Unterschiede
Hier werden die Unterschiede zwischen zwei Versionen angezeigt.
| Beide Seiten der vorigen Revision Vorhergehende Überarbeitung Nächste Überarbeitung | Vorhergehende Überarbeitung | ||
| electrical_engineering_2:the_magnetostatic_field [2022/04/20 22:27] – tfischer | electrical_engineering_2:the_magnetostatic_field [2025/04/29 02:45] (aktuell) – mexleadmin | ||
|---|---|---|---|
| Zeile 1: | Zeile 1: | ||
| - | ====== 3. The magnetostatic Field ====== | + | ====== 3 The magnetostatic Field ====== |
| < | < | ||
| - | For this and the following chapter the online | + | The online |
| </ | </ | ||
| Zeile 8: | Zeile 8: | ||
| < | < | ||
| - | === Goals === | + | === Learning Objectives |
| - | After this lesson | + | By the end of this section, |
| - Know that forces act between magnetic poles and know the direction of the forces. | - Know that forces act between magnetic poles and know the direction of the forces. | ||
| - Know that a magnetic field is formed around a current-carrying conductor. | - Know that a magnetic field is formed around a current-carrying conductor. | ||
| Zeile 20: | Zeile 20: | ||
| < | < | ||
| - | < | + | < |
| + | {{drawio> | ||
| </ | </ | ||
| - | First permanent magnets made of magnetic magnetite ($Fe_{3} O_{4}$) were found in Greece in the region around Magnesia. Besides the iron materials, other elements also show a similar " | + | First, permanent magnets made of magnetic magnetite ($\rm Fe_{3} O_{4}$) were found in Greece in the region around Magnesia. Besides the iron materials, other elements also show a similar " |
| Here now the " | Here now the " | ||
| - | - From the iron ore should now first be separated a handy elongated part. If one is lucky, the found 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. | ||
| - | Of course you already know magnets and also know that there are poles. The considered thought experiment shall clarify, how one could have proceeded at an unknown appearance. In further thought experiments, | + | Of course, you already know magnets and also know that there are poles. The considered thought experiment shall clarify, how one could have proceeded at an unknown appearance. In further thought experiments, |
| ~~PAGEBREAK~~~~CLEARFIX~~ | ~~PAGEBREAK~~~~CLEARFIX~~ | ||
| The result here is: | The result here is: | ||
| - | - There are two poles. These are called the north pole and the south pole. The north pole is coloured | + | - There are two poles. These are called the north pole and the south pole. The north pole is colored |
| - Poles with the same name repel each other. Unequal poles attract each other. This is similar to the electric field (opposite charges attract). | - Poles with the same name repel each other. Unequal poles attract each other. This is similar to the electric field (opposite charges attract). | ||
| - So magnets experience a force in the vicinity of other magnets. | - So magnets experience a force in the vicinity of other magnets. | ||
| - A compass is a small rotating " | - A compass is a small rotating " | ||
| - The naming of the magnetic poles was done by the part of the compass which points to the geographic north pole. The reason for this is that the magnetic south pole is found at the geographic north pole. | - The naming of the magnetic poles was done by the part of the compass which points to the geographic north pole. The reason for this is that the magnetic south pole is found at the geographic north pole. | ||
| - | - Magnetic poles are not isolatable. Even the smallest fraction of a magnet shows either no magnetism, or both north and south poles. | + | - Magnetic poles are not isolatable. Even the smallest fraction of a magnet shows either no magnetism or both north and south poles. |
| < | < | ||
| - | < | + | < |
| + | {{drawio> | ||
| </ | </ | ||
| - | An interesting aspect is that even non-magnetised, ferromagnetic materials experience a force effect in the magnetic field. A non-magnetic nail is attracted by a permanent magnet. This even happens independently of the magnetic pole. This also explains the visualization about iron filings (= small ferromagnetic parts), see <imgref BildNr02> | + | Interestingly, |
| <callout icon=" | <callout icon=" | ||
| Zeile 69: | Zeile 71: | ||
| < | < | ||
| - | < | + | < |
| + | {{drawio> | ||
| </ | </ | ||
| - | In 1820, Christian Ørsted discovered by chance during a lecture that current-carrying conductors also have an effect on 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 |
| <callout icon=" | <callout icon=" | ||
| - | * If the technical direction of current is considered, the magnetic field lines surround the current in the sense of a right-hand screw. (" | + | * If the technical direction of the current is considered, the magnetic field lines surround the current in the sense of a right-hand screw. (" |
