Motional Emf using Faraday's Law: Difference between revisions

Revision as of 15:17, 30 November 2015

Claimed by Chelsea Calhoun

Motional Emf Using Faraday's Law

The Main Idea

State, in your own words, the main idea for this topic

A Mathematical Model

What are the mathematical equations that allow us to model this topic. For example [math]\displaystyle{ {\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net} }[/math] where p is the momentum of the system and F is the net force from the surroundings.

A Computational Model

How do we visualize or predict using this topic. Consider embedding some vpython code here Teach hands-on with GlowScript

Examples

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Simple

Middling

Difficult

Connectedness

How is this topic connected to something that you are interested in?
How is it connected to your major?
Is there an interesting industrial application?

History

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@@ Line 3: / Line 3: @@
 ==Motional Emf Using Faraday's Law==
-The potential difference between two locations does not depend on the path taken between the locations chosen.
+==The Main Idea==
+State, in your own words, the main idea for this topic
 ===A Mathematical Model===
-In order to find the potential difference between two locations, we use this formula <math> dV = -\left(E_x*dx + E_y*dy + E_z*dz\right) </math>,  where '''E''' is the electric field with components in the x, y, and z directions. Delta x, y, and z are the components of final location minus to the components of the initial location.
+What are the mathematical equations that allow us to model this topic.  For example <math>{\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}</math> where '''p''' is the momentum of the system and '''F''' is the net force from the surroundings.
 ===A Computational Model===
@@ Line 13: / Line 15: @@
 How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]
-=Simple Example=
+==Examples==
-[[File:pathindependence.png]]
-In this example, the electric field is equal to <math> E = \left(E_x, 0, 0\right)</math>. The initial location is A and the final location is C. In order to find the potential difference between A and C, we use <math>dV = V_C - V_A </math>.
-Since there are no y and z components of the electric field, the potential difference is <math> dV = -\left(E_x*\left(x_1 - 0\right) + 0*\left(-y_1 - 0\right) + 0*0\right)  = -E_x*x_1</math>
-[[File:BC.png]]
-Let's say there is a location B at <math> \left(x_1, 0, 0\right) </math>. Now in order to find the potential difference between A and C, we need to find the potential difference between A and B and then between B and C.
-The potential difference between A and B is <math>dV = V_B - V_A = -\left(E_x*\left(x_1 - 0\right) + 0*0 + 0*0\right) = -E_x*x_1</math>.
-The potential difference between B and C is <math>dV = V_C - V_B = -\left(E_x*0 + 0*\left(-y_1 - 0\right) + 0*0\right) = 0</math>.
+Be sure to show all steps in your solution and include diagrams whenever possible
-Therefore, the potential difference A and C is <math>V_C - V_A = \left(V_C - V_B\right) + \left(V_B - V_A\right) = E_x*x_1 </math>, which is the same answer that we got when we did not use location B.
+===Simple===
+===Middling===
+===Difficult===
 ==Connectedness==

Motional Emf using Faraday's Law: Difference between revisions

Revision as of 15:17, 30 November 2015

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