Dies ist eine alte Version des Dokuments!

Block 16 - Ampère's Law and Magnetomotive Force (MMF)

Learning objectives

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:

• …

90-minute plan

1. Warm-up (x min):
   1. ….
2. Core concepts & derivations (x min):
   1. …
3. Practice (x min): …
4. Wrap-up (x min): Summary box; common pitfalls checklist.

Conceptual overview

Core content

Generalization of the Magnetic Field Strength

So far, only the rotational symmetric problem on a single wire was considered in formula, when the current $I$ ant the length $s$ of a magnetic field line around it is given:

\begin{align*} \quad H_\varphi ={I\over{s}} = {{I}\over{2 \cdot \pi \cdot r}} \quad \Leftrightarrow \quad I = H_\varphi \cdot {s} \quad \quad \quad | \quad \text{applies only to the long, straight conductor} \end{align*}

Now, this shall be generalized. For this purpose, we will 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*} U = E \cdot s \quad \quad | \quad \text{applies to capacitor only} \end{align*}

In words: The potential difference is given by adding up the field strength along the path of a probe from one plate to the other.
This was extended to the vltage between to points $1$ and $2$. Additionally, we know by Kirchhoff's voltage law that the voltage on a closed path is „0“.

\begin{align*} U_{12} &= \int_1^2 \vec{E} \; {\rm d}\vec{s} \\ U &= \oint \vec{E} \; {\rm d}\vec{s} =0 \\ \end{align*}

We can now try to look for similarities. Also for the magnetic field, the magnitude of the field strength is summed up along a path to arrive at another field-describing quantity.
Because of the similarity the so-called magnetic potential difference $V_m$ between point $1$ and $2$ is introduced:

\begin{align*} V_m &= H \cdot s \quad \quad | \quad \text{applies to rotational symmetric problems only} \\ \end{align*}

\begin{align*} \boxed{ V_m = V_{m, 12} = \int_1^2 \vec{H} \; {\rm d}\vec{s} \\ V_m = \oint \vec{H} {\rm d}\vec{s} = \theta } \end{align*}

We need to take a loser look here. Any closed path in the static electric field leads to a potential difference of $U = \oint \vec{E} \; {\rm d}\vec{s} =0$.
BUT: closed paths in the static magnetic field leads to a magnetic potential difference which is not mandatorily $0$! $ V_m = \oint \vec{H} {\rm d}\vec{s} = \theta$

Another new quantity is introduced: the magnetic voltage $\theta$:

1. The magnetic voltage $\theta$ is the magnetic potential difference on a closed path.
2. Since the magnetic voltage $\theta$ is valid for exactly one turn along our single wire, $\theta$ is also equal to the current through the wire:
   \begin{align*} \theta = H \cdot s = I \quad \quad | \quad \text{applies only to the long, straight conductor} \end{align*}
3. The magnetic potential difference can take a fraction or a multiple of one turn and is therefore not mandatorily equal to $I$.
4. The magnetic voltage is generalized in the following box.

Common pitfalls

• …

Exercises

Exercise E2 Magnetic Field Lines
(written test, approx. 6 % of a 120-minute written test, SS2024)

The following setup shall be given:

• Four conductors are located perpendicular to the plane of the diagram
• All of them conduct a current with the same magnitude, but not in the same direction.
• A permanent magnet is located in between the conductors.

1. Do not consider the permanent magnet at first. Draw at least 10 field lines of the H-field qualitatively. Give a a correct representation of their direction, and density for the shown area.

Result

2. Discuss how the permanent magnet affects the H-field, based on the fundamental definition of the H-field.

Result

• The H-field is defined by currents $\sum I = \int H {\rm d}s$ .
• In the permanent magnet, there are no free currents.
• The bound currents (of the permanent magnet) create also an H field.
• This exits on the north pole and enters the magnet on the south pole (similar to the B-field)_
• $H = B/\mu$
• The H-field from task 1 gets distracted

Exercise E1 Fields of an coax Cable
(written test, approx. 12 % of a 120-minute written test, SS2024)

A $0.5 ~\rm m$ long coax cable is used for signal transmission. The diagram shows the cross-section of the coax cable with the origin in the center of the coax cable. Due to the given load, the following situation appears:

• Inner conductor: $+3.3 ~\rm mA$, $+10 ~\rm nC$ (current into the plane of the diagram)
• Outer conductor: $-3.3 ~\rm mA$, $ 0 ~\rm nC$ (current out of the plane of diagram)