| * This rule can also be remembered in another way: If the thumb of the **__r__**ight hand points in the (technical) cu**__rr__**ent direction, the fingers of the hand surround the conductor like the magnetic field lines. Likewise, if the thumb of the **__l__**eft hand points in the E**__l__**ectron flow direction, the fingers of the hand surround the conductor like the magnetic field lines. | * This rule can also be remembered in another way: If the thumb of the **__r__**ight hand points in the (technical) cu**__rr__**ent direction, the fingers of the hand surround the conductor like the magnetic field lines. Likewise, if the thumb of the **__l__**eft hand points in the E**__l__**ectron flow direction, the fingers of the hand surround the conductor like the magnetic field lines. | ||
| </ | </ | ||
| Zeile 86: | Zeile 89: | ||
| ^ Property | ^ Property | ||
| - | | Field line images | + | | Field line images |
| | sample for the field | positive sample charge | | sample for the field | positive sample charge | ||
| | field lines | - start on a positive charge | | field lines | - start on a positive charge | ||
| Zeile 97: | Zeile 100: | ||
| < | < | ||
| - | === Goals === | + | === Learning Objectives |
| - | After this lesson, you should: | + | By the end of this section, you will be able to: |
| - know the two field-describing quantities of the magnetic field. | - know the two field-describing quantities of the magnetic field. | ||
| Zeile 124: | Zeile 127: | ||
| </ | </ | ||
| - | On the right side the magnetic field of a single current-carrying conductor is shown. This was already derived | + | On the right side, the magnetic field of a single current-carrying conductor is shown. This was already derived |
| ~~PAGEBREAK~~~~CLEARFIX~~ | ~~PAGEBREAK~~~~CLEARFIX~~ | ||
| Zeile 132: | Zeile 135: | ||
| </ | </ | ||
| - | If there is another current-carrying conductor near the first conductor, the fields overlap. In the simulation | + | If there is another current-carrying conductor near the first conductor, the fields overlap. In the simulation below, the current of both conductors is directed in the same direction. The field between the conductors overlaps just enough to weaken. This can also be deduced by previous knowledge if just the middle point between both conductors is considered: There, for the left conductor |
| ~~PAGEBREAK~~~~CLEARFIX~~ | ~~PAGEBREAK~~~~CLEARFIX~~ | ||
| Zeile 142: | Zeile 145: | ||
| Using the nomenclature from the previous chapter, it is also possible to assign north and south poles locally. Towards the outside, one pole appears to be located in front of the two conductors and another one behind. | Using the nomenclature from the previous chapter, it is also possible to assign north and south poles locally. Towards the outside, one pole appears to be located in front of the two conductors and another one behind. | ||
| - | in both simulations, | + | in both simulations, |
| ~~PAGEBREAK~~ ~~CLEARFIX~~ | ~~PAGEBREAK~~ ~~CLEARFIX~~ | ||
| - | ==== magnetic | + | ==== Magnetic |
| < | < | ||
| Zeile 154: | Zeile 157: | ||
| So far the magnetic field was defined quite pragmatically by the effect on a compass. | So far the magnetic field was defined quite pragmatically by the effect on a compass. | ||
| For a deeper analysis of the magnetic field, the field is now to be considered again - as with the electric field - from __two__ directions. | For a deeper analysis of the magnetic field, the field is now to be considered again - as with the electric field - from __two__ directions. | ||
| - | The magnetic field will also be considered | + | The magnetic field will also be considered a " |
| This chapter will first discuss the acting magnetic field. For this, it is convenient to consider the effects inside a toroidal coil (= donut-like setup). | This chapter will first discuss the acting magnetic field. For this, it is convenient to consider the effects inside a toroidal coil (= donut-like setup). | ||
| This can be seen in <imgref BildNr04> | This can be seen in <imgref BildNr04> | ||
| - | In an experiment, a magnetic needle inside the toroidal coil is now to be aligned | + | In an experiment, a magnetic needle inside the toroidal coil is now to be aligned perpendicular to the field lines. |
| Then, the magnetic field will generate a torque $M$ which tries to align the magnetic needle in the field direction. | Then, the magnetic field will generate a torque $M$ which tries to align the magnetic needle in the field direction. | ||
| Zeile 164: | Zeile 167: | ||
| < | < | ||
| </ | </ | ||
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| It now follows: | It now follows: | ||
| - | - $M = const. \neq f(\varphi)$ : For the same distance from the axis of symmetry, the torque $M$ is independent of the angle $\phi$. | + | - $M = {\rm const.} \neq f(\varphi)$: |
| - | - $M \sim I$ : The stronger the current flowing through a winding, the stronger the effect, i.e. the stronger the torque. | + | - $M \sim I$: The stronger the current flowing through a winding, the stronger the effect, i.e. the stronger the torque. |
| - | - $M \sim N$ : The greater the number $N$ of windings, the stronger the torque $M$. | + | - $M \sim N$: The greater the number $N$ of windings, the stronger the torque $M$. |
| - | - $M \sim {1 \over l}$ : The smaller the average coil circumference $l$ the greater the torque. The average coil circumference $l$ is equal to the average field line length. | + | - $M \sim {1 \over l}$ : The smaller the average coil circumference $l$ the greater the torque. The average coil circumference $l$ is equal to the **mean magnetic path length** (=average field line length). |
