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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 [2024/04/16 01:34] mexleadmin [Bearbeiten - Panel] |
electrical_engineering_2:the_magnetostatic_field [2024/04/23 02:24] mexleadmin |
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Zeile 644: | Zeile 644: | ||
Explanation of diamagnetism and paramagnetism | Explanation of diamagnetism and paramagnetism | ||
- | {{youtube> | + | < |
+ | <WRAP column half>{{ youtube> | ||
+ | <WRAP column half>{{ youtube> | ||
+ | </ | ||
==== Ferromagnetic Materials ==== | ==== Ferromagnetic Materials ==== | ||
Zeile 743: | Zeile 746: | ||
\begin{align*} | \begin{align*} | ||
H(r) &= {{I_0}\over{2\pi \cdot r}} \\ | H(r) &= {{I_0}\over{2\pi \cdot r}} \\ | ||
- | &= {{100~\rm A}\over{2\pi \cdot 0.01 ~\rm m}} \\ | + | &= {{100~\rm A}\over{2\pi \cdot 0.1 ~\rm m}} \\ |
\end{align*} | \end{align*} | ||
Zeile 750: | Zeile 753: | ||
# | # | ||
\begin{align*} | \begin{align*} | ||
- | H(10~\rm cm) & | + | H(10~\rm cm) & |
- | \rightarrow H(10~\rm cm) & | + | \rightarrow H(10~\rm cm) & |
\end{align*} | \end{align*} | ||
Zeile 943: | Zeile 946: | ||
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*} | ||
+ | |||
+ | # | ||
</ | </ |