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electrical_engineering_and_electronics_1:block16 [2025/11/22 15:54] – mexleadminelectrical_engineering_and_electronics_1:block16 [2025/11/23 12:21] (aktuell) – mexleadmin
Zeile 33: Zeile 33:
 ===== Generalization of the Magnetic Field Strength ===== ===== Generalization of the Magnetic Field Strength =====
  
-So far, only the rotational symmetric problem on a single wire was considered in formula, when the current $I$ ant the length $s$ of a magnetic field line around it is given:+So far, only the rotational symmetric problem of a single wire was considered in formula. I.e a current $I$ and the length $s$ of a magnetic field line around the wire was given to calculate the magnetic field strength $H$:
  
 \begin{align*} \begin{align*}
Zeile 43: Zeile 43:
  
 \begin{align*} \begin{align*}
-U = E \cdot s \quad \quad | \quad \text{applies to capacitor only}+U = E \cdot s \quad \quad | \quad \text{applies to plate capacitor only}
 \end{align*} \end{align*}
  
Zeile 72: Zeile 72:
 Another new quantity is introduced: the **magnetic voltage $\theta$**: Another new quantity is introduced: the **magnetic voltage $\theta$**:
   - The magnetic voltage $\theta$ is the magnetic potential difference on a closed path.   - The magnetic voltage $\theta$ is the magnetic potential difference on a closed path.
-  - Since the magnetic voltage $\theta$ is valid for exactly __one turn__ along our single wire, $\theta$ is also equal to the current through the wire: \\ \begin{align*} \theta = \oint \vec{H} {\rm d}\vec{s} = I  \quad \quad | \quad \text{applies only to the long, straight conductor} \end{align*} +  - Since the magnetic voltage $\theta$ is valid for exactly __one turn__ along our single wire, $\theta$ is also equal to the current through the wire: \\ \begin{align*} \theta = H \cdot s = I  \quad \quad | \quad \text{applies only to the long, straight conductor} \end{align*} 
   - The  magnetic potential difference can take a fraction or a multiple of one turn and is therefore  **not mandatorily** equal to $I$.   - The  magnetic potential difference can take a fraction or a multiple of one turn and is therefore  **not mandatorily** equal to $I$.
   - The magnetic voltage is generalized in the following box.   - The magnetic voltage is generalized in the following box.
Zeile 81: Zeile 81:
 The magnetic voltage $\theta$ (and therefore the current) is the cause of the magnetic field strength. \\ The magnetic voltage $\theta$ (and therefore the current) is the cause of the magnetic field strength. \\
  
- 
-From the chapter [[block15]] the general representation of the current through a surface is known. \\ 
 This leads to the **{{wp>Ampere's Circuital Law}}** This leads to the **{{wp>Ampere's Circuital Law}}**
  
-|  \begin{align*} \boxed{\oint_{s} \vec{H} \cdot {\rm d} \vec{s} = \theta } \end{align*}| The magnetic voltage $\theta$ can be given as \\ (nbsp)(nbsp) •  $\theta = I \quad \quad \quad \ $ for a single conductor \\  (nbsp)(nbsp) • $\theta = N \cdot I \quad \:\; \, $ for a coil\\  (nbsp)(nbsp) • $\theta = \sum_n \cdot I_n \quad$ for multiple conductors\\  (nbsp)(nbsp) • $\theta = \iint_A \; \vec{S} {\rm d}\vec{A}$ for any spatial distribution|+|  \begin{align*} \boxed{\oint_{s} \vec{H} {\rm d} \vec{s} = \theta } \end{align*}| The magnetic voltage $\theta$ can be given as \\ (nbsp)(nbsp) •  $\theta = I \quad \quad \quad \ $ for a single conductor \\  (nbsp)(nbsp) • $\theta = N \cdot I \quad \:\; \, $ for a coil\\  (nbsp)(nbsp) • $\theta = \sum_n \cdot I_n \quad$ for multiple conductors\\  (nbsp)(nbsp) • $\theta = \iint_A \; \vec{S} {\rm d}\vec{A}$ for any spatial distribution (see [[block15]])|
  
-${\rm d}\vec{s}$ and ${\rm d}\vec{A}$ build a right-hand system. \\ +The unit of the magnetic voltage $\theta$ is **Ampere** (or **Ampere-turns**).  
 + 
 +In the english literature the magnetic voltage is called **{{wp>Magnetomotive force}}** 
 + 
 +</callout> 
 +<callout icon="fa fa-exclamation" color="red" title="Notice:"> 
 +${\rm d}\vec{s}$ and ${\rm d}\vec{A}$ in $\oint_{s} \vec{H} {\rm d} \vec{s} = \theta = \iint_A \; \vec{S} {\rm d}\vec{A}$ build a right-hand system. \\ 
   - Once the thumb of the right hand is pointing along ${\rm d}\vec{A}$, the fingers of the right hand show the correct direction for ${\rm d}\vec{s}$ for positive $\vec{H}$ and $\vec{S}$   - Once the thumb of the right hand is pointing along ${\rm d}\vec{A}$, the fingers of the right hand show the correct direction for ${\rm d}\vec{s}$ for positive $\vec{H}$ and $\vec{S}$
   - Currents into the direction of the right hand's thumb count positive. Currents antiparallel to it count negative.   - Currents into the direction of the right hand's thumb count positive. Currents antiparallel to it count negative.
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 </callout> </callout>
  