| To summarize: | To summarize: | ||
| Zeile 183: | Zeile 186: | ||
| \end{align*} | \end{align*} | ||
| - | For the unit of the magnetic field strength $H$ we get $[H] = {{[I]}\over{[l]}}= 1{{A}\over{m}}$ | + | For the unit of the magnetic field strength $H$ we get $[H] = {{[I]}\over{[l]}}= |
| - | ==== magnetic | + | ==== Magnetic |
| - | The previous derivation from the toroidal coil is now to be used to derive the field strength around a long, straight conductor. For a single conductor the parte $N \cdot I$ of the formula can be reduced to $ N \cdot I = 1 \cdot I = I$, since there is only one conductor. For the toroidal coil, the magnetic field strength was given by this current(s) divided by the (average) field line length. Because of the (same rotational) symmetry, this is also true for the single conductor. Also here the field line length has to be taken into account. | + | The previous derivation from the toroidal coil is now used to derive the field strength around a long, straight conductor. For a single conductor the part $N \cdot I$ of the formula can be reduced to $ N \cdot I = 1 \cdot I = I$ since there is only one conductor. For the toroidal coil, the magnetic field strength was given by this current(s) divided by the (average) field line length. Because of the (same rotational) symmetry, this is also true for the single conductor. Also here the field line length has to be taken into account. |
| The length of a field line around the conductor is given by the distance $r$ of the field line from the conductor: $l = l(r) = 2 \cdot \pi \cdot r$. \\ For the magnetic field strength of the single conductor we then get: | The length of a field line around the conductor is given by the distance $r$ of the field line from the conductor: $l = l(r) = 2 \cdot \pi \cdot r$. \\ For the magnetic field strength of the single conductor we then get: | ||
| \begin{align*} | \begin{align*} | ||
| - | \boxed{H ={{\theta}\over{l}} = {{I}\over{2 \cdot \pi \cdot r}}} \quad \quad | \quad \text{applies only to the long, straight conductor} | + | \boxed{H ={I\over{l}} = {{I}\over{2 \cdot \pi \cdot r}}} \quad \quad | \quad \text{applies only to the long, straight conductor} |
| \end{align*} | \end{align*} | ||
| Zeile 204: | Zeile 207: | ||
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| {{url> | {{url> | ||
| Zeile 217: | Zeile 220: | ||
| ==== Magnetic Voltage ==== | ==== Magnetic Voltage ==== | ||
| - | The cause of the magnetic field is the current | + | The cause of a magnetic field is a current. As seen for the coil, sometimes |
| \begin{align*} | \begin{align*} | ||
| - | \boxed{\theta = N \cdot I} | + | \boxed{\theta |
| \end{align*} | \end{align*} | ||
| - | The unit of $\theta$ is: $[\theta]= | + | The unit of $\theta$ is: $[\theta]= |
| < | < | ||
| < | < | ||
| </ | </ | ||
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| Thus, the magnetic field strength $H$ of the toroidal coil is then given by: $H= {{\theta}\over{l}}$ | Thus, the magnetic field strength $H$ of the toroidal coil is then given by: $H= {{\theta}\over{l}}$ | ||
| - | **Important** One also have here to consider the orientation of the current and way on the enclosing path. The <imgref BildNr05> | + | <callout icon=" |
| - | This is also given as the right-hand-rule (see < | + | |
| + | | ||
| < | < | ||
| < | < | ||
| </ | </ | ||
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| - | In the English literature often the name **{{wp> | + | </ |
| + | |||
| + | In the English literature often the name **{{wp> | ||
| - | ==== magnetic | + | ==== Magnetic |
| - | So far, only rotationally | + | So far, only rotational |
| + | For this purpose, we will have a look back at the electric field. For the electric field strength $E$ of a capacitor with two plates at a distance of $s$ and the potential difference $U$ holds: | ||
| \begin{align*} | \begin{align*} | ||
| Zeile 252: | Zeile 258: | ||
| \end{align*} | \end{align*} | ||
| - | In words: The potential difference | + | In words: The potential difference |
| + | If we compare | ||
| \begin{align*} | \begin{align*} | ||
| Zeile 258: | Zeile 265: | ||
| \end{align*} | \end{align*} | ||
| - | Can you see the similarities? | + | Can you see the similarities? |
| + | Because of the similarity - which continues below - the so-called **magnetic potential difference $V_m$** is introduced: | ||
| \begin{align*} | \begin{align*} | ||
| Zeile 264: | Zeile 272: | ||
| \end{align*} | \end{align*} | ||
| - | 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 $ds$ along a field line, the resulting infinitesimally small magnetic potential difference $dV_m = H \cdot ds$ 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: |
| \begin{align*} | \begin{align*} | ||
| - | V_{m12} = V_m(\vec{s_1}, \vec{s_2}) = \int_\vec{s_1}^\vec{s_2} H(\vec{s}) | + | V_{\rm m12} = V_{\rm m}(\vec{s_1}, \vec{s_2}) |
| + | | ||
| \end{align*} | \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. 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 $d \vec{s}$. This is made possible by scalar multiplication. Thus, it is generally valid: | + | 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. |
