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Block 17 — Magnetic Flux Density and Forces
Learning objectives
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Preparation at Home
Well, again
- read through the present chapter and write down anything you did not understand.
- Also here, there are some clips for more clarification under 'Embedded resources' (check the text above/below, sometimes only part of the clip is interesting).
For checking your understanding please do the following exercises:
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90-minute plan
- Warm-up (x min):
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- Core concepts & derivations (x min):
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- Practice (x min): …
- Wrap-up (x min): Summary box; common pitfalls checklist.
Conceptual overview
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Core content
We know from block11 that a static charge $Q_1$ generate a static electric field $D$.
Before in block09, we developed that a static electric field $E= {{1}\over{\varepsilon}} D$ effects a force $F_C$ on a static charge $Q_2$
From the last chapter (block16) we got, that moving charges ${{d}\over{dt}}Q_1 = I_1$ generate a static magnetic field $H$.
So, how does an acting magnetic field effects a force on a moving charge ${{d}\over{dt}}Q_2 = I_2$?
Definition of the Magnetic Flux Density
To derive the forces, we do a step back to the images of field lines.
In Abbildung 1 1) the field lines of a single current-carrying wire is shown.
Abbildung 1 2) depicts the homogenous field of a coil.
When a current-carrying wire is within the homogenous field, we get the superimposed picture of both fields ( Abbildung 1 3) ).
This leads to an enrichment of magnetic field on the left and an depletion on the right.
With the knowledge, that the field lines usually do not like to stay next to each other, one can conclude that there will be a force to the right.
When no current is flowing through the conductor the force is equal to zero.
The following is detectable:
- $|\vec{F}| \sim I$ : The stronger the current, the stronger the force $F$.
- $|\vec{F}| \sim l$ : As longer the conductor length $l$, as stronger the force $F$ gets.
- $|\vec{F}| \sim {H}$ : As more current through the coil, as stronger the $H$-field. And a stronger the $H$-field leads to stronger force $F$.
To summarize: \begin{align*} F \sim H \cdot I \cdot l \end{align*}
The proportionality factor is $\mu_0$, the magnetic field constant, permeability or vacuum permeability: $\mu_0 = 4\pi \cdot 10^{-7} {\rm {Vs}\over{Am}}$.
\begin{align*} F = \mu_0 \cdot H \cdot I \cdot l \end{align*}
When adding an iron core into the coil the force $F$ gets stronger. Therefore, we include a material-dependent constant $\mu_r$, the so-called relative permeability
\begin{align*} F = \mu_0 \mu_r \cdot H \cdot I \cdot l \\ \end{align*}
The new field quantity is $B$ the magnetic flux density:
\begin{align*} \boxed{ \vec{B} = \mu \cdot \vec{H} } \quad | \text{with } \mu = \mu_0 \mu_r \end{align*}
Investigating the vectorial behaviour leads to the cross-product, and to the so-called Lorentz force
\begin{align*} \boxed{ \vec{F}_L = I \cdot \vec{l} \times \vec{B} } \end{align*}
With \vec{l} pointing in the direction of the positive current $I$. The absolute value can be calculated by
\begin{align*} \boxed{|\vec{F_L}| = I \cdot |l| \cdot |B| \cdot \sin(\angle \vec{l},\vec{B} )} \end{align*}
For the orientation of the vectors, another right-hand rule can be applied.
Notice:
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 Abbildung 2.
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 $\vec{l}$)
Materials
The material can be divided into different types by looking at its relative permeability. Abbildung 3 shows the relative permeability in the magnetization curve (also called $B$-$H$-curve). In this diagram, the different effect ($B$-field on $y$-axis) based on the causing external $H$-field (on $x$-axis) for different materials is shown. The three most important material types shall be discussed shortly.
Diamagnetic Materials
- Diamagnetic materials weaken the magnetic field, compared to the vacuum.
- The weakening is very low (see Tabelle 1).
