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--- electrical_engineering_and_electronics_1:block16 [2025/11/22 15:13] – mexleadmin
+++ electrical_engineering_and_electronics_1:block16 [2025/11/23 12:21] (aktuell) – mexleadmin
@@ Zeile 33: / Zeile 33: @@
 ===== 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*}
@@ Zeile 43: / Zeile 43: @@
 \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*}
@@ Zeile 57: / Zeile 57: @@
 We can now try to look for similarities. Also for the magnetic field, the magnitude of the field strength is summed up along a path to arrive at another field-describing quantity. \\
-Because of the similarity the so-called **magnetic potential difference $V_m$** is introduced:
+Because of the similarity the so-called **magnetic potential difference $V_m$** between point $1$ and $2$ is introduced:
 \begin{align*}
@@ Zeile 70: / Zeile 70: @@
 We need to take a loser look here. Any closed path in the static electric field leads to a potential difference of $U = \oint \vec{E} \; {\rm d}\vec{s} =0$. \\
 BUT: closed paths in the static magnetic field leads to a magnetic potential difference which is **not mandatorily** $0$! $ V_m = \oint \vec{H} {\rm d}\vec{s} = \theta$ \\ \\
-Another new quantity is introduced: the **magnetic voltage $\theta$**. The magnetic voltage is the magnetic potential difference on a closed path.
+Another new quantity is introduced: the **magnetic voltage $\theta$**:
+  - 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 = 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 voltage is generalized in the following box.
+<callout icon="fa fa-exclamation" color="red" title="Notice:">
-Now, what is the difference between the magnetic potential difference $V_m$ and the magnetic voltage $\theta$?
+The path integral of the magnetic field strength along an arbitrary closed path is equal to the free currents (= current density) through the surface enclosed by the path.
-  - The first equation of the toroidal coil ($\theta = H \cdot l$) is valid for exactly __one turn__ along a field line with the length $l$. In addition, the magnetic voltage was equal to the current times the number of windings: $\theta = N \cdot I$.
-  - The second equation ($V_{\rm m} = H \cdot s$) is independent of the length of the field line $l$. Only if $s = l$ is chosen, the magnetic voltage equals the magnetic potential difference. The path length $s$ can be a fraction or multiple of a single revolution $l$ for the magnetic potential difference.
-Thus, for each infinitesimally small path ${\rm d}s$ along a field line, the resulting infinitesimally small magnetic potential difference ${\rm d}V_{\rm m} = H \cdot {\rm d}s$ can be determined. If now along the field line the magnetic field strength $H = H(\vec{s})$ changes, then the magnetic potential difference from point $\vec{s_1}$ to point $\vec{s_2}$ results to:
+The magnetic voltage $\theta$ (and therefore the current) is the cause of the magnetic field strength. \\
-\begin{align*}
+This leads to the **{{wp>Ampere's Circuital Law}}**
-V_{\rm m12} = V_{\rm m}(\vec{s_1}, \vec{s_2})
-            = \int_\vec{s_1}^\vec{s_2} H(\vec{s}) {\rm d}s
-\end{align*}
-Up to now, only the situation was considered that one always walks along one single field line. $\vec{s}$ therefore always arrived at the same spot of the field line.
+|  \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]])|
-If one wants to extend this to arbitrary directions (also perpendicular to field lines), then only that part of the magnetic field strength $\vec{H}$ may be used in the formula, which is parallel to the path ${\rm d} \vec{s}$. This is made possible by scalar multiplication. Thus, it is generally valid:
-\begin{align*}
+The unit of the magnetic voltage $\theta$ is **Ampere** (or **Ampere-turns**).
-\boxed{V_{\rm m12} = \int_\vec{s_1}^\vec{s_2} \vec{H} \cdot {\rm d} \vec{s}}
-\end{align*}
-The magnetic voltage $\theta$ (and therefore the current) is the cause of the magnetic field strength.
+In the english literature the magnetic voltage is called **{{wp>Magnetomotive force}}**
-From the chapter [[electrical_engineering_2:The stationary Electric Flow]] the general representation of the current through a surface is known.
-This leads to the **{{wp>Ampere's Circuital Law}}**
-\begin{align*}
+</callout>
-\boxed{\oint_{s} \vec{H} \cdot {\rm d} \vec{s} = \iint_A \; \vec{S} {\rm d}\vec{A} = \theta}
+<callout icon="fa fa-exclamation" color="red" title="Notice:">
-\end{align*}
+${\rm d}\vec{s}$ 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 $\vec{S}$
-  * The path integral of the magnetic field strength along an arbitrary closed path is equal to the free currents (= current density) through the surface enclosed by the path.
+  - Currents into the direction of the right hand's thumb count positive. Currents antiparallel to it count negative.
-  * The magnetic voltage $\theta$ can be given as
-    * for a single conductor: $\theta = I$
-    * for a coil: $\theta = N \cdot I$
-    * for multiple conductors: $\theta = \sum_n \cdot I_n$
-    * for spatial distribution: $\theta = \iint_A \; \vec{S} {\rm d}\vec{A}$
-  * ${\rm d}\vec{s}$ and ${\rm d}\vec{A}$ 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 $\vec{S}$
 <WRAP>
@@ Zeile 111: / Zeile 100: @@
 {{drawio>Righthandrule.svg}}
 </WRAP>
+</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>
@@ Zeile 117: / Zeile 152: @@
 ===== 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 =====