| + | 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*} | \begin{align*} | ||
| - | \boxed{V_{m12} = \int_\vec{s_1}^\vec{s_2} \vec{H} \cdot d \vec{s}} | + | \boxed{V_{\rm m12} = \int_\vec{s_1}^\vec{s_2} \vec{H} \cdot {\rm d} \vec{s}} |
| \end{align*} | \end{align*} | ||
| - | The magnetic voltage $\theta$ (and therefore the current) is the cause of the magnetic field strength. From the chapter [[electrical_engineering_2: | + | The magnetic voltage $\theta$ (and therefore the current) is the cause of the magnetic field strength. |
| + | From the chapter [[electrical_engineering_2: | ||
| + | This leads to the **{{wp> | ||
| \begin{align*} | \begin{align*} | ||
| - | \boxed{\int_{closed \\ path} \vec{H} \cdot d \vec{s} = \iint_{enclosed \\ surface | + | \boxed{\oint_{s} \vec{H} \cdot {\rm d} \vec{s} = \iint_A |
| \end{align*} | \end{align*} | ||
| * 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 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 magnetic voltage $\theta$ can be gib | + | * The magnetic voltage $\theta$ can be given as |
| * for a single conductor: $\theta = I$ | * for a single conductor: $\theta = I$ | ||
| * 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: |
| - | * $d\vec{s}$ and $d\vec{A}$ build a right-hand system: once the thumb of the right hand is pointing along $d\vec{A}$, the fingers of the right hand show the correct direction for $d\vec{s}$ 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> |
| </ | </ | ||
| - | ==== Recap: Application of magn. Field Strength ==== | + | ==== Recap: Application of magnetic |
| - | The Ampere' | + | Ampere' |
| < | < | ||
| < | < | ||
| </ | </ | ||
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| - | * The closed path $C$ is on 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 |
| This leads to: | This leads to: | ||
| \begin{align*} | \begin{align*} | ||
| - | \int_C \vec{H} \cdot d \vec{s} &= \iint_{A} | + | \oint_s |
| \end{align*} | \end{align*} | ||
| - | Since $\vec{H} \uparrow \uparrow d \vec{s}$ the term $\vec{H} \cdot d \vec{s}$ can be substituted by $H ds$: | + | Since $\vec{H} \uparrow \uparrow |
| \begin{align*} | \begin{align*} | ||
| - | \int_C H \cdot ds &= \iint_{A} | + | \oint_s |
| \end{align*} | \end{align*} | ||
| - | The magnetic voltage is the current through the surface and given as $N\cdot I$: | + | The magnetic voltage is the current through the surface and is given as $N\cdot I$: |
| \begin{align*} | \begin{align*} | ||
| - | \int_C H \cdot ds &= N\cdot I | + | \oint_{\rm s} H \cdot {\rm d}s &= N\cdot I |
| \end{align*} | \end{align*} | ||
| Zeile 334: | Zeile 346: | ||
| \begin{align*} | \begin{align*} | ||
| - | H \cdot \int_C | + | H \cdot \int_C |
| \end{align*} | \end{align*} | ||
| Zeile 343: | Zeile 355: | ||
| \end{align*} | \end{align*} | ||
| - | Therefore, the magnetic field strength in the toroidel | + | Therefore, the magnetic field strength in the toroidal |
| \begin{align*} | \begin{align*} | ||
| H &= {{N\cdot I}\over{ \pi \cdot d }} | H &= {{N\cdot I}\over{ \pi \cdot d }} | ||
| Zeile 349: | Zeile 361: | ||
| Similarly, the magnetic field strength $H$ in the distance $r$ to a single conductor with the current $I$ can be derived. | Similarly, the magnetic field strength $H$ in the distance $r$ to a single conductor with the current $I$ can be derived. | ||
| - | In this situation the result is: | + | In this situation, the result is: |
| \begin{align*} | \begin{align*} | ||
| Zeile 357: | Zeile 369: | ||
| ~~PAGEBREAK~~~~CLEARFIX~~ | ~~PAGEBREAK~~~~CLEARFIX~~ | ||
| - | + | ===== 3.3 Magnetic | |
| - | <panel type=" | + | |
| - | + | ||
| - | The current $I = 100A$ flows in a long straight conductor with a round cross-section. The radius of the conductor is $r_{L}= 4mm$. | + | |
| - | + | ||
| - | * 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? | + | |
| - | + | ||
| - | </ | + | |
| - | + | ||
| - | <panel type=" | + | |
| - | + | ||
| - | < | + | |
| - | < | + | |
| - | </ | + | |
| - | {{drawio> | + | |
| - | </ | + | |
| - | + | ||
| - | Three long straight conductors are arranged in a vacuum so that they lie at the vertices of an equilateral triangle (see <imgref BildNr01> | + | |
| - | + | ||
| - | - 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 the magnetic field strength $H(P)$ changes? | + | |
| - | </ | + | |
| - | + | ||
| - | <panel type=" | + | |
| - | + | ||
| - | < | + | |
| - | < | + | |
| - | </ | + | |
| - | {{drawio> | + | |
| - | </ | + | |
| - | + | ||
| - | Given are the adjacent closed trajectories in the magnetic field of current-carrying conductors (see <imgref BildNr05> | + | |
| - | + | ||
| - | In each case, the magnetic voltage $V_m$ along the drawn path is sought. | + | |
| - | + | ||
| - | </ | + | |
| - | + | ||
| - | ~~PAGEBREAK~~~~CLEARFIX~~ | + | |
| - | + | ||
| - | ===== 3.3 Ampere' | + | |
| < | < | ||
| - | === Goals === | + | === Learning Objectives |
| - | After this lesson, you should: | + | By the end of this section, you will be able to: |
| - know the force law for current-carrying conductors. | - know the force law for current-carrying conductors. | ||