1. What is the magnitude of the magnetic field strength $H$ at $\rm (0.1 ~mm | 0)$ and $\rm (0.55 ~mm | 0)$?

Path

The magnitude of the magnetic field strength $H$ can be calculated by: $H = {{I}\over{2 \pi \cdot r}} $
So, we get for $H_{\rm i}$ at $\rm (0.1 ~mm | 0)$, and $H_{\rm o}$ at $\rm (0.55 ~mm | 0)$:

\begin{align*} H_{\rm i} &= {{I}\over{2 \pi \cdot r_{\rm i}}} \\ &= {{+3.3 A}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m}}} \\ H_{\rm o} &= {{I}\over{2 \pi \cdot r_{\rm o}}} \\ &= {{+3.3 A}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}}} \\ \end{align*}

Hint: For the direction, one has to consider the right-hand rule. By this, we see that the $H$-field on the right side points downwards.
Therefore, the sign of the $H$-field is negative.
But here, only the magnitude was questioned!

Result

• for $(0.1 ~/rm mm | 0)$ : $H_{\rm i} = 5.25... ~\rm A/m$
• for $(0.55 ~/rm mm| 0)$ : $H_{\rm o} = 0.955... ~\rm A/m$

2. Plot the graph of the magnitude of $H(x)$ with $x \in \rm [-0.6~mm, +0.6~mm]$ from $\rm (-0.6 ~mm | 0)$ to $\rm (0.6 ~mm | 0)$ in one diagram. Use proper dimensions and labels for the diagram!

Path

• In general, the $H$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors.
• For $H(x)$ with $x$ within the inner conductor, one has to consider how much current is conducted within a circle with the radius $x$.
  This is proportional to the area within this radius. Therefore, Ther formula $H = {{I}\over{2 \pi \cdot r}} $ gets $H(x) = {{I}\over{2 \pi \cdot x}} \cdot {{\pi x^2}\over{\pi (0.1~\rm mm)^2}}$. This leads to a formula proportional to $x$.
• For $x$ within the outer conductor one also gets a linear proportionality with a similar approach.

Result

3. What is the magnitude of the electric displacement field $D$ at $\rm (-0.1 ~mm | 0)$ and $\rm (0.55 ~mm | 0)$?

Path

The magnitude of the electric displacement field $D$ can be calculated by: $\int D {\rm d}A = Q$.
Here, for any position radial to the center, the surrounding area is the surface of a cylindrical shape (here for simplicity without the round endings).
This leads to: \begin{align*} D(x) &= {{Q}\over{A}} \\ &= {{Q}\over{l \cdot 2\pi \cdot x}} \\ \end{align*}

So, we get for $D_{\rm i}$ at $\rm (0.1 ~mm | 0)$, and $D_{\rm o}$ at $\rm (0.55 ~mm | 0)$:

\begin{align*} D_{\rm i} &= {{Q }\over{2 \pi \cdot r_{\rm i} \cdot l}} \\ &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.1 \cdot 10^{-3}~\rm m} \cdot 0.5 ~\rm m}} \\ D_{\rm o} &= {{Q }\over{2 \pi \cdot r_{\rm o} \cdot l}} \\ &= {{10 \cdot 10^{-9} C}\over{2 \pi \cdot { 0.55 \cdot 10^{-3}~\rm m}\cdot 0.5 ~\rm m}} \\ \end{align*}

Hint: For the direction, one has to consider the sign of the enclosed charge. By this, we see that the $D$-field is positive.
But here, again only the magnitude was questioned!

Result

• for $(0.1 ~\rm mm | 0)$ : $D_{\rm i} = 31.8... ~\rm uC/m^2$
• for $(0.55 ~\rm mm| 0)$ : $D_{\rm o} = 5.78... ~\rm uC/m^2$

4. Plot the graph of the magnitude of $D(x)$ with $x \in \rm [-0.6~mm, +0.6~mm]$ from $\rm (-0.6 ~mm | 0)$ to $\rm (0.6 ~mm | 0)$ in one diagram. Use proper dimensions and labels for the diagram!

Path

• In general, the $D$-field is proportional to ${{1}\over{r}}$ for the situation between both conductors.
• Since the charges are on the surface of the conductor, there is no $D$-field within the conductor.

Result

Embedded resources