- For diamagnetic materials applies $0<\mu_{\rm r}<1$
- The principle behind the effect is based on quantum mechanics (see Abbildung 4):
- 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).
Paramagnetic Materials
- Paramagnetic materials strengthen the magnetic field, compared to the vacuum.
- The strengthening is very low (see Tabelle 2).
- For paramagnetic materials applies $\mu_{\rm r}>1$
- The principle behind the effect is again based on quantum mechanics (see Abbildung 5):
- Without the external field no counteracting field is generated by the matter.
- With an external magnetic field internal „tiny magnets“ based on the electrons in their orbitals are orientated similarly.
- This reaction strengthens the external field.
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 (see Abbildung 6).
- For ferromagnetic materials applies $\mu_{\rm r} \gg 1$
- Ferromagnetic materials are characterized by the magnetization curve (see Abbildung 7)
- Non-magnetized ferromagnets are located in the origin.
- With an external field $H$ the initial magnetization curve (in German: Neukurve, dashed in Abbildung 7) is passed.
- 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 $H_{\rm sat}$)
Applications of the Lorentz Force
We want to apply the Lorentz force for two common situations.
Two parallel Conductors
The Lorentz force can be applied to two parallel conductors.
The experiment consists of a part $l$ of two very long1) and parallel conductors with the currents $I_1, I_2$ and the distance $r$ (see Abbildung 2).
Here, we get for the $B$ field caused by $I_2$: \begin{align*} B_2 &= \mu \cdot H_2 \\ &= \mu \cdot {{ I_2 }\over{2\pi \cdot r}} \end{align*}
We insert this into the formula of the Lorenz force \begin{align*} \vec{F}_L = I \cdot \vec{l} \times \vec{B} \end{align*}
This leads to the so-called Ampere's Force Law, applied on long and parallel conductors: \begin{align*} \boxed{ |\vec{F}_{12}| = {{\mu}\over{2 \pi}} \cdot {{I_1 \cdot I_2 }\over{r}} \cdot l } \end{align*}
Moving single Charge
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 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*} \vec{{\rm d}F}_{\rm L} = I \cdot {\rm d}\vec{l} \times \vec{B} \end{align*}
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*} \vec{{\rm d}F}_{\rm L} = {{{\rm d}Q}\over{{\rm d}t}} \cdot {\rm d}\vec{l} \times \vec{B} \end{align*}
Mathematically not quite correct, but in a physical way true the following rearrangement can be done:
\begin{align*} \vec{{\rm d}F}_{\rm L} &= {{{\rm d}Q \cdot {\rm d}\vec{l}}\over{{\rm d}t}} \times \vec{B} \\ &= {\rm d}Q \cdot {{{\rm d}\vec{l}}\over{{\rm d}t}} \times \vec{B} \\ &= {\rm d}Q \cdot {{{\rm d}\vec{l}}\over{{\rm d}t}} \times \vec{B} \\ \end{align*}
Here, the part ${{{\rm d}\vec{l}}\over{{\rm d}t}}$ represents the speed $\vec{v}$ of the small charge packet ${\rm d}Q$.
\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*}
Notice:
- A charge $Q$ moving with a velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences a force of $\vec{F_{\rm L}}$.
- The direction of the force is given by the right-hand rule.
