How does a curled magnetic field produce a time-varying magnetic field?

[jbunten]

Hi,

With regards to the Faraday Maxwell Equation form of Farday's Law.

if

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

Then a curled magnetic field (say for instance a loop of wire with direct current passing through it) would produce a time-varying magnetic field? however as we know it does not..

 

[Troels]

if

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

Then a curled magnetic field (say for instance a loop of wire with direct current passing through it) would produce a time-varying magnetic field? however as we know it does not..

It seems to me that your conclusion arises from a notion that the curl of the electric field driving the current somehow depends on the geometry of the wire in which the current runs - it doesn't

 

[Defennder]

Hi,

With regards to the Faraday Maxwell Equation form of Farday's Law.

if

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

Then a curled magnetic field (say for instance a loop of wire with direct current passing through it) would produce a time-varying magnetic field? however as we know it does not..

The equation doesn't say that. It says that the curl of an E-field gives you the time varying form of a B-field. In the case of a DC current, you get a constant B-field, so \frac{\partial \mathbf{B}}{\partial t} = 0. Note that the curl of a constant E-field is 0, so there's no contradiction here.

 

[Ben Niehoff]

The equation doesn't say that. It says that the curl of an E-field gives you the time varying form of a B-field. In the case of a DC current, you get a constant B-field, so \frac{\partial \mathbf{B}}{\partial t} = 0. Note that the curl of a constant E-field is 0, so there's no contradiction here.

The confusion is that in a DC circuit, the current clearly travels in a loop. Due to Ohm's law,

\vec J = \sigma \vec E

the E field clearly must circle in a loop, too.

The catch is that just because the flux lines of a vector field form closed loops, does NOT mean that the curl of the field is non-zero! For example, consider the vector field (in cylindrical coordinates)

\vec E = \frac{1}{\rho} \hat \phi

The flux lines of this field are circles centered around the z-axis. But the curl is zero everywhere!

If this vector field represented the velocity of a fluid, then a small object co-moving with the velocity field would NOT rotate, but it would maintain its orientation while traveling around the z-axis. Velocity fields such as this can occur in a free vortex in fluid dynamics.

 

[Defennder]

The catch is that just because the flux lines of a vector field form closed loops, does NOT mean that the curl of the field is non-zero! For example, consider the vector field (in cylindrical coordinates)

\vec E = \frac{1}{\rho} \hat \phi

The flux lines of this field are circles centered around the z-axis. But the curl is zero everywhere!

I believe the OP was referring to the wrong version of the equation. His/her original assertions seems more relevant to the one in integral form:

\oint \mathbf{E} \cdot d\mathbf{L} = - \int_S \frac{\partial \mathbf{B}}{\partial t} \cdot d\mathbf{S}

My post would make more sense if I were referring to the above one instead. Since curl is evaluated at a point it doesn't seem to make sense to talk about whether E is in a loop or not.

 

[jbunten]

Thank you for all the replies. I now see that where I was getting confused was my incomplete understanding of curl.