Unterschiede
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Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung Nächste Überarbeitung Beide Seiten der Revision | ||
electrical_engineering_2:the_magnetostatic_field [2023/04/09 07:41] ott |
electrical_engineering_2:the_magnetostatic_field [2024/04/16 18:49] mexleadmin [Bearbeiten - Panel] |
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Zeile 1: | Zeile 1: | ||
- | ====== 3. The magnetostatic Field ====== | + | ====== 3 The magnetostatic Field ====== |
< | < | ||
Zeile 207: | Zeile 207: | ||
< | < | ||
- | < | + | < |
</ | </ | ||
{{url> | {{url> | ||
Zeile 236: | Zeile 236: | ||
<callout icon=" | <callout icon=" | ||
- | * For the sign of the magnetic voltage, one has to consider the orientation of the current and way on the enclosing path. The <imgref BildNr05> | + | * For the sign of the magnetic voltage, one has to consider the orientation of the current and way on the enclosing path. The <imgref BildNr05> |
* This is again given as the right-hand rule (see <imgref BildNr76> | * This is again given as the right-hand rule (see <imgref BildNr76> | ||
Zeile 599: | Zeile 599: | ||
< | < | ||
< | < | ||
- | ^ Material | + | ^ Material |
- | | Antimon | + | | Antimon |
- | | Copper | + | | Copper |
- | | Mercury | + | | Mercury |
- | | Silver | + | | Silver |
- | | Water | $H_2O$ | + | | Water |
- | | Bismut | + | | Bismut |
</ | </ | ||
</ | </ | ||
Zeile 688: | Zeile 688: | ||
* {{wp> | * {{wp> | ||
- | === magnetically | + | === Magnetically |
* low permeability | * low permeability | ||
Zeile 725: | Zeile 725: | ||
<panel type=" | <panel type=" | ||
- | The current $I = 100~\rm A$ flows in a long straight conductor with a round cross-section. The radius of the conductor is $r_{\rm L}= 4~\rm mm$. | + | The current $I_0 = 100~\rm A$ flows in a long straight conductor with a round cross-section. |
+ | The conductor shall have constant electric properties everywhere. | ||
+ | The radius of the conductor is $r_{\rm L}= 4~\rm mm$. | ||
+ | |||
+ | 1. What is the magnetic field strength $H_1$ at a point $P_1$, which is __outside__ the conductor at a distance of $r_1 = 10~\rm cm$ from the conductor axis? | ||
+ | |||
+ | # | ||
+ | |||
+ | * The $H$-field is given as: the current $I$ through an area divided by the " | ||
+ | * The relevant current is the given $I_0$. | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | |||
+ | The $H$-field is given as: | ||
+ | \begin{align*} | ||
+ | H(r) &= {{I_0}\over{2\pi \cdot r}} \\ | ||
+ | &= {{100~\rm A}\over{2\pi \cdot 0.1 ~\rm m}} \\ | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | H(10~\rm cm) &= 159.15... ~\rm{{A}\over{m}} \\ | ||
+ | \rightarrow H(10~\rm cm) &= 159 ~\rm{{A}\over{m}} | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | 2. What is the magnetic field strength $H_2$ at a point $P_2$, which is __inside__ the conductor at a distance of $r_2 = 3~\rm mm$ from the conductor axis? | ||
+ | |||
+ | # | ||
+ | |||
+ | * Again, the $H$-field is given as: the current $I$ through an area divided by the " | ||
+ | * Here, the relevant current is **not** the given one. There is only a fraction of the current flowing through the part of the conductor inside the $r_2$ | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | |||
+ | The $H$-field is given as: | ||
+ | \begin{align*} | ||
+ | H(r) &= {{I}\over{2\pi \cdot r}} | ||
+ | \end{align*} | ||
+ | |||
+ | But now $I$ is not $I_0$ anymore, but only a fraction, so $\Delta I$. | ||
+ | $I_0$ is evenly distributed over the cross-section $A$ of the conductor. | ||
+ | The cross-sectional area is given as $A= r^2 \cdot \pi$ | ||
+ | |||
+ | So the current $\Delta I$ is given as: current divided by the full area and then times the fractional area: | ||
+ | \begin{align*} | ||
+ | \Delta I &= I_0 \cdot {{r_2^2 \cdot \pi}\over{r_{\rm L}^2 \cdot \pi}} \\ | ||
+ | & | ||
+ | \end{align*} | ||
+ | |||
+ | Therefore, the $H$-field is: | ||
+ | \begin{align*} | ||
+ | H(r) &= {{\Delta I}\over{2\pi \cdot r_2}} | ||
+ | && | ||
+ | & | ||
+ | && | ||
+ | \end{align*} | ||
+ | |||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | H(3~\rm mm) &= 2984.1... ~\rm{{A}\over{m}} \\ | ||
+ | \rightarrow H(3~\rm mm) &= 3.0 ~\rm{{kA}\over{m}} | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
