Unterschiede
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electrical_engineering_2:the_time-dependent_magnetic_field [2023/03/16 16:26] mexleadmin |
electrical_engineering_2:the_time-dependent_magnetic_field [2023/09/19 23:51] (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 |
| </ | </ | ||
| 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, | ||
| Zeile 44: | Zeile 44: | ||
| This definition leads to a magnetic flux similar to the electric flux studied earlier: | This definition leads to a magnetic flux similar to the electric flux studied earlier: | ||
| - | \begin{align*} \Phi_m = \iint_A \vec{B} \cdot 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. | ||
| Zeile 54: | Zeile 54: | ||
| <imgref ImgNr05> depicts a circuit and an arbitrary surface $S$ that it bounds. Notice that $S$ is an open surface: The planar area bounded by the circuit is not part of the surface, so it is not fully enclosing a volume. | <imgref ImgNr05> depicts a circuit and an arbitrary surface $S$ that it bounds. Notice that $S$ is an open surface: The planar area bounded by the circuit is not part of the surface, so it is not fully enclosing a volume. | ||
| - | Since the magnetic field is a source-free vortex field, the flux over a closed area is always zero: $\Phi_{\rm m} = \iint_{A} \vec{B} \cdot {\rm d} \vec{A} = 0$. \\ 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 [[: | + | 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 [[: | ||
| + | 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 Weber (Wb), \begin{align*} [\Phi_{\rm m}] = [B] \cdot [A] = 1 ~\rm |
| - | Occasionally, | + | 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=" | ||
| 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 $|U_{\rm ind}|$ 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=" | ||
| Zeile 123: | Zeile 141: | ||
| - 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 |
| - Now determine the direction of the induced magnetic field $\vec{B_{\rm ind}}$. The induced magnetic field tries to reinforce a magnetic flux that is decreasing or opposes a magnetic flux that is increasing. Therefore, the induced magnetic field adds or subtracts from the applied magnetic field, depending on the change in magnetic flux. | - Now determine the direction of the induced magnetic field $\vec{B_{\rm ind}}$. The induced magnetic field 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 the right-hand rule to determine the direction of the induced current $i_{\rm ind}$ that is responsible for the induced magnetic field $\vec{B}_{\rm ind}$. | - Use the right-hand rule to determine the direction of the induced current $i_{\rm ind}$ that is responsible for the induced magnetic field $\vec{B}_{\rm ind}$. | ||
| Zeile 165: | Zeile 183: | ||
| * The charges in the rod experience the Lorentz force $\vec{F}_{\rm L}$. | * 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 separated charges create a potential difference and by this a Coulomb force $\vec{F}_{\rm C}$ onto the charges within the rod. | + | * The separated charges create a potential difference and by this, a 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 215: | Zeile 233: | ||
| Furthermore, | Furthermore, | ||
| - | The situation of the single rod can be interpreted in the following way: We can calculate a motionally induced potential difference with Faraday’s law even when an actual closed circuit is not present. We simply imagine an enclosed area whose boundary includes the moving conductor, calculate $\Phi_m$, and then find the potential difference from Faraday’s law. For example, we can let the moving rod of <imgref ImgNr10> be one side of the imaginary rectangular area represented by the dashed lines. The area of the rectangle is $A = l \cdot x$, so the magnetic flux through it is $\Phi= B\cdot l \cdot x$. Differentiating this equation, we obtain | + | The situation of the single rod can be interpreted in the following way: We can calculate a motionally induced potential difference with Faraday’s law even when an actual closed circuit is not present. We simply imagine an enclosed area whose boundary includes the moving conductor, calculate $\Phi_{\rm m}$, and then find the potential difference from Faraday’s law. For example, we can let the moving rod of <imgref ImgNr10> be one side of the imaginary rectangular area represented by the dashed lines. The area of the rectangle is $A = l \cdot x$, so the magnetic flux through it is $\Phi= B\cdot l \cdot x$. Differentiating this equation, we obtain |
| \begin{align*} | \begin{align*} | ||
| Zeile 270: | Zeile 288: | ||
| From geometry, the area of the loop $\rm OPSO$ 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_{\rm 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*} U_{\rm ind} &= |{{\rm d}\over{{\rm d}t}} \cdot \Phi_{\rm 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} | \\ | ||
| + | | ||
| + | \end{align*} | ||
| When divided by the resistance $R$ of the loop, this yields 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_{\rm ind} &= {{|U_{\rm ind}|}\over{R}} \\ &= B\cdot {{r^2\omega}\over{2R}} \end{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 286: | 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 294: | 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 simplify quick. The induced potential difference is written out using Faraday’s law. </ | + | 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. </ |