Common pitfalls
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Exercises
Exercise E1 Cylindrical Coil
(written test, approx. 6 % of a 120-minute written test, SS2021)
A cylindrical coil with the following information is given:
- Length $𝑙 = 30 {~\rm cm}$,
- Winding diameter $𝑑 = 390 {~\rm mm}$,
- Number of windings $𝑤 = 240$ ,
- Current through the conductor $𝐼 = 500 {~\rm mA}$,
- Material inside: Air
- $\mu_0 = 4\pi\cdot 10^{-7} {{\rm Vs}\over{\rm Am}}$
The proportion of the magnetic voltage outside the coil can be neglected. Determine the following for the inside of the coil:
a) the magnetic field strength (2 points)
\begin{align*} H &= {{N \cdot I}\over{l}} = {{w \cdot I}\over{l}} \end{align*}
Putting in the numbers: \begin{align*} H &= {{240 \cdot 0.5 {~\rm A}}\over{0.3 {~\rm m}}} \end{align*}
b) the magnetic flux density (2 points)
The magnetic field strength is $B = \mu_0 \mu_{\rm r} \cdot H$:
\begin{align*} B = \mu_0 \mu_{\rm r} H \end{align*}
Putting in the numbers: \begin{align*} B &= 4\pi\cdot 10^{-7} {{\rm Vs}\over{\rm Am}} \cdot 400 ~\rm {{A}\over{m}} \\ &= 0.0005026... {{\rm Vs}\over{\rm m^2}} \end{align*}
c) the magnetic flux (2 points)
The magnetic flux is given as:
\begin{align*} \Phi &= B \cdot A \end{align*}
Since the coil is cylindrical, the cross-sectional area is given as
\begin{align*} A = \pi r^2 = \pi \left( {{d}\over{2}} \right)^2 \end{align*}
Therefore: \begin{align*} \Phi &= B \cdot \pi \left( {{d}\over{2}} \right)^2 \end{align*}
Putting in the numbers: \begin{align*} \Phi &= 0.0005026... {{\rm Vs}\over{\rm m^2}} \cdot \pi \left( {{0.39{\rm m}}\over{2}} \right)^2 \\ &= 0.00006004... {\rm Vs} \end{align*}
Exercise E2 Magnetic Flux Density
(written test, approx. 6 % of a 120-minute written test, SS2021)
An electric motor is operated for experiments in the laboratory. An alternating current with an amplitude of $\hat{I} = 100~\rm A$ is operated.
You stand next to it and think about whether you have any health problems to worry about. The figure below shows the top view of the laboratory with the supply line between $\rm A$ and $\rm B$.
$\mu_{0} = 4\pi\cdot 10^{-7} {{\rm Vs}\over{\rm Am}}$, $\mu_{r}=1$
a) What is the highest magnetic flux density through the line in your body? (3 points)
The magnetic field strength for a conducting wire is given as:
\begin{align*} H &= {{I}\over{2\pi \cdot r}} \end{align*}
The magnetic flux density $B$ is given as: $B = \mu_0 \mu_r H$
Here, the maximum current is $\hat{I} = 100~\rm A$ and the distance to the cable is $r = \sqrt{(0.1 {~\rm m})^2 + (0.4 {~\rm m})^2}= 0.412... ~\rm m$.
Therefore: \begin{align*} B &= 4\pi\cdot 10^{-7} {{\rm Vs}\over{\rm Am}} \cdot 1 \cdot {{100 ~\rm A}\over{2\pi \cdot 0.412... ~\rm m}} \end{align*}
b) The limit value for the magnetic flux density at the frequency used is $B_0 = 100~\rm \mu T$.
At what distance around the conductor is this value exceeded? (3 points, independent)
The formula for the magnetic field strength can be rearranged: \begin{align*} H &= {{I}\over{2\pi \cdot r}} \\ r &= {{I}\over{2\pi \cdot H}} \\ \end{align*}
Again, the magnetic flux density $B$ is given as: $B = \mu_0 \mu_r H$
Therefore:
\begin{align*}
r &= \mu_0 \mu_r {{ I }\over{2\pi \cdot B}} \\
&= 4\pi\cdot 10^{-7} {{\rm Vs}\over{\rm Am}} {{100 ~\rm A}\over{2\pi \cdot 100\cdot 10^{-6} {~\rm T}}} \\
\end{align*}
Exercise E3 Toroidal Coil
(written test, approx. 5 % of a 120-minute written test, SS2021)
A magnetic field with a flux density of at least $50 ~\rm mT$ is to be achieved in a ring-shaped coil (toroidal coil).
The coil has 60 turns, wound around soft iron with $\mu_{\rm r} = 1200$.
The average field line length in the coil should be $l = 12 ~\rm cm$.
