Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
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electrical_engineering_2:the_magnetostatic_field [2023/03/16 13:53] 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 ====== |
| < | < | ||
| - | For this and the following chapter the online | + | The online |
| </ | </ | ||
| Zeile 30: | Zeile 30: | ||
| - From the iron ore should now first be separated a handy elongated part. If one is lucky, the given iron ore is already magnetic by itself. This case will be considered in the following. The elongated piece is now to be cut into two small pieces. | - From the iron ore should now first be separated a handy elongated part. If one is lucky, the given iron ore is already magnetic by itself. This case will be considered in the following. The elongated piece is now to be cut into two small pieces. | ||
| - As soon as the two pieces are removed from each other, one notices that the two pieces attract each other again directly at the cut surface. | - As soon as the two pieces are removed from each other, one notices that the two pieces attract each other again directly at the cut surface. | ||
| - | - If one of the two parts is turned (the upper part in the picture | + | - If one of the two parts is turned (the upper part in the picture below), a repulsive force acts on the two parts. |
| So it seems that there is a directed force around each of the two parts. If you dig a little deeper you will find that this force is focused on one part of the outer surface. | So it seems that there is a directed force around each of the two parts. If you dig a little deeper you will find that this force is focused on one part of the outer surface. | ||
| Zeile 76: | Zeile 76: | ||
| - | In 1820, Christian Ørsted discovered by chance during a lecture that current-carrying conductors also affect a compass. This experiment is illustrated in <imgref BildNr03> | + | In 1820, Christian Ørsted discovered by chance during a lecture that current-carrying conductors also affect a compass. This experiment is illustrated in <imgref BildNr03> |
| <callout icon=" | <callout icon=" | ||
| Zeile 207: | Zeile 207: | ||
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| {{url> | {{url> | ||
| Zeile 220: | Zeile 220: | ||
| ==== Magnetic Voltage ==== | ==== Magnetic Voltage ==== | ||
| - | The cause of the magnetic field is the current. As seen for the coil, sometimes the current has to be counted up ($N \cdot I$), e.g. by the number of windings of the coil. If this current $I$ and/or the number $N$ of windings is increased, the effect is amplified. To make this easier to handle, we introduce the **Magnetic Voltage**. The magnetic voltage $\theta$ is defined as | + | The cause of a magnetic field is a current. As seen for the coil, sometimes the current has to be counted up ($N \cdot I$), e.g. by the number of windings of the coil. If this current $I$ and/or the number $N$ of windings is increased, the effect is amplified. To make this easier to handle, we introduce the **Magnetic Voltage**. The magnetic voltage $\theta$ is defined as |
| \begin{align*} | \begin{align*} | ||
| 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 247: | Zeile 247: | ||
| </ | </ | ||
| - | In the English literature often the name **{{wp> | + | In the English literature often the name **{{wp> |
| ==== Magnetic Field Strength part 3: Generalization ==== | ==== Magnetic Field Strength part 3: Generalization ==== | ||
| Zeile 274: | Zeile 274: | ||
| Now, what is the difference between the magnetic potential difference $V_m$ and the magnetic voltage $\theta$? | Now, what is the difference between the magnetic potential difference $V_m$ and the magnetic voltage $\theta$? | ||