| <button size=" | <button size=" | ||
| Zeile 300: | 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 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_{ind} &= U_{ind,0} \cdot \sin \omega t \end{align*} |
| - | where $U_{ind,0} = NBA\omega$. </ | + | where $U_{\rm ind,0} = NBA\omega$. </ |
| </ | </ | ||
| Zeile 316: | 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? | ||
| Zeile 326: | 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 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: | + | 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 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*} </ | + | \begin{align*} |
| + | u_{\rm ind} &= N B \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 $N=200$, $B=0.80~\rm T$ , $\varphi = 90°$ , $d\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 344: | 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 352: | Zeile 399: | ||
| ==== Linked Flux ==== | ==== Linked Flux ==== | ||
| - | 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)). | + | 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)). | ||
| < | < | ||
| Zeile 358: | Zeile 407: | ||
| The resulting electric voltage in such a situation is 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}} \\ | ||
| + | | ||
| + | \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 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}} \\ | ||
| + | | ||
| + | \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 376: | Zeile 434: | ||
| < | < | ||
| - | In order to understand 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> |
| < | < | ||
| Zeile 382: | 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. Besides 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*} | + | 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 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. | ||
| < | < | ||
| Zeile 418: | 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 &+ & | + | \begin{align*} |
| + | u_0 &= u_R &+ &u_L \\ | ||
| + | | ||
| + | | ||
| + | \end{align*} | ||
| ==== Inductance of different Components ==== | ==== Inductance of different Components ==== | ||
| Zeile 424: | Zeile 518: | ||
| === 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 analyzed in the last chapter(see [[: | + | The toroidal coil was analyzed in the last chapter(see [[: |
| + | Here, a rectangular intersection a assumed (see <imgref ImgNr16> | ||
| < | < | ||
| Zeile 438: | 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} & | ||
| + | &= {{ 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 458: | Zeile 564: | ||
| A change of magnetic flux is passing a coil with a single winding. The following pictures <imgref ImgNrEx01> | A change of magnetic flux is passing a coil with a single winding. The following pictures <imgref ImgNrEx01> | ||
| - | * 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 $u_{\rm ind}(t)$ |
| < | < | ||
| Zeile 465: | Zeile 571: | ||
| <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} |
| - | * $t= 1.5 ... 2.1s$: $u_{ind} = -{{0Vs}\over{0.6s}}= 0$ | + | * $t= 1.5 ... 2.1 ~\rm s$: $u_{\rm ind} = -{{0 ~\rm Vs}\over{0.6 ~\rm s}}= 0$ |
| </ | </ | ||
| Zeile 481: | Zeile 591: | ||
| <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 494: | Zeile 605: | ||
| <panel type=" | <panel type=" | ||
| - | A single winding is located in a homogenous 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 $u_{ind}(t)$, | + | * 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 angle between the surface vector of the winding and 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? |
| < | < | ||
| Zeile 503: | Zeile 615: | ||
| <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 514: | Zeile 626: | ||
| <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 homogeneous 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 | + | * 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 528: | 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: | + | 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 polarization 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 |
| + | So, the course of the voltage when entering or exiting is not uniquely given. | ||
| - | < | + | < |
| Zeile 561: | 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 series | - 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 ($\mu_{\rm r}=1000$) |
| </ | </ | ||
| Zeile 570: | Zeile 685: | ||
| <panel type=" | <panel type=" | ||
| - | A cylindrical air coil (length $l=40cm$, diameter $d=5cm$, and a 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 a number of windings $N=300$) passes a current of $30 ~\rm A$. The current shall be reduced linearly in $2 ~\rm ms$ down to $0 ~\rm A$. |
| - | What is the amount of the induced voltage $u_{ind}$? </ | + | What is the amount of the induced voltage $u_{\rm ind}$? </ |
| <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> |
| - | * What 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 $u_{\rm ind}(t)$! |
| < | < | ||