Faraday's Law and Stokes Theorem

Click For Summary
SUMMARY

This discussion focuses on the application of Stokes' Theorem in the context of Faraday's Law, specifically addressing the cancellation of integrals in the equation involving electric fields and magnetic fields. The participant successfully transforms the integral of ∫E⋅dr into ∫∇×E dA, while also manipulating the left side to represent ∫dB/dt dA. The key insight provided by the teaching assistant involves justifying the cancellation of integrals by appealing to the continuity of vector fields and the properties of suitable regions in calculus.

PREREQUISITES
  • Understanding of Stokes' Theorem
  • Familiarity with Faraday's Law of Induction
  • Knowledge of vector calculus
  • Basic principles of continuity in mathematical functions
NEXT STEPS
  • Study the proof of Stokes' Theorem in detail
  • Explore the implications of Faraday's Law in electromagnetic theory
  • Learn about the continuity of vector fields and its applications
  • Review integral calculus concepts related to multi-dimensional integrals
USEFUL FOR

Students of physics and mathematics, particularly those studying electromagnetism and vector calculus, as well as educators looking to clarify the application of Stokes' Theorem in practical scenarios.

CGI
Messages
74
Reaction score
1

Homework Statement


upload_2018-2-21_14-52-9.png


Homework Equations


Stokes Theorem

The Attempt at a Solution


I'm having a tough time "cancelling" out integrals from both sides of an equation (if possible). On the right hand of the equation, we know since it is a closed curve, that Stoke's Theorem applies and we can change the integral of ∫E⋅dr to ∫∇×E dA. On the left hand of the equation, we can insert d/dt into the integrand since it is independent of dA, giving us ∫dB/dt dA.

Overall, After moving the negative side over, we have:
-∫dB/dt dA = ∫∇×E dA

Now, from what my TA told me, he said there was a way to justify the "cancellation" of the integrals. I'm assuming they have the same bounds, but he gave me a hint and said it was something related to what I learned in Calc I, but having done so in only 1D.

Anyone have any ideas? It would be much appreciated!
 

Attachments

  • upload_2018-2-21_14-52-9.png
    upload_2018-2-21_14-52-9.png
    11.3 KB · Views: 1,744
Physics news on Phys.org
CGI said:

Homework Statement


View attachment 220817

Homework Equations


Stokes Theorem

The Attempt at a Solution


I'm having a tough time "cancelling" out integrals from both sides of an equation (if possible). On the right hand of the equation, we know since it is a closed curve, that Stoke's Theorem applies and we can change the integral of ∫E⋅dr to ∫∇×E dA. On the left hand of the equation, we can insert d/dt into the integrand since it is independent of dA, giving us ∫dB/dt dA.

Overall, After moving the negative side over, we have:
-∫dB/dt dA = ∫∇×E dA

Now, from what my TA told me, he said there was a way to justify the "cancellation" of the integrals. I'm assuming they have the same bounds, but he gave me a hint and said it was something related to what I learned in Calc I, but having done so in only 1D.

Anyone have any ideas? It would be much appreciated!

Basically, you have that ##\int \int_S \mathbf{U} \cdot d\mathbf{A}= \int \int_S \mathbf{V} \cdot d\mathbf{A}## for every "suitable" region ##S##, and then need to show that this gives ##\mathbf{U} = \mathbf{V}## at all points.

The standard way to try proving such things is to apply it so an "infinitesimal" region ##A## surrounding a point ##\mathbf{r}_0 = (x_0,y_0,z_0)## and appeal to continuity of the vector fields ##\mathbf{U}## and ##\mathbf{V}## around ##\mathbf{r}_o## to show that we must have ##\mathbf{U}(\mathbf{r}_0) = \mathbf{V}(\mathbf{r}_0)##.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K