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## Ampère's Law and Symmetry Argument (20 minutes)

This lecture is the introduction to the Ampère's Law Activity.

Start with an introduction to both Ampère's Law and total current:

- Total current as the flux of current density through a gate (see Acting Out Current Density Activity).
- Ampère's law relates the circulation of the magnetic field around a closed loop and the total current enclosed by the loop.

$$\int{\vec{B}\cdot{\vec{dr}}}=\mu_{0}I_{encl}$$

Refresher about *Proof by Contradiction* (“Little Observer” arguments):

- Finding the magnetic field using Ampère's Law is an inverse problem and can only be done when there is a sufficient amount of symmetry in the current distribution so that the field be pulled out of the flux integral.
- In order to pull the field out of the integral, one must use the symmetry to make assumptions about the components and the dependence of the field (
*e.g.*that there is only an azimuthal component and it is only dependent on the radius). To justify these assumptions, we build on the idea of proof by contradiction, where one assumes the opposite of what we want to show and demonstrates a contradiction. - The idea is to make arguments about the functional dependence and direction of the magnetic field by comparing the assumptions made about the field and what an observer placed near the current density in various locations and facing various directions would expect to see based on how she perceives the current distribution to change.

*Example:* Infinite sheet of current that extends infinitely in the $x$ and $y$ directions and has finite thickness in the $z$ direction.

(can be done as a SWBQ where students are asked to Draw the Amperian loop)*What Amperian loop do we want?*- One that is parallel or perpendicular to field at every point and where the magnitude is constant ($\vec{B}=\pm B(z)\, \hat{y}$ from symmetry arguments below).
- constant $z$, same infinitesimal distance ($dz$) above and below sheet

*Symmetry arguments for directional components*- Magnetic field must be perpendicular to the current at all points (cross product in Biot-Savart Law): $\hat{J}=\hat{x} \Rightarrow \hat{B}\perp\hat{x}$
- Magnetic field does not have component in the $z$ direction.
- If an observer standing on a point above the surface facing in the direction of the current were to close her eyes, reverse the direction of the current ($-\hat{x}$), rotate 180$^\circ$ and open her eyes, there would be nothing to indicate that she has turned. Therefore, every measurement that she takes must be the same in both orientations.
- If she were to measure the magnetic field in the $+\hat{z}$ direction, close her eyes, reverse the direction of the current, rotate, open her eyes, and measure again, the current would have reversed direction and thus, the magnetic field would measure in the $-z$ direction.
- This is a contradiction, thus the magnetic field cannot have a component in the $z$ direction: $\hat{B}\neq\hat{z}$

- Therefore, the magnetic field is purely in the $y$ direction.

*Symmetry arguments for coordinate dependence*- Magnetic field does not have a dependence on $x$ or $y$.
- If an observer were to close her eyes, move in the $x$ or $y$ direction, and open her eyes, the current distribution would look the same.
- Suppose she measures the magnitude of the magnetic field to be $B=3$, closes her eyes, moves in the $x$ or $y$ direction, opens her eyes, and measures again. If the magnetic field depended on either $x$ or $y$, she would measure a different magnitude.
- This is a contradiction, thus the magnitude of the magnetic field cannot depend on $x$ or $y$: $B\neq B(x)\neq(y)$

- If she were to move up or down ($z$ direction), the world may or may not look the same. Therefore, the magnitude of the magnetic field only depends on $z$.

This problem can drag on if followed through all the way to the end, so it is better to stop after the symmetry arguments and the Amperian loop is chosen. This takes them from an inverse problem to a point where you can pull B out of the integral. If students are really struggling with the integral or with finding the current, you can stop the class and continue working on the example.