| - Be able to determine the direction of the forces using given current directions and, if applicable, flux density. | - Be able to determine the direction of the forces using given current directions and, if applicable, flux density. | ||
| - | - be able to represent the acting force vectors in a sketch. | + | - be able to represent the vectors |
| - be able to determine a force vector by superimposing several force vectors using vector calculus. | - be able to determine a force vector by superimposing several force vectors using vector calculus. | ||
| - | - be able to state the following quantities for a force vector: | + | - be able to state the following quantities for a force vector |
| - | - Force vector in coordinate representation | + | - the force vector in coordinate representation |
| - | - magnitude of the force vector | + | - the magnitude of the force vector |
| - | - Angle of the force vector | + | - the angle of the force vector |
| </ | </ | ||
| - | In the last subchapter | + | ==== Definition of the Magnetic Flux Density ==== |
| - | 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: | + | |
| + | In the last sub-chapter, | ||
| + | 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: | ||
| < | < | ||
| < | < | ||
| </ | </ | ||
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| Zeile 444: | 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*} | ||
| - | This leads to: | + | This leads to the **Ampere' |
| \begin{align*} | \begin{align*} | ||
| |\vec{F}_{12}| = {{\mu}\over{2 \pi}} \cdot {{I_1 \cdot I_2 }\over{r}} \cdot l | |\vec{F}_{12}| = {{\mu}\over{2 \pi}} \cdot {{I_1 \cdot I_2 }\over{r}} \cdot l | ||
| \end{align*} | \end{align*} | ||
| - | Since we want to investigate the effect | + | Since we want to investigate the effect |
| \begin{align*} | \begin{align*} | ||
| - | |\vec{F}_{12}| &= {{\mu}\over{2 \pi}} \cdot {{I_2 }\over{r}} \cdot I_1 \cdot l \\ | + | |\vec{F}_{12}| &= {{\mu}\over{2 \pi}} \cdot {{I_2 }\over{r}} |
| - | |\vec{F}_{12}| | + | |
| \end{align*} | \end{align*} | ||
| - | The properties of the field from $I_2$ acting on $I_1$ is 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*} | ||
| - | ==== Generalization ==== | + | 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 $\vec{F}$ using vector multiplication too: |
| + | \begin{align*} | ||
| + | \boxed{\vec{F_L} = I \cdot \vec{l} \times \vec{B}} | ||
| + | \end{align*} | ||
| + | |||
| + | The absolute value can be calculated by | ||
| + | |||
| + | \begin{align*} | ||
| + | \boxed{|\vec{F_L}| = I \cdot |l| \cdot |B| \cdot \sin(\angle \vec{l}, | ||
| + | \end{align*} | ||
| + | |||
| + | The force is often called **Lorentz Force** $F_L$. For the orientation, | ||
| + | |||
| + | <callout icon=" | ||
| + | Right-hand rule for the Lorentz Force: | ||
| + | * The causing current $I$ is on the thumb. Since the current is not a vector, the direction is given by the direction of the conductor $\vec{l}$ | ||
| + | * The mediating external magnetic field $\vec{B}$ is on the index finger | ||
| + | * The resulting force $\vec{F}$ on the conductor is on the middle finger | ||
| + | This is shown in <imgref BildNr06> | ||
| + | A way to remember the orientation is the mnemonic **FBI** (from middle finger to thumb): | ||
| + | * $\vec{F}$orce on middle finger | ||
| + | * $\vec{B}$-Field on index finger | ||
| + | * Current $I$ on thumb (direction with length $\vec{l}$) | ||
| + | \\ \\ | ||
| + | < | ||
| + | < | ||
| + | \\ | ||
| < | < | ||
| < | < | ||
| </ | </ | ||
| - | {{drawio> | + | {{drawio> |
| </ | </ | ||
| + | </ | ||
| - | <panel type=" | + | ==== Lorentz Law and Lorentz Force ==== |
| - | A NdFeB-magnet | + | The true Lorentz force is not the force on the whole conductor but the single force onto an (elementary) charge. \\ |
| + | To find this force the previous force onto a conductor | ||
| - | </ | + | \begin{align*} |
| + | \vec{{\rm d}F}_{\rm L} = I \cdot {\rm d}\vec{l} \times \vec{B} | ||
| + | \end{align*} | ||
| - | ===== 3.4 Lorentz Force ===== | + | 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*} |
| - | === Goals === | + | \vec{{\rm d}F}_{\rm L} = {{{\rm d}Q}\over{{\rm d}t}} \cdot {\rm d}\vec{l} \times \vec{B} |
| + | \end{align*} | ||
| - | After this lesson, you should: | + | Mathematically not quite correct, but in a physical way true the following rearrangement can be done: |
| - | | + | \begin{align*} |
| - | - be able to determine the resulting vector of magnetic flux density by superimposing several vectors using vector calculus. | + | \vec{{\rm d}F}_{\rm L} &= {{{\rm d}Q \cdot |
| - | - Be able to determine the force on a current-carrying conductor in a magnetostatic field by applying the force law for current-carrying conductors in a magnetic field: | + | &= {\rm d}Q \cdot {{{\rm d}\vec{l}}\over{{\rm d}t}} \times \vec{B} \\ |
| - | - Force vector in coordinate representation | + | &= {\rm d}Q \cdot {{{\rm d}\vec{l}}\over{{\rm d}t}} \times \vec{B} \\ |
| - | - magnitude of the force vector | + | \end{align*} |
| - | - Angle of the force vector | + | |
| - | | + | |
| - | </ | + | |
| - | ==== Video ==== | + | Here, the part ${{{\rm d}\vec{l}}\over{{\rm d}t}}$ represents the speed $\vec{v}$ of the small charge packet ${\rm d}Q$. |
