# Electromagnetic induction of an uniform magnetic field

EDIT: Thanks to the first answer, I may see that there is differences between a giant solenoid and a completely uniform magnetic field. Therefore it would be great if one can explain to me both the case.

Let's say there is an uniform magnetic field inside a very long and large solenoid. The magnetic field is slowly decreased. How can we find the electric field?

From the Maxwell-Faraday Equation we have $\nabla \times E=-\frac{\partial B}{\partial t}$ and it's can be solved by a curly electric field with the value of electric field is constant (proportional to $|\frac{\partial B}{\partial t}|$) and directed to circles around a centre. But the centre is arbitrarily (one may say it's the central of the solenoid but I find it uncompelling since any centre will satisfy Maxwell equation). So how can one find the electric field in this situation?

• "can be solved by a curly electric field with the value of electric field is constant (proportional to $|\frac{\partial B}{\partial t}|$) and directed to circles around a centre." That's not true. The electric field has magnitude $E(r) = \frac{k}{2}r$, proportional to distance $r$ from the center axis. Commented Aug 16, 2014 at 7:07
• @JánLalinský Your solution is correct inside the solenoid. But there is also an outside, lol... Commented Aug 16, 2014 at 8:13

Uniformity of $\mathbf B$ in space and the equation $$\nabla \times \mathbf E = -\frac{\partial\mathbf B}{\partial t}$$ implies uniformity of $\nabla \times \mathbf E$, but not uniformity of $\mathbf E$, because these conditions do not determine electric field uniquely - there is infinity of solutions. It is an artificial situation which lacks further conditions and no unique electric field can selected.

In case we are interested in the fields of an isolated cylindrical solenoid, there is a new condition: the electric field has the cylindrical symmetry determined by the solenoid. Together with some other conditions, like continuity of the electric field and limiting magnitude 0 at infinity, unique electric field may be selected, which is centered on the solenoid.

Uniqueness of this solution above all other solutions of the above differential equation can be easily understood in terms of retarded fields; if we assume that the EM fields are determined by the charge and current density, since these can be non-zero only in the winding of the solenoid, the EM field has to have cylindrical symmetry of the solenoid.

• There is no need to impose boundary for $\mathbf E$. It is derived to be $\propto\frac{1}{r}$ outside of the infinitely long solenoid from the fact the magnetic field vanishes outside the solenoid.
– Hans
Commented Aug 16, 2014 at 13:54
• @Hans, I've changed the answer. Commented Aug 17, 2014 at 0:44
• That is better.
– Hans
Commented Aug 17, 2014 at 4:56

Because of the cylindrical symmetry of the problem, we know the electric field has to have cylindrical symmetry with respect to the central axis. Integrate both sides of the equation over the cross sectional area and use Stokes theorem.

The cylindrical symmetry will give you the electrical field lies in the plane perpendicular to the central axis and its radial component is zero.

Is this clear?

• why should I need cylindrical symmetry in this situation? In every other time I use symmetry to solve Maxwell equations, the symmetry conditions of the boundary REQUIRE the symmetry of the solution to satisfy Maxwell equations. However in this situation the unsymmetric solution satisfies the Maxwell equation as well, and I see symmetry dispensable. Commented Aug 16, 2014 at 5:45
• @Mr.T: What do you mean by "REQUIRE the symmetry of the solution to satisfy Maxwell equations"? Any solution, even without symmetry, should satisfy Maxwell's equation. Which "unsymmetric solution satisfies the Maxwell equation"? Asymmetric around the axis? Could you give such an asymmetric solution?
– Hans
Commented Aug 16, 2014 at 13:58
• @Mr.T: "why should I need cylindrical symmetry in this situation?" The solution can be seen to be symmetric. Are you saying there is no need to use cylindrical symmetry to solve the problem or are you saying the solution should not be cylindrically symmetric? If your answer is the former, then yes, you do not have to use it to solve the problem. You can compute directly. It is only that the computation would be much more cumbersome. The utility of symmetry is to save labor. If your answer is the latter, then you are wrong. The solution will be cylindrically symmetric around the axis.
– Hans
Commented Aug 16, 2014 at 14:09