| - 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 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 ($V_m = H \cdot s$) 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. | + | - 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: | 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: | ||
| Zeile 295: | Zeile 295: | ||
| \begin{align*} | \begin{align*} | ||
| - | \boxed{\int_{closed \\ path} \vec{H} \cdot {\rm d} \vec{s} = \iint_{enclosed \\ surface | + | \boxed{\oint_{s} \vec{H} \cdot {\rm d} \vec{s} = \iint_A |
| \end{align*} | \end{align*} | ||
| Zeile 303: | Zeile 303: | ||
| * for a coil: $\theta = N \cdot I$ | * for a coil: $\theta = N \cdot I$ | ||
| * for multiple conductors: $\theta = \sum_n \cdot I_n$ | * for multiple conductors: $\theta = \sum_n \cdot I_n$ | ||
| - | * for spatial distribution: | + | * for spatial distribution: |
| * ${\rm d}\vec{s}$ 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 $\vec{S}$ | * ${\rm d}\vec{s}$ 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 $\vec{S}$ | ||
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| {{drawio> | {{drawio> | ||
| Zeile 314: | Zeile 314: | ||
| ==== Recap: Application of magnetic Field Strength ==== | ==== Recap: Application of magnetic Field Strength ==== | ||
| - | Ampere' | + | Ampere' |
| < | < | ||
| Zeile 322: | Zeile 322: | ||
| </ | </ | ||
| - | * The closed path $C$ is on a revolution of a field line in the center of the coil | + | * The closed path ${\rm s}$ is on a revolution of a field line in the center of the coil |
| * The surface $A$ is the enclosed surface | * The surface $A$ is the enclosed surface | ||
| Zeile 328: | Zeile 328: | ||
| \begin{align*} | \begin{align*} | ||
| - | \int_C \vec{H} \cdot {\rm d} \vec{s} &= \iint_{A} | + | \oint_s |
| \end{align*} | \end{align*} | ||
| Zeile 334: | Zeile 334: | ||
| \begin{align*} | \begin{align*} | ||
| - | \int_C H \cdot {\rm d}s &= \iint_{A} | + | \oint_s |
| \end{align*} | \end{align*} | ||
| Zeile 340: | Zeile 340: | ||
| \begin{align*} | \begin{align*} | ||
| - | \int_C H \cdot {\rm d}s &= N\cdot I | + | \oint_{\rm s} H \cdot {\rm d}s &= N\cdot I |
| \end{align*} | \end{align*} | ||
| Zeile 391: | Zeile 391: | ||
| In the last sub-chapter, | In the last sub-chapter, | ||
| - | In order to do so, the effect between two parallel conductors has to be examined closer. The experiment consists of a part $l$ of two very long((ideally: | + | To do so, the effect between two parallel conductors has to be examined closer. The experiment consists of a part $l$ of two very long((ideally: |
| < | < | ||
| Zeile 418: | Zeile 418: | ||
| Here $\mu$ is the magnetic permeability and for vacuum ({{wp> | Here $\mu$ is the magnetic permeability and for vacuum ({{wp> | ||
| \begin{align*} | \begin{align*} | ||
| - | \mu = \mu_0 = 4\pi \cdot 10^{-7} {{Vs}\over{Am}} = 1.257 \cdot 10^{-7} {{Vs}\over{Am}} | + | \mu = \mu_0 = 4\pi \cdot 10^{-7} {{\rm Vs}\over{\rm Am}} = 1.257 \cdot 10^{-7} {{\rm Vs}\over{\rm Am}} |
| \end{align*} | \end{align*} | ||
| Zeile 432: | Zeile 432: | ||
| \end{align*} | \end{align*} | ||
| - | The properties of the field from $I_2$ acting on $I_1$ are summarized to $B$ - the **magnetic flux density**. $B$ has the unit: | + | The properties of the field from $I_2$ acting on $I_1$ are summarized to $B$ - the **magnetic flux density**. |
| + | $B$ has the unit: | ||
| \begin{align*} | \begin{align*} | ||
| - | [B] &= {{[F]}\over{[I]\cdot[l]}} = 1 {{N}\over{Am}} = 1 {{{VAs}\over{m}}\over{Am}} = 1 {{Vs}\over{m^2}} \\ | + | [B] &= {{[F]}\over{[I]\cdot[l]}} = 1 \rm {{N}\over{Am}} = 1 {{{VAs}\over{m}}\over{Am}} = 1 {{Vs}\over{m^2}} \\ |
| - | & | + | & |
| \end{align*} | \end{align*} | ||
