# Why is the curl of this time varying magnetic field still 0?

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• #1
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I'm trying to become reacquainted with basic electromagnetics. From my understanding a changing magnetic field induces a changing electric field and visa versa, through the equation:
$$\overrightarrow{\bigtriangledown } \times \overrightarrow{B} = \mu_{0} \left [ \widehat{J} + \varepsilon _{0} \frac{\partial \overrightarrow{E} }{\partial t}\right ]$$

So calculating the magnetic field B, around an infinite line of current, where the current flows in the positive z direction, using cylindrical coordinates, I get:
$$\overrightarrow{B} = \frac{\mu _{0} I}{2\pi \rho }\widehat{\phi}$$
Then to find the change of the electric field in the vicinity, take the curl of B, and we get:
$$\overrightarrow{\bigtriangledown } \times \overrightarrow{B} = \left [ \frac{1}{\rho } \frac{\partial Bz }{\partial \phi } - \frac{\partial B\phi }{\partial z} \right ]\widehat{\rho } + \left [ \frac{\partial B\rho }{\partial z} - \frac{\partial Bz }{\partial \rho } \right ]\widehat{\phi } + \frac{1}{\rho } \left [ \frac{\partial (\rho B\phi) }{\partial \rho } - \frac{\partial B\rho }{\partial \phi } \right ]\widehat{z } = 0$$ which seems like it makes sense because the current and magnetic field are not changing. But, if i use a time varying current: where $$I= I_{0}cos(\omega t) \Rightarrow \overrightarrow{B} = \frac{\mu _{0} I_{0}cos(\omega t)}{2\pi \rho }\widehat{\phi}$$ but i still get:
$$\overrightarrow{\bigtriangledown } \times \overrightarrow{B} = \left [ \frac{1}{\rho } \frac{\partial Bz }{\partial \phi } - \frac{\partial B\phi }{\partial z} \right ]\widehat{\rho } + \left [ \frac{\partial B\rho }{\partial z} - \frac{\partial Bz }{\partial \rho } \right ]\widehat{\phi } + \frac{1}{\rho } \left [ \frac{\partial (\rho B\phi) }{\partial \rho } - \frac{\partial B\rho }{\partial \phi } \right ]\widehat{z } = 0$$
So even though I have a time varying current, which seems to produce a time varying magnetic field around the infinite wire, I don't seem to be getting a time varying electric field as I would expect. I can understand why mathematically, seeing that I0cos(ωt) is not a function of ρ or z, so that in the curl, the partial derivatives with respect to ρ and z are still 0.
So I'm not sure where I went wrong. I can see that if I included a phase term in the current, where I0cos(ωt-βz) the curl of B would not equal 0, but I thought that just a time varying current would cause a changing magnetic field which in turn would cause a changing electric field. Can someone point my in the right direction about where I'm going wrong? Thanks

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• #2
BiGyElLoWhAt
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The partial of B sub phi w.r.t. rho is zero?

• #3
jtbell
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If you have a time-varying magnetic field, and you want to find the electric field that it induces, you need to use the Maxwell equation that contains the time derivative of ##\vec B##, that is: $$\vec \nabla \times \vec E = - \frac{\partial \vec B}{\partial t}$$ or the integral version $$\oint {\vec E \cdot d \vec l} = -\frac{d}{dt}\int {\vec B \cdot d \vec a}$$ Griffiths does what is basically your problem as Example 7.9, using the integral version and taking advantage of symmetries in the fields. I expect most any E&M textbook will also discuss this situation as an example, in the chapter on electromagnetic induction.

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• #4
BiGyElLoWhAt
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Hmmm... My phone wouldn't view the last half of your curl last night, and I didn't realize there was a rho in the numerator. I don't see how that comes in though. If the Gradient operator is ##\frac{\partial}{\partial \rho} \hat{\rho} +\frac{1}{\rho}\frac{\partial}{\partial \phi} \hat{\phi} +\frac{\partial}{\partial z} \hat{z}##, crossing that with a vector field doesn't (alone) give you that rho in the numerator within the partial derivative. Would anyone care to explain how this came to be? That's exactly how it's listed on wikipedia, but it seems inconsistent.

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• #5
BiGyElLoWhAt
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Ahh... I think I may have an explaination for this.

The current density is going to be in one direction, whereas the derivative of the electric field (the field driving the electrons) will be negative. This doesn't say you have a time constant E field, it says the sum of the two is 0. This seems to make a little bit more sense. I'm going to look into this cylindrical coordinate's thing a little more. (It's not something I've done a lot with, in terms of del and other things.)
If you take the approach jtbell suggested, you'll end up with an exact PDE, which is readily solveable. This one should be pretty simple (one integral to solve)

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