Differences
This shows you the differences between two versions of the page.
| Both sides previous revision Previous revision Next revision | Previous revision | ||
| electrical_engineering_2:the_magnetostatic_field [2023/03/16 13:48] – mexleadmin | electrical_engineering_2:the_magnetostatic_field [2025/04/29 02:45] (current) – mexleadmin | ||
|---|---|---|---|
| Line 1: | Line 1: | ||
| - | ====== 3. The magnetostatic Field ====== | + | ====== 3 The magnetostatic Field ====== |
| < | < | ||
| - | For this and the following chapter the online | + | The online |
| </ | </ | ||
| Line 30: | Line 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. | ||
| Line 39: | Line 39: | ||
| 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 colored red, and the south pole green. | + | - There are two poles. These are called the north pole and the south pole. The north pole is colored red, and the south pole is green. |
| - 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. | ||
| Line 51: | Line 51: | ||
| </ | </ | ||
| - | An interesting aspect is that even non-magnetized, | + | Interestingly, |
| <callout icon=" | <callout icon=" | ||
| Line 76: | Line 76: | ||
| - | 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=" | ||
| Line 135: | Line 135: | ||
| </ | </ | ||
| - | 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 the right-hand rule results in a vector directed towards the observer. For the right conductor, it results in a vector that is directed away from the observer. These just cancel each other out. Further outward field lines go around both conductors. North and south poles here are not fixed localized | + | 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 the right-hand rule results in a vector directed towards the observer. For the right conductor, it results in a vector that is directed away from the observer. These just cancel each other out. Further outward field lines go around both conductors. |
| ~~PAGEBREAK~~~~CLEARFIX~~ | ~~PAGEBREAK~~~~CLEARFIX~~ | ||
| Line 190: | Line 190: | ||
| ==== Magnetic Field Strength part 2: straight conductor ==== | ==== Magnetic Field Strength part 2: straight conductor ==== | ||
| - | 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 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 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. |
| Line 207: | Line 207: | ||
| < | < | ||
| - | < | + | < |
| </ | </ | ||
| {{url> | {{url> | ||
| Line 220: | Line 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*} | ||
| Line 236: | Line 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> | ||
| Line 247: | Line 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 ==== | ||
| Line 274: | Line 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: | ||
| Line 283: | Line 283: | ||
| \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. | + | 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: | 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: | ||
| Line 295: | Line 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*} | ||
| Line 303: | Line 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> | ||
| Line 314: | Line 314: | ||
| ==== Recap: Application of magnetic Field Strength ==== | ==== Recap: Application of magnetic Field Strength ==== | ||
| - | Ampere' | + | Ampere' |
| < | < | ||
| Line 322: | Line 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 | ||
| Line 328: | Line 328: | ||
| \begin{align*} | \begin{align*} | ||
| - | \int_C \vec{H} \cdot {\rm d} \vec{s} &= \iint_{A} | + | \oint_s |
| \end{align*} | \end{align*} | ||
| Line 334: | Line 334: | ||
| \begin{align*} | \begin{align*} | ||
| - | \int_C H \cdot {\rm d}s &= \iint_{A} | + | \oint_s |
| \end{align*} | \end{align*} | ||
| Line 340: | Line 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*} | ||
| Line 391: | Line 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: |
| < | < | ||
| Line 418: | Line 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*} | ||
| Line 432: | Line 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*} | ||
| Line 447: | Line 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*} | ||
| Line 463: | Line 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 $\vec{l}$) | ||
| + | \\ \\ | ||
