Faraday's law and a uniform magnetic field

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Homework Help Overview

The discussion revolves around Faraday's law in the context of a uniform magnetic field that is changing over time. The original poster is exploring the implications of this change on the direction of the electric field generated.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the reasoning behind the assertion that the electric field is circumferential, drawing an analogy between Faraday's Law and Ampere's Law. Other participants engage in discussing the mathematical similarities between the two laws and question the implications of the changing magnetic field on the electric field's direction.

Discussion Status

The discussion is active, with participants questioning the logic behind the circumferential nature of the electric field in relation to the changing magnetic field. Some guidance is offered regarding the mathematical relationships, but there is no explicit consensus on the reasoning yet.

Contextual Notes

The original poster references a specific example from Griffiths, indicating a reliance on established texts for understanding the concepts involved. There may be assumptions about the participants' familiarity with the laws of electromagnetism that are being examined.

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[SOLVED] faraday's law

Homework Statement


A uniform magnetic field \mathbf{B}(t) in the z-direction, fills a circular region in the x-y plane. If B is changing with time, what is the direction of \mathbf{E}/


Homework Equations





The Attempt at a Solution


My book says it is circumferential, just like the magnetic field inside a long straight wire carrying a uniform current density.

Apparently they are using the analogy between Faraday's Law and Ampere's Law. But I do not see the logic at all.

This is Griffiths Example 7.7.
 
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They are mathematically very similar. Faraday's law:

\nabla \times \vec E = -\frac{\partial \vec B}{\partial t}

And Ampere's Law (for electrostatics):

\nabla \times \vec B = \mu_0 \vec J
 
I know, but why does that imply that the E-field is circumferential?
 
Because \partial \vec B / \partial t is vertical.
 

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