Why does a closed current carrying coil experience no force when placed in a uniform electric field, whereas a charge moving with constant velocity through a uniform electric field does?
Electromagnetism – Force on a Current-Carrying Loop Due to Uniform Electric Field
electric-fieldselectricityelectromagnetismelectrostatics
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For the magnetic field, the currents are one source of the magnetic, but this problem is more linked to the source of the current in the wire. For a conductor with finite conductivity, an electric field is needed in order to drive a current in the wire.
If we assume your wire is straight, this required field is uniform. One way to realize this field is by taking two oppositely charged particles and send them to infinity while increasing the magnitude of their charge to maintain the correct magnitude of electric field. In this limit, you will obtain a uniform electric field through all space.
Now, put your conductor in place along the axis between the voltage sources--a current will flow. In the DC case, this gives rise to the magnetic field outside of the wire. As for the electric field, a conductor is a material with electrons that can move easily in response to electric fields and their tendency is to shield out the electric field to obtain force balance. Because the electrons can't just escape the conductor, they can only shield the field inside the conductor and not outside the conductor. With this model, we see that the electric field is entirely set up by the source and placing the conductor in the field really just establishes a current. Note here that if you bend the wire or put it at an angle relative to the field, surface charges will form because you now have a field component normal to the surface.
For the limit of an ideal conductor, no electric field is needed to begin with to drive the current and so there isn't one outside the wire.
For the AC case, solving for the fields becomes wildly complicated very fast as now the electric field driving particle currents has both a voltage source and a time-varying magnetic source through the magnetic vector potential. The essential physics is the same, though, as the source will establish the fields (in zeroth order), and the addition of the conductor really just defines the path for particle currents to travel. In the next order, the current feeds back and produces electromagnetic fields in addition to the source(s) and will affect the current at other locations in the circuit.
I guess a short answer to your question is that there are always fields outside of the current-carrying wire and the electric field outside disappears only in the ideal conductor limit. Conductors generally do not require very strong fields to drive currents anyway so that the electric field outside is usually negligible, but don't neglect it for very large potentials in small circuits.
This depends on exactly what you mean by non-uniform, or (equivalently) on how big the loop is. In particular, the important criterion is whether the field changes appreciably over distances that are about the same size as the loop.
If the field changes throughout space, but the loop is small enough that the field doesn't change much from point to point on the loop, then the uniform-field formula $\boldsymbol\tau=\boldsymbol\mu\times\mathbf B(\mathbf r)$ still applies. In essence, the field is locally uniform, though the direction and magnitude it's uniform on can change from place to place.
If the loop is big enough that the field changes appreciably over its span then there's nothing for it but to integrate the local torque on each bit of circuit and add them up, which gives you $$\boldsymbol\tau =\oint_\mathcal{C}\mathbf r\times\mathbf F(\mathbf r)\:\mathrm d l =\oint_\mathcal{C}\mathbf r\times(\hat{\mathbf t}I\times\mathbf B(\mathbf r))\:\mathrm d l =\oint_\mathcal{C}\mathbf r\times(I\mathrm d \mathbf l\times\mathbf B(\mathbf r)). $$ There really isn't much you can do to simplify it beyond that without special assumptions. The integral is a line integral, of exactly the same sort you use to calculate the magnetic dipole moment $\boldsymbol \mu$ itself.
The following is a bit more technical and uses a standard amount of vector calculus as used in undergraduate electromagnetism courses.
