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Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
electrical_engineering_2:the_magnetostatic_field [2023/05/02 13:03] mexleadmin |
electrical_engineering_2:the_magnetostatic_field [2024/04/28 17:43] (aktuell) mexleadmin |
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Zeile 1: | Zeile 1: | ||
- | ====== 3. The magnetostatic Field ====== | + | ====== 3 The magnetostatic Field ====== |
< | < | ||
Zeile 464: | Zeile 464: | ||
* $\vec{B}$-Field on index finger | * $\vec{B}$-Field on index finger | ||
* Current $I$ on thumb (direction with length $\vec{l}$) | * Current $I$ on thumb (direction with length $\vec{l}$) | ||
+ | \\ \\ | ||
+ | < | ||
+ | < | ||
+ | \\ | ||
< | < | ||
< | < | ||
Zeile 470: | Zeile 473: | ||
{{drawio> | {{drawio> | ||
</ | </ | ||
- | |||
</ | </ | ||
+ | |||
+ | |||
==== Lorentz Law and Lorentz Force ==== | ==== Lorentz Law and Lorentz Force ==== | ||
Zeile 644: | Zeile 648: | ||
Explanation of diamagnetism and paramagnetism | Explanation of diamagnetism and paramagnetism | ||
- | {{youtube> | + | < |
+ | <WRAP column half>{{ youtube> | ||
+ | <WRAP column half>{{ youtube> | ||
+ | </ | ||
==== Ferromagnetic Materials ==== | ==== Ferromagnetic Materials ==== | ||
Zeile 725: | Zeile 732: | ||
<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 819: | ||
</ | </ | ||
- | 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 908: | ||
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 950: | ||
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 835: | Zeile 1072: | ||
<WRAP group> <WRAP half column> | <WRAP group> <WRAP half column> | ||
- | <quizlib id=" | + | <quizlib id=" |
< | < | ||
The right hand| | The right hand| |