How does the differential form of Faraday's law relate to induced current?

[PhysicsInNJ]

Homework Statement

I'm in a class where we have to essentially learn E&M ourselves and I'm challenged by Maxwells equations. I'm studying out of Purcell's E&M.

The differential form of Faradays law for Maxwells equations is curl E = -∂B/∂t

Im having trouble interpreting what to make of this. I know that the law is supposed to state that a changing magnetic field (right term) will induce a current. How does curlE represent current? I know that curl is a way of determining how much a vector field rotates but can't seem to merge this with the idea of induced current.

 

[phyzguy]

The way that helped me to visualize things is the following. Imagine the vector field as a flow pattern in a fluid, with the arrows representing the flow of the fluid.

(1) For $\nabla \cdot E$, imagine a small volume at each point in the vector field. $\nabla \cdot E$ measures how much of the flow enters or leaves the small volume. If $\nabla \cdot E = 0$, it means that the same amount of flow that enters the volume leaves. This also means that the flow lines form closed curves, never starting or ending anywhere. Regions where the flow lines terminate are regions of non-zero $\nabla \cdot E$.

(1) For $\nabla \times E$, imagine a small paddle wheel at each point in the vector field. $\nabla \times E$ tells you how much the paddle wheel will turn. The magnitude of $\nabla \times E$ tells you how fast it will turn, and the direction tells you the axis of rotation.

For EM, $\nabla \times E$ having a non-zero value means that the current is flowing around in a loop.

 

[vela]

How does curl E represent current?

Faraday's law says a changing magnetic field induces an electric field. The E field can cause charges to move, giving rise to a current, but if there are no charges around, there's no current.

 

[PhysicsInNJ]

The way that helped me to visualize things is the following. Imagine the vector field as a flow pattern in a fluid, with the arrows representing the flow of the fluid.

(1) For $\nabla \cdot E$, imagine a small volume at each point in the vector field. $\nabla \cdot E$ measures how much of the flow enters or leaves the small volume. If $\nabla \cdot E = 0$, it means that the same amount of flow that enters the volume leaves. This also means that the flow lines form closed curves, never starting or ending anywhere. Regions where the flow lines terminate are regions of non-zero $\nabla \cdot E$.

(1) For $\nabla \times E$, imagine a small paddle wheel at each point in the vector field. $\nabla \times E$ tells you how much the paddle wheel will turn. The magnitude of $\nabla \times E$ tells you how fast it will turn, and the direction tells you the axis of rotation.

For EM, $\nabla \times E$ having a non-zero value means that the current is flowing around in a loop.

I see. Then I would ask, is the rotation an artifact of using a wire in a loop and thus it can only travel in that way, or is it more fundamental than that?

 

[phyzguy]

I see. Then I would ask, is the rotation an artifact of using a wire in a loop and thus it can only travel in that way, or is it more fundamental than that?

No, it's more fundamental than that. I'm sorry if my earlier post implied the necessity of a wire or an actual current flow. Around any loop, the line integral of E around the loop is equal to the minus rate of change of the flux of B through the loop. It does not matter whether there is a physical wire present or not, and it does not matter if any current is flowing or not.

 

[PhysicsInNJ]

No, it's more fundamental than that. I'm sorry if my earlier post implied the necessity of a wire or an actual current flow. Around any loop, the line integral of E around the loop is equal to the minus rate of change of the flux of B through the loop. It does not matter whether there is a physical wire present or not, and it does not matter if any current is flowing or not.

Ok, just wanted to make sure. thank you!