If the field were constant, then you can manipulate it more freely to get \begin{align} \boldsymbol\tau =\oint_\mathcal{C}\mathbf r\times(I\mathrm d \mathbf l\times\mathbf B) =I\oint_\mathcal{C}\left[(\mathbf B·\mathbf r)\mathrm d\mathbf l-(\mathbf r·\mathrm d\mathbf l)\mathbf B\right], \end{align} using a vector triple product. Here the second integral vanishes, because Stokes' theorem implies that $$ \oint_\mathcal{C}\mathbf r·\mathrm d\mathbf l=\int_\mathcal{S}(\nabla\times\mathbf r)·\mathrm d\mathbf S=0. $$ You're left with the first integral, $\boldsymbol\tau=I\oint_\mathcal{C}(\mathbf B·\mathbf r)\mathrm d\mathbf l$, which in component notation reads $\tau_i=IB_j\oint_\mathcal{C} x_j \mathrm dx_i$ (I use Einstein summations throughout). As it turns out, the indices on that last integral are antisymmetric, i.e. $$\oint_\mathcal{C} x_j \mathrm dx_i=-\oint x_i \mathrm dx_j,$$ which you can again prove via Stokes' theorem since \begin{align} \oint_\mathcal{C}\left[ x_j \mathrm dx_i+x_i\mathrm dx_j\right] =\oint_\mathcal{C}\left[ x_j (\nabla x_i)·\mathrm d\mathbf l+x_i(\nabla x_j)·\mathrm d\mathbf l\right] =\oint_\mathcal{C}\nabla(x_ix_j)·\mathrm d\mathbf l =\int_\mathcal{S}\nabla\times\nabla(x_ix_j)·\mathrm d\mathbf S =0. \end{align} What this means is that you can "fold" the old integral into two antisymmetric parts, as $$\tau_i=IB_j\oint_\mathcal{C} \frac{x_j \mathrm dx_i-x_i \mathrm dx_j}{2},$$ and the antisymmetrized integral is now exactly the magnetic dipole moment $$ \boldsymbol\mu=\frac I2\oint_\mathcal C\mathbf r\times\mathrm d\mathbf l $$ which in component notation reads $$ \mu_k=I\oint_\mathcal C\varepsilon_{kij}x_i\mathrm dx_j ,\quad\text{or in other words}\quad \mu_k\varepsilon_{kij}=I\oint_\mathcal C \left[x_i\mathrm dx_j-x_j\mathrm dx_i\right]. $$ Back to the torque this means $$\tau_i=\frac12 B_j\varepsilon_{kji}\mu_k,$$ or in vector notation $$ \boldsymbol\tau=\boldsymbol\mu\times\mathbf B. $$ So what is it I've done? All of this work has been to take an integral that had the magnetic field inside it, and factorize it into a system-dependent part and a field dependent part: $$ \boldsymbol\tau =\oint_\mathcal{C}\mathbf r\times(I\mathrm d \mathbf l\times\mathbf B) =\left(\frac I2\oint_\mathcal C\mathbf r\times\mathrm d\mathbf l\right)\times\mathbf B. $$ That's really what the magnetic dipole moment $\boldsymbol\mu$ really is.
OK, sorry, that got out of hand, but I'll get back on topic now. What does this have to do with nonuniform magnetic fields? Well, if your field varies very slowly with respect to the dimensions of the loop, you can suppose that $\mathbf B$ is constant in your integral, and you get the calculation above. The next thing you might try is suppose that $\mathbf B$ is almost constant throughout the extent of the loop, but that you do need to consider the first-order term of its Taylor series. Thus, you could suppose that $$ \mathbf B(\mathbf r)=\mathbf B(\mathbf r_0)+(\mathbf r·\nabla)\mathbf B(\mathbf r_0), $$ or more clearly in component notation $$ B_i(\mathbf r)=B_i(\mathbf r_0)+x_k\frac{\partial B_i}{\partial x_k}(\mathbf r_0). $$
You can then do the same game I've done above with the second term, with the added complication that you have an extra factor of $x_k$ in your integral. The result will be something which depends on the first-order derivative of the field, $\frac{\partial B_i}{\partial x_k}(\mathbf r_0)$, and if you're clever you can factorize out the dependence on the loop into a single factor. This factor will in general be (i) a matrix, or in fancy-speak a tensor, (ii) the integral of a homogeneous second-degree polynomial over the loop, and (iii) it is called the quadrupole moment of the loop.
In general, such calculations are long and messy, but if you're feeling brave I encourage you to have a go at deriving an expression for the quadrupole moment and seeing how simple of an expression you can get for the moment itself and its interaction with the magnetic field's gradient to get the torque. As a start, though, here's one expression for the quadrupolar part of the torque, which I encourage you to derive $$ \tau_i=\frac{\partial B_m}{\partial x_k}·I\oint_\mathcal{C}\left[ x_kx_m\mathrm d x_i-\delta_{im}x_k x_j\mathrm dx_j \right]. $$
Of course, there's only so much you can do with only first-order derivatives. If your field varies slightly faster than that - or if your circuit is somewhat bigger - then the next thing you can try is a second-order Taylor expansion of the field. This gives you a third term which depends on the second derivatives of the magnetic field and on a system-dependent term which is (i) an unwieldy object with three different indices, called a rank-3 tensor, (ii) the integral of a homogeneous third-degree polynomial over the loop, and (iii) is called the octupole moment of the loop. And after that, you can go to even higher orders, and on and on it goes until you feel your calculation is accurate enough, or you give up through exhaustion.
I can tell you, though - none of the formulas there are pretty. That's why they're hard to find online.
Best Answer
A current carrying coil experiences no net force in a uniform electric field because it is electrically neutral. For each positive charge in the coil with a force $\vec F = q \vec E$ there is a corresponding negative charge with force $-\vec F =-q \vec E$. The isolated charge is not electrically neutral.
The motion of the isolated charge is irrelevant as is the motion of the charges in the current.