MyOpenMath/Solutions/Quizbank EM

User:Guy vandegrift/T/Title Here are posted solutions to three Quizbank quizzes that have been transported to MyOpenMath.

Ampere's Law Symmetry

File:Quizbank Ampere's law with H solutions.pdf

Click here to expand. Or click image to access file.

• This quiz may be found at special:permalink/1863400

${ϵ}_0\partial \overrightarrow{E}/\partial t$ is called the displacement current because it replaces the current density when using Ampère's circuital law to calculate the line integral of the magnetic field around a closed loop. To keep this problem simple, we can use H instead of B: By defining B=μ₀H, we can express Ampere's law without any constants:

${\oint }_C𝐇\cdot \mathrm{d}𝐥=I_{\text{enclosed}}\text{ where }𝐁={\mu }_0𝐇.$

It should be noted that another way to remove the annoying μ₀ is to use a different set of variables. By considering an infinitely long wire viewed top-down along the z-axis, the line integral associated with a single line charge reduces to an integral over the angle:

$\underset{a}{\overset{b}{\int }}𝐇\cdot \mathrm{d}𝐥=\frac{\mathrm{\Delta }\theta }{2\pi }{I}_{\text{enclosed}}$

Note the similarity between this discussion of Ampere's law for a single infinite line charge to the three-dimensional discussion of Gauss's law for a single point source at [[../Gauss_law_(TF)/Proof/]]. And to keep the questions even simpler for those who understand the concept, the solution posted to the right uses shapes where $\mathrm{\Delta }\theta$ can be obtained by inspection of a sketch of the graph.

Similar links to Wikiversity pdf files for two other quizzes are shown below

Ampere's Law

File:22-AmpereLaw-answers.pdf

https://upload.wikimedia.org/wikiversity/en/0/0c/22-AmpereLaw-answers.pdf

• This quiz may be found at special:permalink/1828922

These are more conventional problems, except that one uses a known formula for magnetic field to calculate the line integral. My philosophy in constructing this OER quizbank is that we need so many ways to ask the same question that students will it easier difficult to understand the physics and then glance over the publicly posted solutions, than to attempt to memorize all the solutions.

Displacement current

• For more help with these questions visit Physics_equations/24-Electromagnetic_Waves

File:24-displmntCurrentCalc.pdf

https://upload.wikimedia.org/wikiversity/en/f/f2/24-displmntCurrentCalc.pdf

• This quiz may be found at special:permalink/1863401

This was a much more difficult quiz than I anticipated. The problem is that there are many was to do the same problem, and some are quite tedious.

Appendix:Two annoying constants μ₀ and ε₀

Another way to avoid the constants μ₀ and ε₀ is to use a different system of units. From Wikipedia, we see that Maxwell's equations can be described in a wide variety of units. The two most common are:

SI units

${\oint }_S𝐄\cdot \mathrm{d}𝐀=\frac{1}{{ϵ}_0}\underset{V}{\int }\rho \mathrm{d}V$	${\oint }_C𝐄\cdot \mathrm{d}𝐥=-\underset{S}{\int }\frac{\partial 𝐁}{\partial t}\cdot \mathrm{d}𝐀$	$\overrightarrow{F}=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$
${\oint }_S𝐁\cdot \mathrm{d}𝐀=0$	${\oint }_C𝐁\cdot \mathrm{d}𝐥={\mu }_0\underset{S}{\int }𝐉\cdot \mathrm{d}𝐀+{ϵ}_0{\mu }_0\underset{S}{\int }\frac{\partial 𝐄}{\partial t}\cdot \mathrm{d}𝐀$	$c^2=\frac{1}{{ϵ}_0{\mu }_0}$

These equations have the practical advantage that if $\overrightarrow{E}=-\mathrm{\nabla }V$, then $V$ is measured in volts. And the units of $dQ/dt$ are coulombs per second, or ampheres ("amps").

Gaussian units

The use of Gaussian units permits the removal of the annoying constants $({\epsilon }_0,{\mu }_0)$, but at the expense of not being able to measure V in volts, Q in coulombs, I in amps, or B in tesla. As before we have:

$\underset{V}{\int }\rho \mathrm{d}V=Q_{\text{enc}}$ and $\underset{S}{\int }𝐉\cdot \mathrm{d}𝐀=\overrightarrow{I},$

but now $(Q,V,I,B)$ are measured in statC (or Fr), statV, statC/s, and gauss (respectively). Length is measured in centimeters and mass is measured in grams.

${\oint }_S𝐄\cdot \mathrm{d}𝐀=4\pi \underset{V}{\int }\rho \mathrm{d}V$	${\oint }_C𝐄\cdot \mathrm{d}𝐥=-\frac{1}{c}\underset{S}{\int }\frac{\partial 𝐁}{\partial t}\cdot \mathrm{d}𝐀$	$\overrightarrow{F}=q(\overrightarrow{E}+\frac{\overrightarrow{v}}{c}\times \overrightarrow{B})$
${\oint }_S𝐁\cdot \mathrm{d}𝐀=0$	${\oint }_C𝐁\cdot \mathrm{d}𝐥=\frac{1}{c}\underset{S}{\int }𝐉\cdot \mathrm{d}𝐀+\frac{1}{c}\underset{S}{\int }\frac{\partial 𝐄}{\partial t}\cdot \mathrm{d}𝐀$