| - | Please have a look at the contents (text, videos, exercises) on the page of the [[https://lx3.mint-kolleg.kit.edu/ | + | \begin{align*} |
| + | \vec{{\rm d}F}_{\rm L} &= {\rm d}Q \cdot \vec{v} \times \vec{B} | ||
| + | \end{align*} | ||
| + | |||
| + | The **Lorenz Force** on a finite charge packet is the integration: | ||
| + | |||
| + | \begin{align*} | ||
| + | \boxed{\vec{F}_{\rm L} = Q \cdot \vec{v} \times \vec{B}} | ||
| + | \end{align*} | ||
| + | |||
| + | |||
| + | <callout icon=" | ||
| + | |||
| + | * A charge $Q$ moving with a velocity $\vec{v}$ | ||
| + | * The direction of the force is given by the right-hand rule. | ||
| + | |||
| + | </ | ||
| + | Please have a look at the German contents (text, videos, exercises) on the page of the [[https:// | ||
| - | ===== 3.5 Matter in the magnetic field ===== | + | ===== 3.4 Matter in the Magnetic Field ===== |
| < | < | ||
| - | === Goals === | + | === Learning Objectives |
| - | After this lesson, you should: | + | By the end of this section, you will be able to: |
| - know the two field-describing quantities of the magnetostatic field. | - know the two field-describing quantities of the magnetostatic field. | ||
| - be able to describe and apply the relationship between these two quantities via the material law. | - be able to describe and apply the relationship between these two quantities via the material law. | ||
| - know the classification of magnetic materials. | - know the classification of magnetic materials. | ||
| - | - be able to read the relevant data from a magnetisation | + | - be able to read the relevant data from a magnetization |
| </ | </ | ||
| + | The magnetic interaction is also represented via two fields in the previous subchapters similar to the electric field. | ||
| + | Since the magnetic field strength $\vec{H}$ is only dependent on the field causing current $I$, this field is independent of surrounding matter. \\ | ||
| + | The magnetic flux density $\vec{B}$ on the other hand, showed with the magnetic permeability a factor that can contain the material effects. | ||
| - | ===== 3.6 Poynting Vector (not part of the curriculum) ===== | + | Both fields are connected. The Ampere' |
| - | | + | \begin{align*} |
| - | | + | |\vec{F}_{12}| &= {{\mu}\over{2 \pi}} \cdot {{I_2 }\over{r}} &\cdot I_1 \cdot l \\ |
| - | * Very detailed view of the energy flow in an electric circuit: http:// | + | &= B &\cdot I_1 \cdot l \\ |
| + | \end{align*} | ||
| - | Force effect on dia- and paramagnetic | + | Therefore, the magnetic flux density $\vec{B}$ is equal to: |
| - | {{youtube>u36QpPvEh2c}} | + | |
| + | \begin{align*} | ||
| + | B &= {{\mu}\over{2 \pi}} \cdot {{I_2 }\over{r}} | ||
| + | \end{align*} | ||
| + | |||
| + | $I_2$ was here the field-causing current. \\ | ||
| + | For the field strength of the straight conductor, we had: | ||
| + | |||
| + | \begin{align*} | ||
| + | H &= {{I}\over{2 \pi \cdot r }} | ||
| + | \end{align*} | ||
| + | |||
| + | The connection between the two fields is also $B = \mu \cdot H$. | ||
| + | Since the field lines of both fields are always parallel to each other it results to | ||
| + | |||
| + | \begin{align*} | ||
| + | \boxed{ | ||
| + | \vec{B} = \mu \cdot \vec{H} | ||
| + | } | ||
| + | \end{align*} | ||
| + | |||
| + | This is similar to the $\vec{D} = \varepsilon \cdot \vec{E}$. Similarly, the permeability is separated into two parts: | ||
| + | |||
| + | \begin{align*} | ||
| + | \mu &= \mu_0 \cdot \mu_{\rm r} | ||
| + | \end{align*} | ||
| + | |||
| + | where | ||
| + | * $\mu_0 | ||
| + | * $\mu_{\rm r}$ is the relative permeability, | ||
| + | |||
| + | The material can be divided into different types by looking at its relative permeability. | ||
| + | <imgref BildNr00> | ||
| + | In this diagram, the different | ||
| + | The three most important material types shall be discussed shortly. | ||
| + | |||
| + | < | ||
| + | < | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | ==== Diamagnetic Materials ==== | ||
| + | |||
| + | * Diamagnetic materials weaken | ||
| + | * The weakening is very low (see <tabref tab01> | ||
| + | * For diamagnetic materials applies $0<\mu_{\rm r}<1$ | ||
| + | * 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. | ||
| + | * With an external magnetic field an antiparallel-orientated magnet is induced. | ||
| + | * The reaction weakens the external field. This is similar to the weakening of the electric field by the dipoles of materials. | ||
| + | * Due to the repulsion of the outer magnetic field the material tends to move out of a magnetic field. \\ For very strong magnetic fields small objects can be levitated (see clip). | ||
| + | |||
| + | < | ||
| + | < | ||
| + | ^ Material | ||
| + | | Antimon | ||
| + | | Copper | ||
| + | | Mercury | ||
| + | | Silver | ||
| + | | Water | $\rm H_2O$ | $0.999 946$ | | ||
| + | | Bismut | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | < | ||
| + | < | ||
| + | {{drawio> | ||
| + | </ | ||
| - | A living | + | A living |
| {{youtube> | {{youtube> | ||
| + | |||
| + | ==== Paramagnetic Materials ==== | ||
| + | |||
| + | * Paramagnetic materials strengthen the magnetic field, compared to the vacuum. | ||
| + | * The strengthening is very low (see <tabref tab02>). | ||