- | * What is the magnetic field strength $H_1$ at a point $P_1$, which is __outside__ the conductor at a distance of $r_1 = 10~\rm cm$ from the conductor axis? | ||
- | * What is the magnetic field strength $H_2$ at a point $P_2$, which is __inside__ the conductor at a distance of $r_2 = 3~\rm mm$ from the conductor axis? | ||
</ | </ | ||
Zeile 740: | Zeile 812: | ||
</ | </ | ||
- | Three long straight conductors are arranged in a vacuum | + | Three long straight conductors are arranged in a vacuum |
+ | |||
+ | 1. What is the magnetic field strength $H({\rm P})$ at the center of the equilateral triangle? | ||
+ | |||
+ | # | ||
+ | |||
+ | * The formula for a single wire can calculate the field of a single conductor. | ||
+ | * For the resulting field, the single wire fields have to be superimposed. | ||
+ | * Since it is symmetric the resulting field has to be neutral. | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | |||
+ | In general, the $H$-field of the single conductor is given as: | ||
+ | \begin{align*} | ||
+ | H &= {{I}\over{2\pi \cdot r}} \\ | ||
+ | &= {{2~\rm A}\over{2\pi \cdot 0.02 ~\rm m}} \\ | ||
+ | \end{align*} | ||
+ | |||
+ | * However, even without calculation, | ||
+ | * By the symmetry of the conductor, the angles of the $H$-field vectors are defined and evenly distributed on the revolution: | ||
+ | < | ||
+ | < | ||
+ | </ | ||
+ | {{drawio> | ||
+ | </ | ||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | H &= 0 ~\rm{{A}\over{m}} | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | 2. Now, the current in one of the conductors is reversed. To which value does the magnetic field strength $H({\rm P})$ change? | ||
+ | |||
+ | # | ||
+ | |||
+ | * Now, the formula for a single wire has to be used to calculate the field of a single conductor. | ||
+ | * For the resulting field, the single wire fields again have to be superimposed. | ||
+ | * The symmetry and the result of question 1 give a strong hint about how much stronger the resulting field has to be compared to the field of the reversed one. | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | |||
+ | The $H$-field of the single reversed conductor $I_3$ is given as: | ||
+ | \begin{align*} | ||
+ | H(I_3) &= {{I}\over{2\pi \cdot r}} \\ | ||
+ | &= {{2~\rm A}\over{2\pi \cdot 0.02 ~\rm m}} \\ | ||
+ | \end{align*} | ||
+ | |||
+ | Once again, one can try to sketch the situation of the field vectors: | ||
+ | < | ||
+ | < | ||
+ | </ | ||
+ | {{drawio> | ||
+ | </ | ||
+ | |||
+ | Therefore, it is visible, that the resulting field is twice the value of $H(I_3)$: \\ | ||
+ | The vectors of $H(I_1)$ plus $H(I_2)$ had in the task 1 just the length of $H(I_3)$. | ||
+ | |||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | H &= 31.830... ~\rm{{A}\over{m}} \\ | ||
+ | \rightarrow H &= 31.8 ~\rm{{A}\over{m}} \\ | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
- | - What is the magnetic field strength $H({\rm P})$ at the center of the equilateral triangle? | ||
- | - Now, the current in one of the conductors is reversed. To which value does the magnetic field strength $H({\rm P})$ change? | ||
</ | </ | ||
Zeile 757: | Zeile 901: | ||
In each case, the magnetic voltage $V_{\rm m}$ along the drawn path is sought. | In each case, the magnetic voltage $V_{\rm m}$ along the drawn path is sought. | ||
+ | |||
+ | |||
+ | # | ||
+ | |||
+ | * The magnetic voltage 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. | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | a) $\theta_\rm a = - I_1 = - 2~\rm A$ \\ | ||
+ | # | ||
+ | |||
+ | # | ||
+ | b) $\theta_\rm b = - I_2 = - 4.5~\rm A$ \\ | ||
+ | # | ||
+ | |||
+ | # | ||
+ | c) $\theta_\rm c = 0 $ \\ | ||
+ | # | ||
+ | |||
+ | # | ||
+ | d) $\theta_\rm d = + I_1 - I_2 = 2~\rm A - 4.5~\rm A = - 2.5~\rm A$ \\ | ||
+ | # | ||
+ | |||
+ | # | ||
+ | e) $\theta_\rm e = + I_1 = + 2~\rm A$ \\ | ||
+ | # | ||
+ | |||
+ | # | ||
+ | f) $\theta_\rm f = 2 \cdot (- I_1) = - 4~\rm A$ \\ | ||
+ | # | ||
</ | </ | ||
Zeile 767: | Zeile 943: | ||
A $\rm NdFeB$ magnet can show a magnetic flux density up to $1.2 ~\rm T$ near the surface. | A $\rm NdFeB$ magnet can show a magnetic flux density up to $1.2 ~\rm T$ near the surface. | ||
- | - For comparison, the same flux density shall be created | + | 1. For comparison, the same flux density shall be created inside a toroidal coil with $10' |