| + | < | ||
| + | < | ||
| + | \\ | ||
| < | < | ||
| < | < | ||
| Line 469: | Line 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*} | ||
| Line 481: | Line 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*} | ||
| Line 515: | Line 520: | ||
| </ | </ | ||
| - | Please have a look at the German contents (text, videos, exercises) on the page of the [[https://lx3.mint-kolleg.kit.edu/ | + | Please have a look at the German contents (text, videos, exercises) on the page of the [[https://obkp.mint-kolleg.kit.edu/ |
| ===== 3.4 Matter in the Magnetic Field ===== | ===== 3.4 Matter in the Magnetic Field ===== | ||
| Line 568: | Line 573: | ||
| \begin{align*} | \begin{align*} | ||
| - | \mu &= \mu_0 \cdot \mu_r | + | \mu &= \mu_0 \cdot \mu_{\rm r} |
| \end{align*} | \end{align*} | ||
| Line 589: | Line 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. | ||
| Line 598: | Line 603: | ||
| < | < | ||
| < | < | ||
| - | ^ Material | + | ^ Material |
| - | | Antimon | + | | Antimon |
| - | | Copper | + | | Copper |
| - | | Mercury | + | | Mercury |
| - | | Silver | + | | Silver |
| - | | Water | $H_2O$ | + | | Water |
| - | | Bismut | + | | Bismut |
| </ | </ | ||
| </ | </ | ||
| Line 643: | Line 648: | ||
| Explanation of diamagnetism and paramagnetism | Explanation of diamagnetism and paramagnetism | ||
| - | {{youtube> | + | < |
| + | <WRAP column half>{{ youtube> | ||
| + | <WRAP column half>{{ youtube> | ||
| + | </ | ||
| ==== Ferromagnetic Materials ==== | ==== Ferromagnetic Materials ==== | ||
| Line 659: | Line 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 $H_{\rm sat}$) | ||
| Line 677: | Line 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 | ||
| Line 687: | Line 695: | ||
| * {{wp> | * {{wp> | ||
| - | === magnetically | + | === Magnetically |
| * low permeability | * low permeability | ||
| Line 693: | Line 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: | ||
| Line 724: | Line 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? | ||
| </ | </ | ||
| Line 739: | Line 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? | ||
| </ | </ | ||
| - | <panel type=" | + | <panel type=" |
| < | < | ||
| Line 753: | Line 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 | + | In each case, the magnetic |
| + | |||
| + | |||
| + | # | ||
| + | |||
| + | * 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) $V_{\rm m,f} = 2 \cdot (- I_1) = - 4~\rm A$ \\ | ||
| + | # | ||
| </ | </ | ||
| Line 764: | Line 948: | ||
| <panel type=" | <panel type=" | ||
| - | A $\rm NdFeB$ magnet can show a magnetic flux density up to $1.2 ~\rm T$ near to 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*} | ||
| + | |||
| + | # | ||
| </ | </ | ||
| Line 775: | Line 1013: | ||
| <panel type=" | <panel type=" | ||
| - | An electron shall move with the velocity $\vec{v}$ in a plate capacitor parallel to the plates, which have a potential difference $U$ and a distance $d$. | + | An electron |
| - | In the vacuum in between the plates | + | It shall move with the velocity $\vec{v}$ in the plate capacitor parallel to the plates. |
| + | The plates | ||
| + | In the vacuum in between the plates, there is also a magnetic field $\vec{B}$ | ||
| < | < | ||
| Line 785: | Line 1025: | ||
| Calculate the velocity depending on the other parameters $\vec{v} = f(U, |\vec{B}|, d) $! | 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=" | <button size=" | ||
| Line 800: | Line 1046: | ||
| \end{align*} | \end{align*} | ||
| - | The absolute value of both forces must be equal in order to compensate each other: | + | The absolute value of both forces must be equal to compensate each other: |
| \begin{align*} | \begin{align*} | ||
| Line 812: | Line 1058: | ||
| Additionally, | Additionally, | ||
| - | Therefore, it leads to: | + | Therefore, it leads to the following: |
| \begin{align*} | \begin{align*} | ||
| Line 834: | Line 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? | ||
| ++++ | ++++ | ||
| Line 851: | Line 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 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? | ||
| ++++ | ++++ | ||
| Line 861: | Line 1132: | ||
| ++++ | ++++ | ||
| - | ++++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. | ||