| + | * For paramagnetic materials applies $\mu_{\rm r}> | ||
| + | * The principle behind the effect is again based on quantum mechanics (see <imgref BildNr23> | ||
| + | * Without the external field no counteracting field is generated by the matter. | ||
| + | * With an external magnetic field internal "tiny magnets" | ||
| + | * This reaction strengthens the external field. | ||
| + | |||
| + | < | ||
| + | < | ||
| + | ^ Material | ||
| + | | Aluminum | ||
| + | | Air | | $1.000 000 4$ | | ||
| + | | Oxygen | ||
| + | | Platinum | ||
| + | | Tin | $\rm Sn$ | $1.000 003 8$ | | ||
| + | </ | ||
| + | </ | ||
| + | |||
| + | < | ||
| + | < | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | Explanation of diamagnetism and paramagnetism | ||
| + | < | ||
| + | <WRAP column half>{{ youtube> | ||
| + | <WRAP column half>{{ youtube> | ||
| + | </ | ||
| + | |||
| + | ==== Ferromagnetic Materials ==== | ||
| + | |||
| + | * Ferromagnetic materials strengthen the magnetic field strongly, compared to the vacuum. | ||
| + | * The strengthening can create a field multiple times stronger than in a vacuum. | ||
| + | * For ferromagnetic materials applies $\mu_{\rm r} \gg 1$ | ||
| + | * The principle behind the effect is based on quantum mechanics: | ||
| + | * Also here internal "tiny magnets" | ||
| + | * However, these "tiny magnets" | ||
| + | * Ferromagnetic materials can be permanently magnetized. They build up **permanent magnets**. | ||
| + | * The magnetic flux density of the ferromagnetic matter is depending on the external field strength and the history of the material. | ||
| + | * The dependency between $\mu_{\rm r}$ and the external field strength $H$ is nonlinear. | ||
| + | * Ferromagnetic materials are characterized by the magnetization curve (see <imgref BildNr24> | ||
| + | * Non-magnetized ferromagnets are located in the origin. | ||
| + | * 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** $B(H=0) = B_{\rm R}$ (or remanent magnetization). | ||
| + | * 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 $H_{\rm sat}$) | ||
| + | |||
| + | < | ||
| + | < | ||
| + | {{drawio> | ||
| + | </ | ||
| Explanation of the hysteresis curve | Explanation of the hysteresis curve | ||
| {{youtube> | {{youtube> | ||
| - | Nice illustration of magnetization and demagnetization of soft magnetic material. | + | \\ \\ |
| - | {{youtube> | + | The ferromagnetic materials can again be subdivided into two groups: magnetically |
| - | Wandering magnetic domains in a ferromagnetic material (from [[https:// | + | === magnetically soft ferromagnetic Materials === |
| + | |||
| + | * high permeability | ||
| + | * high saturation flux density $B_{\rm sat}$ (= good magnetic conductor) | ||
| + | * small coercive field strength $H_{\rm C} < 1 ~\rm kA/m$ (= easy to reverse the magnetization) | ||
| + | * low magnetic losses for reversing the magnetization | ||
| + | * Important materials: ferrites | ||
| + | |||
| + | Applications: | ||
| + | * mainly for coil core material | ||
| + | * rotor and stator material for electric motors | ||
| + | * {{wp> | ||
| + | |||
| + | === Magnetically hard ferromagnetic Materials === | ||
| + | |||
| + | * low permeability | ||
| + | * low saturation flux density $B_{\rm sat}$ (= good magnetic conductor) | ||
| + | * high coercive field strength $H_{\rm C}$ up to $2' | ||
| + | * high magnetic losses for reversing the magnetization | ||
| + | * Important materials: ferrites, alloys of iron and cobalt: $\rm AlNiCo$((Aluminum-Nickel-Cobalt)), | ||
| + | |||
| + | Applications: | ||
| + | * mainly for permanent magnets | ||
| + | * permanent electric motors | ||
| + | * loudspeakers and microphones | ||
| + | * mechanical actuators | ||
| + | |||
| + | < | ||
| + | < | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | |||
| + | Wandering magnetic domains in a ferromagnetic material | ||
| {{https:// | {{https:// | ||
| + | |||
| + | |||
| + | ===== 3.5 Poynting Vector (not part of the curriculum) ===== | ||
| + | |||
| + | * Clear picture of the Poynting vector along an electric circuit: https:// | ||
| + | * Good explanation of the Energy flow via a current model: http:// | ||
| + | * Very detailed view of the energy flow in an electric circuit: http:// | ||
| ===== Tasks ===== | ===== Tasks ===== | ||
| + | |||
| + | |||
| + | |||
| + | <panel type=" | ||
| + | |||
| + | 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*} | ||
| + | |||
| + | # | ||
| + | |||
| + | |||
| + | </ | ||
| + | |||
| + | <panel type=" | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | Three long straight conductors are arranged in a vacuum to lie at the vertices of an equilateral triangle (see <imgref BildNr01> | ||
| + | |||
| + | 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*} | ||
| + | |||
| + | # | ||
| + | |||
| + | |||
| + | </ | ||
| + | |||
| + | <panel type=" | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | Given are the adjacent closed trajectories in the magnetic field of current-carrying conductors (see <imgref BildNr05> | ||
| + | |||
| + | In each case, the magnetic potential difference $V_{\rm m}$ along the drawn path is sought. | ||
| + | |||
| + | |||
| + | # | ||
| + | |||
