# I Spatial derivative of Electric Field in Faraday's Law?

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1. Nov 3, 2016

### sawer

According to Faraday's Law, Time-Changing magnetic field creates an induced current in a closed conducting loop.

This is the equation: $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$

1-) Does this current ($\nabla \times \mathbf{E}$) have to be an alternate current(sinusoidal current)?

2-) If it is, then it is not just spatial varying current also time varying current. But why does left side of this equation ($\nabla \times \mathbf{E}$) include spatial derivative of electric field? Can it be written with time derivative of electric field? (I mean time derivative electric field version). So it means time changing magnetic field relates to time changing electric field.

2. Nov 3, 2016

### jasonRF

Not quite. There is no current in this equation. A time varying magnetic field generates an electric field that has a curl given by the above equation. You can think of electromotive force around loops being related to the time changing magnetic flux through the loop as well (eg the integral form of this equation).

The time variation does not have to be sinusoidal. It can be arbitrary. Again, no current here.

Not sure what you are getting at here. Perhaps if you explicitly include the dependencies of the fields it is more clear,
$\nabla \times \mathbf{E}(\mathbf{r},t) = -\frac{\partial} {\partial t} \mathbf{B}(\mathbf{r},t)$
So this equation does indicate that the temporally changing magnetic field does yield a temporally changing electric field. The time derivative of the electric field is related to the curl of the magnetic field and the current density - the relationship is given in Ampere's Law. In free space (and MKS units) it is,
$\nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \frac{1}{c^2} \frac{\partial}{\partial t} \mathbf{E}$.
Here $\mathbf{J}$ is the current density.

Jason

3. Nov 4, 2016

### sawer

If there is a conducting loop, doesn't a current appear?
This equation indicates that the temporally changing magnetic field does yield a "spatially" changing electric field. Right?
I am asking if this spatially changing electric field is also temporally changing, then can we write left side of this equation a time derivative version of the electric field? Would it be wrong?

Thanks...

4. Nov 5, 2016

### jasonRF

Yes, a current appears in a conducting loop. Sorry for my mistake there. But please note that $\nabla \times \mathbf{E}$ is not a current.

If the right hand side of the equation is changing with time, then so is the left hand side. As an example, consider the fields,
$$\mathbf{B} = \mathbf{\hat{z}} B_0 \sin (\omega t)$$
$$\mathbf{E} = (y\mathbf{\hat{x}}- x\mathbf{\hat{y}})\frac{ B_0 \omega }{2} \cos (\omega t)$$
for $B_0$ a constant. These two fields are consistent with,
$$\nabla \times \mathbf{E}(\mathbf{r},t) = -\frac{\partial} {\partial t} \mathbf{B}(\mathbf{r},t)$$.
I would say both the electric and magnetic fields are varying with time, and yes, the electric field varies with position. The magnetic field is spatially uniform in this example.

Yes, the spatially changing electric field would usually be temporally changing, as shown above. One exception would be if you have a case where $\partial \mathbf{B}/\partial t = \mathbf{F}(\mathbf{r})$ so is not a function of time.

No, you cannot in general write the left side as a time derivative of the electric field. There are only three types of derivatives of the electric field that we would need, and all of them are given to us in maxwell's equations. In free space they are,
$$\nabla \times \mathbf{E} = -\frac{\partial} {\partial t} \mathbf{B}$$.
$$\nabla \cdot \mathbf{E} = \rho /\epsilon_0$$
$$\frac{\partial} {\partial t} \mathbf{E} = c^2 \nabla \times \mathbf{B} - \mathbf{J}/\epsilon_0$$

Hope that helps.

Jason