| - | This formula can be generalized with the knowledge of the directions of the conducting wire $\vec{l}$, the magnetic field strength $\vec{B}$ and the force $\vec{F}$ | + | This formula can be generalized with the knowledge of the directions of the conducting wire $\vec{l}$, the magnetic field strength $\vec{B}$ and the force $\vec{F}$ |
| \begin{align*} | \begin{align*} | ||
| Zeile 447: | Zeile 448: | ||
| \begin{align*} | \begin{align*} | ||
| - | \boxed{|\vec{F_L}| = I \cdot |l| \cdot |B| \cdot sin(\angle \vec{l}, | + | \boxed{|\vec{F_L}| = I \cdot |l| \cdot |B| \cdot \sin(\angle \vec{l}, |
| \end{align*} | \end{align*} | ||
| Zeile 463: | 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 469: | Zeile 473: | ||
| {{drawio> | {{drawio> | ||
| </ | </ | ||
| - | |||
| </ | </ | ||
| + | |||
| + | |||
| ==== Lorentz Law and Lorentz Force ==== | ==== Lorentz Law and Lorentz Force ==== | ||
| The true Lorentz force is not the force on the whole conductor but the single force onto an (elementary) charge. \\ | The true Lorentz force is not the force on the whole conductor but the single force onto an (elementary) charge. \\ | ||
| - | In order to find this force the previous force onto a conductor can be used as a start. However, the formula will be investigated infinitesimally for small parts ${\rm d} \vec{l}$ of the conductor: | + | To find this force the previous force onto a conductor can be used as a start. However, the formula will be investigated infinitesimally for small parts ${\rm d} \vec{l}$ of the conductor: |
| \begin{align*} | \begin{align*} | ||
| Zeile 481: | Zeile 486: | ||
| \end{align*} | \end{align*} | ||
| - | The current is now substituted by $I = dQ/dt$, where $dQ$ is the small charge packet in the length $\vec{dl}$ of the conductor. | + | The current is now substituted by $I = {\rm d}Q/{\rm d}t$, where ${\rm d}Q$ is the small charge packet in the length $\vec{{\rm d}l}$ of the conductor. |
| \begin{align*} | \begin{align*} | ||
| Zeile 515: | Zeile 520: | ||
| </ | </ | ||
| - | Please have a look at the German contents (text, videos, exercises) on the page of the [[https:// | + | Please have a look at the German contents (text, videos, exercises) on the page of the [[https:// |
| ===== 3.4 Matter in the Magnetic Field ===== | ===== 3.4 Matter in the Magnetic Field ===== | ||
| Zeile 568: | Zeile 573: | ||
| \begin{align*} | \begin{align*} | ||
| - | \mu &= \mu_0 \cdot \mu_r | + | \mu &= \mu_0 \cdot \mu_{\rm r} |
| \end{align*} | \end{align*} | ||
| Zeile 589: | Zeile 594: | ||
| * Diamagnetic materials weaken the magnetic field, compared to the vacuum. | * Diamagnetic materials weaken the magnetic field, compared to the vacuum. | ||
| * The weakening is very low (see <tabref tab01>). | * The weakening is very low (see <tabref tab01>). | ||
| - | * For diamagnetic materials applies $0<\mu_r< | + | * For diamagnetic materials applies $0<\mu_{\rm r}< |
| * The principle behind the effect is based on quantum mechanics (see <imgref BildNr22> | * The principle behind the effect is based on quantum mechanics (see <imgref BildNr22> | ||
| * Without the external field no counteracting field is generated by the matter. | * Without the external field no counteracting field is generated by the matter. | ||
| Zeile 598: | Zeile 603: | ||
| < | < | ||
| < | < | ||
| - | ^ Material | + | ^ Material |
| - | | Antimon | + | | Antimon |
| - | | Copper | + | | Copper |
| - | | Mercury | + | | Mercury |
| - | | Silver | + | | Silver |
| - | | Water | $H_2O$ | + | | Water |
| - | | Bismut | + | | Bismut |
| </ | </ | ||
| </ | </ | ||
| Zeile 643: | 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 659: | Zeile 667: | ||
| * Non-magnetized ferromagnets are located in the origin. | * Non-magnetized ferromagnets are located in the origin. | ||