$\mu_{0} = 4\pi\cdot 10^{-7} {{\rm Vs}\over{\rm Am}}$
What is the minimum current that must flow through a single winding?
The magnetic field strength of a toroidal coil is given as:
\begin{align*} H &= {{N \cdot I}\over{l}} \end{align*}
Based on the flux density the magnetic field strength can be derived by $B = \mu_0 \mu_{\rm r} \cdot H$.
By this, the formula can be rearranged:
\begin{align*} H &= {{N \cdot I}\over{l}} \\ {{B}\over{ \mu_0 \mu_{\rm r}}} &= {{N \cdot I}\over{l}} \\ I &= {{B \cdot l}\over{ \mu_0 \mu_{\rm r} \cdot N}} \end{align*}
Putting in the numbers: \begin{align*} I &= {{ 0.05 {~\rm T} \cdot 0.12{~\rm m} }\over{ 4\pi\cdot 10^{-7} {{\rm Vs}\over{\rm Am}} \cdot 1'200 \cdot 60}} \\ &= 0.6631... {{\rm T\cdot m}\over{ {{\rm Vs}\over{\rm Am}} }} &= 0.6631... {{\rm {{\rm Vs}\over{\rm m^2}} \cdot m}\over{ {{\rm Vs}\over{\rm Am}} }} &= 0.6631... ~\rm A \end{align*}
Exercise E4 Lorentz Force (hard!)
(written test, approx. 10 % of a 120-minute written test, SS2021)
A $300 ~\rm km$ long, straight high-voltage direct current transmission line shall be analyzed. A current of $I = 1′200 ~\rm A$ flows through it.
A homogeneous geomagnetic field is assumed. The magnetic field strength has a vertical component of $B_{\rm v} = 40 ~\rm \mu T$ and a horizontal component of $B_{\rm h} = 20 ~\rm \mu T$.
The angle between the transmission line and the horizontal component of the field strength is $\alpha = 20°$.
The picture on the right shows the line (black), the field strength components, and the angle in front and top view for illustration purposes.
a) Calculate the force that results from the current flow on the entire conductor.
First, calculate the vertical and horizontal components and combine them accordingly.
The force on the transmission line can be calculated via the Lorentz force $\vec{F}_\rm L$: \begin{align*} \vec{F} = I \cdot (\vec{l} \times \vec{B}) \end{align*}
Here, we have two components for the current - and therefore for the force - to evaluate.
Considering the right-hand rule (and the cross product), the vertical field $B_{\rm v}$ generates a horizontal force $F_{\rm h}$ and vice versa.
The horizontal component is given by
\begin{align*} F_{\rm h} &= I \cdot (l \cdot B_{\rm v}) \\ &= 1′200 {~\rm A} \cdot 300 \cdot 10^3{~\rm m} \cdot 40 \cdot 10^{-6}{~\rm {{Vs}\over{m²}}} \\ &= 14'400 ~\rm {{VAs}\over{m}} = 14'400 ~\rm {{Ws}\over{m}} = 14'400 ~\rm N \end{align*}
For the vertical component the angle &\alpha& has to be considered.
For the maximum $F_{\rm v}$ the angle &\alpha& has to be $90°$, therefore the $\sin$ has to be used.
\begin{align*} F_{\rm v} &= I \cdot l \cdot B_{\rm h} \cdot \sin\alpha \\ &= 1′200 {~\rm A} \cdot 300 \cdot 10^3{~\rm m} \cdot 40 \cdot 10^{-6}{~\rm {{Vs}\over{m²}}} \cdot \sin 20° \\ &= 2'462.545... ~\rm N \end{align*}
For the overall force $F$ the Pythagorean theorem has to be used:
\begin{align*} F &= \sqrt{F_{\rm v}^2 +F_{\rm h}^2} \\ &= \sqrt{({14'400 ~\rm N})^2 +({2'462.545... ~\rm N})^2} \\ &= 14'609.04... ~\rm N \end{align*}
b) The picture below shows the top view again. In which of the directions shown does the horizontal component $F_{\rm h}$ of the resulting force act? (Independent)
- The horizontal component $\vec{F}_{\rm h}$ of the force is based on the vertical component $\vec{B}_{\rm v}$ of the magnetic field.
