=3.0 ~\rm cm$ and length $l=18 ~\rm cm$ |
| - | - similar | + | # |
| - | - two coils as explained in 1. in seriea | + | |
| - | - similar | + | \begin{align*} |
| + | L_1 &= \mu_0 \mu_{\rm r} \cdot N^2 \cdot {{A }\over {l}} \\ | ||
| + | &= 4\pi \cdot 10^{-7} {\rm {{H}\over{m}}} \cdot 1 \cdot (390)^2 \cdot {{\pi \cdot (0.03~\rm m)^2 }\over {0.18 ~\rm m}} | ||
| + | \end{align*} | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_1 &= 3.0 ~\rm mH | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | 2. Similar | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_2 &= \mu_0 \mu_{\rm r} \cdot N_2^2 \cdot {{A }\over {l}} \\ | ||
| + | &= \mu_0 \mu_{\rm r} \cdot (2\cdot N)^2 \cdot {{A }\over {l}} \\ | ||
| + | &= \mu_0 \mu_{\rm r} \cdot 4\cdot N^2 \cdot {{A }\over {l}} \\ | ||
| + | &= 4\cdot L_1 \\ | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_1 &= 12 ~\rm mH | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | 3. Two coils as explained in 1. in series | ||
| + | |||
| + | # | ||
| + | multiple inductances in series just add up. One can think of adding more windings to the first coil in the formula.. | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_1 &= 6.0 ~\rm mH | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | 4. Similar | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_4 &= \mu_0 \mu_{\rm r,4} \cdot N^2 \cdot {{A }\over {l}} \\ | ||
| + | &= \mu_0 \codt 1000 \cdot N^2 \cdot {{A }\over {l}} \\ | ||
| + | &= 1000 \cdot L_4 \\ | ||
| + | \end{align*} | ||
| + | # | ||
| + | |||
| + | # | ||
| + | \begin{align*} | ||
| + | L_4 &= 3.0 ~\rm H | ||
| + | \end{align*} | ||
| + | # | ||
| </ | </ | ||
| Zeile 540: | Zeile 770: | ||
| <panel type=" | <panel type=" | ||
| - | 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 $d=5.0 ~\rm cm$, 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 $u_{ind}$? </ | + | 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=" | ||
| - | 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 | + | * What is the amount of the induced voltage $u_{\rm ind}$? |
| - | * Sketch the course of $u_{ind}(t)$! | + | * Sketch the course of $u_{\rm ind}(t)$! |
| < | < | ||