| * With an external field $H$ the initial magnetization curve (in German: // | * With an external field $H$ the initial magnetization curve (in German: // | ||
| - | * Even without an external field ($H=0$) and the internal field is stable. \\ The stored field without external field is called **remanence** $B(H=0) = B_R$ (or remanent magnetization). | + | * Even without an external field ($H=0$) and the internal field is stable. \\ The stored field without external field is called **remanence** $B(H=0) = B_{\rm R}$ (or remanent magnetization). |
| - | * In order to eliminate the stored field the counteracting **coercive field strength** $H_C$ (also called coercivity) has to be applied. | + | * In order to eliminate the stored field the counteracting **coercive field strength** $H_{\rm C}$ (also called coercivity) has to be applied. |
| * The **saturation flux density** $B_{\rm sat}$ is the maximum possible magnetic flux density (at the maximum possible field strength $H_{\rm sat}$) | * The **saturation flux density** $B_{\rm sat}$ is the maximum possible magnetic flux density (at the maximum possible field strength $H_{\rm sat}$) | ||
| Zeile 677: | Zeile 685: | ||
| * high permeability | * high permeability | ||
| - | * high saturation flux density $B_{sat}$ (= good magnetic conductor) | + | * high saturation flux density $B_{\rm sat}$ (= good magnetic conductor) |
| - | * small coercive field strength $H_C < 1 ~\rm kA/m$ (= easy to reverse the magnetization) | + | * small coercive field strength $H_{\rm C} < 1 ~\rm kA/m$ (= easy to reverse the magnetization) |
| * low magnetic losses for reversing the magnetization | * low magnetic losses for reversing the magnetization | ||
| * Important materials: ferrites | * Important materials: ferrites | ||
| Zeile 687: | Zeile 695: | ||
| * {{wp> | * {{wp> | ||
| - | === magnetically | + | === Magnetically |
| * low permeability | * low permeability | ||
| Zeile 693: | Zeile 701: | ||
| * high coercive field strength $H_{\rm C}$ up to $2' | * high coercive field strength $H_{\rm C}$ up to $2' | ||
| * high magnetic losses for reversing the magnetization | * high magnetic losses for reversing the magnetization | ||
| - | * Important materials: ferrites, alloys of iron and cobalt: $\rm AlNiCo$((Aluminum-Nickel-Cobalt)), | + | * Important materials: ferrites, alloys of iron and cobalt: $\rm AlNiCo$((Aluminum-Nickel-Cobalt)), |
| Applications: | Applications: | ||
| Zeile 724: | 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 $r_1 = 10cm$ 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 $r_2 = 3mm$ from the conductor axis? | ||
| </ | </ | ||
| Zeile 739: | 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 $H(P)$ 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 $H(P)$ change? | ||
| </ | </ | ||
| Zeile 753: | Zeile 905: | ||
| </ | </ | ||
| - | Given are the adjacent closed trajectories in the magnetic field of current-carrying conductors (see <imgref BildNr05> | + | Given are the adjacent closed trajectories in the magnetic field of current-carrying conductors (see <imgref BildNr05> |
| 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 766: | 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 812: | Zeile 1050: | ||
| Additionally, | Additionally, | ||
| - | Therefore, it leads to: | + | Therefore, it leads to the following: |
| \begin{align*} | \begin{align*} | ||
| Zeile 834: | Zeile 1072: | ||
| <WRAP group> <WRAP half column> | <WRAP group> <WRAP half column> | ||
| - | <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 851: | Zeile 1114: | ||
| ++++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 861: | Zeile 1124: | ||
| ++++ | ++++ | ||
| - | ++++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. | ||