 +~~PAGEBREAK~~ ~~CLEARFIX~~
 +==== Recap of the fieldline images ====
  
 +<WRAP group><WRAP half column>
 +=== longitudinal coil ===
 +<WRAP>
 +<imgcaption BildNr04 | Magnetic field in a longitudinal coil></imgcaption> \\
 +{{url>https://www.falstad.com/vector3dm/vector3dm.html?f=SolenoidField&d=streamlines&sl=none&st=3&ld=5&a1=21&a2=30&a3=100&rx=63&ry=1&rz=2&zm=2.396 700,450 noborder}}
 +</WRAP>
  
 +A longitudinal coil can be seen in <imgref BildNr04>. \\ 
 +The magnetic field in a toroidal coil is often considered as homogenious in the inner volume, when the length $l$ is much larger than the diameter: $l \gg d$. \\
 +With a given number $N$ of windings, the magnetic field strength $H$ is
  
 +\begin{align*}
 +\theta = H \cdot l = N \cdot I
 +\end{align*}
 +\begin{align*}
 +\boxed{H = {{N \cdot I}\over{l}}}  \biggr | _\text{longitudinal coil}
 +\end{align*}
 +
 +</WRAP><WRAP half column>
 +=== toroidal coil ===
 +<WRAP>
 +<imgcaption BildNr05 | Magnetic field in a toroidal coil></imgcaption> \\
 +{{url>https://www.falstad.com/vector3dm/vector3dm.html?f=ToroidalSolenoidField&d=streamlines&sl=none&st=1&ld=8&a1=77&a2=26&a3=100&rx=0&ry=0&rz=0&zm=1.8 700,450 noborder}}
 +</WRAP>
 +
 +A toroidal coil has a donut-like setup. This can be seen in <imgref BildNr05>. \\ 
 +The toroidal coil is often defined by: 
 +  * The minor radius $r$: The radius  of the circular cross-section of the coil.
 +  * The major radius $R$: The distance from the center of the entire toroid (the center of the hole) to the center of the circular cross-section of the coil.
 +For reasons of symmetry, it shall get clear that the field lines form concentric circles. \\
 +Also the magnetic field strength $H$ in a toroidal coil is often considered as homogenious, when $R \gg r$. With a given number $N$ of windings, the magnetic field strength $H$ is
 +
 +\begin{align*}
 +\theta = H \cdot 2\pi R = N \cdot I 
 +\end{align*}
 +\begin{align*}
 +\boxed{H = {{N \cdot I}\over{2\pi R}}} \biggr | _\text{toroidal coil}
 +\end{align*}
 +
 +</WRAP></WRAP>
  
  
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 ===== Exercises ===== ===== Exercises =====
-==== Worked examples ==== 
  
-...+<panel type="info" title="Task 3.2.3 Magnetic Potential Difference"> <WRAP group><WRAP column 2%>{{fa>pencil?32}}</WRAP><WRAP column 92%> 
 + 
 +<WRAP> 
 +<imgcaption BildNr05 | different trajectories around current-carrying conductors> 
 +</imgcaption> 
 +{{drawio>Task3MagneticFieldCurrentFlowingConductor.svg}} \\ 
 +</WRAP> 
 + 
 +Given are the adjacent closed trajectories in the magnetic field of current-carrying conductors (see <imgref BildNr05>). Let $I_1 = 2~\rm A$ and $I_2 = 4.5~\rm A$ be valid. 
 + 
 +In each case, the magnetic potential difference $V_{\rm m}$ along the drawn path is sought. 
 + 
 + 
 +#@HiddenBegin_HTML~323100,Path~@# 
 + 
 +  * The magnetic potential difference is given as the **sum of the current through the area within a closed path**. 
 +  * The direction of the current and the path have to be considered with the righthand rule. 
 + 
 +#@HiddenEnd_HTML~323100,Path~@# 
 + 
 +#@HiddenBegin_HTML~323102,Result a)~@# 
 +a) $V_{\rm m,a} = - I_1 = - 2~\rm A$ \\ 
 +#@HiddenEnd_HTML~323102,Result~@# 
 + 
 +#@HiddenBegin_HTML~323103,Result b)~@# 
 +b) $V_{\rm m,b} = - I_2 = - 4.5~\rm A$ \\ 
 +#@HiddenEnd_HTML~323103,Result~@# 
 + 
 +#@HiddenBegin_HTML~323104,Result c)~@# 
 +c) $V_{\rm m,c} = 0 $ \\ 
 +#@HiddenEnd_HTML~323104,Result~@# 
 + 
 +#@HiddenBegin_HTML~323105,Result d)~@# 
 +d) $V_{\rm m,d} = + I_1 - I_2 = 2~\rm A - 4.5~\rm A = - 2.5~\rm A$ \\ 
 +#@HiddenEnd_HTML~323105,Result~@# 
 + 
 +#@HiddenBegin_HTML~323106,Result e)~@# 
 +e) $V_{\rm m,e} = + I_1 = + 2~\rm A$ \\ 
 +#@HiddenEnd_HTML~323106,Result~@# 
 + 
 +#@HiddenBegin_HTML~323107,Result f)~@# 
 +f) $V_{\rm m,f} = 2 \cdot (- I_1) = - 4~\rm A$ \\ 
 +#@HiddenEnd_HTML~323107,Result~@# 
 + 
 +</WRAP></WRAP></panel> 
 + 
 +{{page>electrical_engineering_and_electronics:task_jfzlmsucghsqvop5_with_calculation&nofooter}} 
 +{{page>electrical_engineering_and_electronics:task_kmp8r8y6lvwjnoc3_with_calculation&nofooter}} 
 +{{page>electrical_engineering_and_electronics:task_ddjurcpk494go2q1_with_calculation&nofooter}} 
  
 ===== Embedded resources ===== ===== Embedded resources =====