# Induced electric field from homogeneous magnetic field in free space points in all directions

If there is a homogeneous magnetic field varying with time, say $$\vec{B} = B_0 \sin(t) \hat z$$, then this should, from faradays law of induction, induce an electric field. Since the problem looks symmetric in all directions, the electric field in a circular loop should not change. Thus, if we consider a loop of radius $$r$$, we'd get $$2 \pi r E_\phi = - \pi r^2 B_0 \cos(t) \implies E_\phi = - \frac{r}{2} B_0 \cos(t)$$. But that means that the strength of the E field varies with the radius of the loop chosen. And since it is directed in the $$\hat \phi$$ direction, points in different directions depending on the orientation of the loop chosen. How can this be? What would actually happen?

• It's best to get into the habit of always indicating which quantities are vectors, in particular, a vector equation should have vector notation on both sides. – R. Romero Jan 6 '20 at 16:01

## 1 Answer

Assume your B-field points along the z-axis in cylindrical coordinates. $$\vec{B} = B_o \sin(t)\ \hat{z}$$

Ampere's law (in the absence of any currents and in a vacuum), says $$\nabla \times \vec{B} = \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$$

Now taking the curl of your B-field we have $$\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} = 0,$$ since the curl of a spatially uniform field is zero.

This means that the electric field is a time-independent, This is contrary to your stated E-field.

Now consider Faraday's law, we see that the curl of the electric field must equal $$-B_{0} \cos(t)\ \hat{z}$$. This means that $$\frac{1}{R}\left( \frac{\partial (RE_{\phi})}{\partial R} - \frac{\partial E_r}{\partial \phi} \right) = -B_{0} \cos(t)$$

If we assume there is no charge present, then the divergence of $$\vec{E}$$ must be zero and therefore $$E_r = 0$$ and so $$E_{\phi} = -\frac{R}{2} B_0 \cos(t) + {\rm const}$$

OK, so Faraday's law tells us that a time-varying E-field coexists with the time-varying B-field, but Ampere's law tells us that the E-field is time-independent.

What could be wrong? Well, you have to make assumptions to arrive at this solution. We assumed that there is no charge present and we assumed there are no currents present. I think what you have shown is that you cannot produce such a B-field without having currents present in the system.

If we introduce a current density $$\vec{J}$$ and take the E-field derived from Faraday's law, then Ampere's law tells us that $$\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t} = -\vec{J},$$ $$- \mu_0 \epsilon_0 \frac{R}{2}B_0 \sin(t) \hat{\phi} = -\vec{J}$$ $$\vec{J} = \mu_0 \epsilon_0 \frac{R}{2}B_0 \sin(t) \hat{\phi}$$

Thus whilst the B-field might be homogenous, it takes an inhomogeneous current distribution and E-field to be consistent with that.

• However there are 2 degrees of freedom in this result since the $(x,y)$ of the origin is not specified and $\hat\phi$ is determined by these. All such entries satisfy the OP's question though. – acarturk Jan 6 '20 at 12:55
• @acaturk Indeed the axis of symmetry for the current density can be placed anywhere. But the E-field follows it. – ProfRob Jan 6 '20 at 12:58
• @RobJeffries Okay, so my error was that my assumption of symmetry was wrong? There cannot be a homogenous magnetic field without some current distribution breaking the translational symmetry. Does that mean that knowledge of the magnetic field is not enough to be able to calculate the electric field? – David Hambraeus Jan 6 '20 at 18:12
• @DavidHambraeus The point is that every magnetic field in the absence of charges need to satisfy the wave equation. Your proposed magnetic field doesn't. – Denis Nardin Jan 6 '20 at 19:03
• @DavidHambraeus I don't think there is a unique solution unless you specify the current density. Note that $\vec{J}$ isn't unique. – ProfRob Jan 6 '20 at 20:30