- | - What would be the current $I_{\rm Fe}$, when a iron core with $\varepsilon_{\rm Fe,r} = 10' | + | |
+ | # | ||
+ | |||
+ | | ||
+ | * The $H$-field is given as: the current $I$ through an area divided by the " | ||
+ | * The current is number of windings times $I$. | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | |||
+ | The $B$-field is given as: | ||
+ | \begin{align*} | ||
+ | B &= \mu \cdot H \\ | ||
+ | &= \mu \cdot {{I \cdot N}\over{l}} \\ | ||
+ | \end{align*} | ||
+ | |||
+ | This can be rearranged to the current $I$: | ||
+ | \begin{align*} | ||
+ | I &= {{B \cdot l}\over{\mu \cdot N}} \\ | ||
+ | &= {{1.2 ~\rm T \cdot 1 ~\rm m}\over{4\pi\cdot 10^-7 {\rm{Vs}\over{Am}} | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | I &= 95.49... ~\rm A \\ | ||
+ | \rightarrow I &= 95.5 ~\rm A | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | 2. What would be the current $I_{\rm Fe}$, when a iron core with $\varepsilon_{\rm Fe,r} = 10' | ||
+ | |||
+ | |||
+ | # | ||
+ | |||
+ | Now $\mu$ has to be given as $\mu_r \cdot \mu_0$: | ||
+ | |||
+ | This can be rearranged to the current $I$: | ||
+ | \begin{align*} | ||
+ | I &= {{B \cdot l}\over{\mu \cdot N}} \\ | ||
+ | &= {{1.2 ~\rm T \cdot 1 ~\rm m}\over{10' | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
+ | |||
+ | # | ||
+ | \begin{align*} | ||
+ | I &= 0.009549... ~\rm A \\ | ||
+ | \rightarrow I &= 9.55 ~\rm mA | ||
+ | \end{align*} | ||
+ | |||
+ | # | ||
</ | </ | ||
Zeile 813: | Zeile 1043: | ||
Additionally, | Additionally, | ||
- | Therefore, it leads to: | + | Therefore, it leads to the following: |
\begin{align*} | \begin{align*} | ||
Zeile 836: | Zeile 1066: | ||
<quizlib id=" | <quizlib id=" | ||
- | | + | < |
- | < | + | The right hand| |
- | < | + | The left hand |
- | < | + | </ |
- | < | + | < |
- | < | + | Thumb for current direction, remaining fingers for magnetic field direction | |
+ | Thumb for magnetic field direction, remaining fingers for current direction | | ||
+ | both possibilities are correct | ||
+ | </ | ||
+ | < | ||
+ | none | | ||
+ | The conductors attract | | ||
+ | The conductors repel | ||
+ | </ | ||
+ | < | ||
+ | none | | ||
+ | The conductors attract | | ||
+ | The conductors repel | ||
+ | </ | ||
+ | < | ||
+ | from the magnetic north pole to the south pole | | ||
+ | from the magnetic south pole to the north pole | | ||
+ | the inside is free of field | ||
+ | </ | ||
+ | < | ||
+ | at the magnetic north pole | | ||
+ | at the magnetic south pole | | ||
+ | inside the coil | | ||
+ | at both poles | ||
+ | </ | ||
</ | </ | ||
</ | </ | ||
- | ++++Tip | + | |
+ | ++++Tip | ||
For the current, you use which hand? | For the current, you use which hand? | ||
++++ | ++++ | ||
Zeile 852: | Zeile 1107: | ||
++++Tip for 2| | ++++Tip for 2| | ||
* Imagine a coil with a winding pictorially, | * Imagine a coil with a winding pictorially, | ||
- | * Now think of a generated field through this to it. What direction must the current flow, that cause the field? Does this fit the rule of thumb? | + | * Now think of a generated field through this to it. What direction must the current flow, that causes |
* Then try it the other way round: If a current is given, where do the field lines go in and where out? What poles are created there? | * Then try it the other way round: If a current is given, where do the field lines go in and where out? What poles are created there? | ||
++++ | ++++ | ||
Zeile 862: | Zeile 1117: | ||
++++ | ++++ | ||
- | ++++Tip | + | ++++Tip |
* First imagine the parallel wires again. What happens when the current flows in the same direction and what happens when the current flows in opposite directions? Are the resulting forces equal in magnitude? | * First imagine the parallel wires again. What happens when the current flows in the same direction and what happens when the current flows in opposite directions? Are the resulting forces equal in magnitude? | ||
* The reversal of the direction of the current can now also be produced by turning the wire instead of changing the current - just so that the wires are perpendicular to each other in the meantime when turning. | * The reversal of the direction of the current can now also be produced by turning the wire instead of changing the current - just so that the wires are perpendicular to each other in the meantime when turning. |