- The vertical component $\vec{B}_{\rm v}$ of the magnetic field is not shown in the image but is pointing into the ground.
- It has to be perpendicular to $\vec{B}_{\rm v}$ and to $\vec{l}$. The right-hand rule has to be applied.
Task 3.2.1 Magnetic Field Strength around a horizontal straight Conductor
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 „specific“ length $l$ of the closed path around the area. This shall give you the formula (when not in already known)
- 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*}
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 „specific“ length $l$ of the closed path around the area.
- 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}} \\ &= I_0 \cdot {{r_2^2 }\over{r_{\rm L}^2 }} \end{align*}
Therefore, the $H$-field is: \begin{align*} H(r) &= {{\Delta I}\over{2\pi \cdot r_2}} &&= {{I_0 \cdot {{ r_2^2}\over{r_{\rm L}^2}} }\over{2\pi \cdot r_2}} \\ &= {{I_0 \cdot {{ r_2}\over{r_{\rm L}^2}} }\over{2\pi}} &&= {{1}\over{2\pi}} I_0 \cdot {{ r_2}\over{r_{\rm L}^2}} \end{align*}
Task 3.2.2 Superposition
Three long straight conductors are arranged in a vacuum to lie at the vertices of an equilateral triangle (see Abbildung 1). The radius of the circumcircle is $r = 2 ~\rm cm$; the current is given by $I = 2 ~\rm A$.
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, the constant distance between point $\rm P$ and the three conductors dictates, that the $H$-field has a similar magnitude.
- By the symmetry of the conductor, the angles of the $H$-field vectors are defined and evenly distributed on the revolution:
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:
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 $H(I_3)$.
Task 3.2.3 Magnetic Potential Difference
Given are the adjacent closed trajectories in the magnetic field of current-carrying conductors (see Abbildung ##). Let $I_1 = 2~\rm A$ and $I_2 = 4.5~\rm A$ be valid.
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.
Task 3.3.1 magnetic Flux Density
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 $10'000$ windings and a toroidal diameter for the average field line of $d = 1~\rm m$.
How much current $I$ is necessary for one of the windings of the toroidal coil?
- 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 „specific“ length $l$ of the closed path around the area. This shall give you the formula (when not already known)
- 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}} \cdot 10'000}} \end{align*}
2. What would be the current $I_{\rm Fe}$, when a iron core with $\varepsilon_{\rm Fe,r} = 10'000$?
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{10'000 \cdot 4\pi\cdot 10^{-7} {\rm{Vs}\over{Am}} \cdot 10'000}} \end{align*}
Task 3.3.2 Electron in Plate Capacitor with magnetic Field
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.
Calculate the velocity depending on the other parameters $\vec{v} = f(U, |\vec{B}|, d) $!
- 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 $U$?
\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| &= |\vec{F}_L|\\ |q_e \cdot \vec{E}| &= |q_e \cdot \vec{v} \times \vec{B}| \\ q_e \cdot |\vec{E}| &= q_e \cdot |\vec{v} \times \vec{B}| \\ |\vec{E}| &= |\vec{v} \times \vec{B}| \\ \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, for the plate capacitor $|\vec{E}|= U/d$.
Therefore, it leads to the following:
\begin{align*} {{U}\over{d}} &= v \cdot B \\ v &= {{U}\over{B \cdot d}} \end{align*}
Task 1
Embedded resources
Please have a look at the German contents (text, videos, exercises) on the page of the KIT-Brückenkurs >> Lorentz-Kraft. The last part „Magnetic field within matter“ can be skipped.
A living insect („diamagnet“) floats in a very strong magnetic field
Explanation of diamagnetism and paramagnetism