| + | * 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. | ||
| + | |||
| + | # | ||
| + | |||
| + | # | ||
| + | a) $V_{\rm m,a} = - I_1 = - 2~\rm A$ \\ | ||
| + | # | ||
| + | |||
| + | # | ||
| + | b) $V_{\rm m,b} = - I_2 = - 4.5~\rm A$ \\ | ||
| + | # | ||
| + | |||
| + | # | ||
| + | c) $V_{\rm m,c} = 0 $ \\ | ||
| + | # | ||
| + | |||
| + | # | ||
| + | d) $V_{\rm m,d} = + I_1 - I_2 = 2~\rm A - 4.5~\rm A = - 2.5~\rm A$ \\ | ||
| + | # | ||
| + | |||
| + | # | ||
| + | e) $V_{\rm m,e} = + I_1 = + 2~\rm A$ \\ | ||
| + | # | ||
| + | |||
| + | # | ||
| + | f) $V_{\rm m,f} = 2 \cdot (- I_1) = - 4~\rm A$ \\ | ||
| + | # | ||
| + | |||
| + | </ | ||
| + | |||
| + | ~~PAGEBREAK~~~~CLEARFIX~~ | ||
| + | |||
| + | |||
| + | <panel type=" | ||
| + | |||
| + | A $\rm NdFeB$ magnet can show a magnetic flux density up to $1.2 ~\rm T$ near the surface. | ||
| + | |||
| + | 1. For comparison, the same flux density shall be created inside a toroidal coil with $10' | ||
| + | |||
| + | # | ||
| + | |||
| + | * The $B$-field can be calculated by the $H$-field. | ||
| + | * 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*} | ||
| + | |||
| + | # | ||
| + | |||
| + | </ | ||
| + | |||
| + | <wrap #task3_3_2 /> | ||
| + | |||
| + | <panel type=" | ||
| + | |||
| + | An electron enters a plate capacitor on a trajectory parallel to the plates. | ||
| + | It shall move with the velocity $\vec{v}$ in the plate capacitor parallel to the plates. | ||
| + | The plates have a potential difference $U$ and a distance $d$. | ||
| + | In the vacuum in between the plates, there is also a magnetic field $\vec{B}$ present. | ||
| + | |||
| + | < | ||
| + | < | ||
| + | </ | ||
| + | {{drawio> | ||
| + | </ | ||
| + | |||
| + | Calculate the velocity depending on the other parameters $\vec{v} = f(U, |\vec{B}|, d) $! | ||
| + | |||
| + | <button size=" | ||
| + | * Think about the two forces on the electron from the fields - gravity is ignored. \\ Write their definitions down. | ||
| + | * With which relationship between these two forces does the electron moves through the plate capacitor __parallel__ to the plates? \\ So the trajectory neither get bent up nor down. | ||
| + | * What is the relationship between the $E$-field in the plate capacitor and the electric voltage $U$? | ||
| + | </ | ||
| + | |||
| + | <button size=" | ||
| + | |||
| + | Within the electric field, the Coulomb force acts on the electron: | ||
| + | |||
| + | \begin{align*} | ||
| + | \vec{F}_C = q_e \cdot \vec{E} | ||
| + | \end{align*} | ||
| + | |||
| + | Within the magnetic field, also the Lorentz force acts on the electron: | ||
| + | |||
| + | \begin{align*} | ||
| + | \vec{F}_L = q_e \cdot \vec{v} \times \vec{B} | ||
| + | \end{align*} | ||
| + | |||
| + | The absolute value of both forces must be equal to compensate each other: | ||
| + | |||
| + | \begin{align*} | ||
| + | |\vec{F}_C| | ||
| + | |q_e \cdot \vec{E}| | ||
| + | q_e \cdot |\vec{E}| | ||
| + | |\vec{E}| | ||
| + | \end{align*} | ||
| + | |||
| + | Since $\vec{v}$ is perpendicular to $\vec{B}$ the right side is equal to $|\vec{v}| \cdot |\vec{B}| = v \cdot B$. \\ | ||
| + | Additionally, | ||
| + | |||
| + | Therefore, it leads to the following: | ||
| + | |||
| + | \begin{align*} | ||
| + | {{U}\over{d}} | ||
| + | v & | ||
| + | \end{align*} | ||
| + | |||
| + | </ | ||
| + | |||
| + | <button size=" | ||
| + | \begin{align*} | ||
| + | v = {{U}\over{B\cdot d}} | ||
| + | \end{align*} | ||
| + | |||
| + | </ | ||
| + | |||
| + | </ | ||
| | | ||
| Zeile 546: | Zeile 1080: | ||
| <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 563: | Zeile 1122: | ||
| ++++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 | + | * Now think of a generated field through this to it. What direction must the current |
| * 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? | ||
| ++++ | ++++ | ||
| ++++Tip for 3| See 3rd video. | ++++Tip for 3| See 3rd video. | ||
| - | * Picture the two wires, or draw it on. | + | * Picture the two wires, or draw them on. |
| * In which direction would the outer field run in each case? | * In which direction would the outer field run in each case? | ||
| * The field is a linear vector field. So the total field can be created from several individual fields by adding them together. Does adding the field in between make it larger, or smaller? | * The field is a linear vector field. So the total field can be created from several individual fields by adding them together. Does adding the field in between make it larger, or smaller? | ||
| ++++ | ++++ | ||
| - | ++++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. | ||
| - | * If now with parallel wires and different current-direction | + | * With parallel wires and different current |
| * But then there must be a point at which the force becomes 0. | * But then there must be a point at which the force becomes 0. | ||
| ++++ | ++++ | ||
| Zeile 582: | Zeile 1141: | ||
| ++++Tip for 5| | ++++Tip for 5| | ||
| * The magnetic field lines must be closed. | * The magnetic field lines must be closed. | ||
| - | * Compare the field curve between coil and permanent magnet. | + | * Compare the field curve between |
| ++++ | ++++ | ||
| ++++Tip for 6| | ++++Tip for 6| | ||