Consider the two Mawell equations $$∇×E=-∂B/∂t \tag{1}$$ $$∇×B=μ_0 ε_0 ∂E/∂t \,\,\, \mathrm{if} \,\, i=0 \tag{2}$$

Consider the following situation with cilindrycal symmetry. A magnetic field $B$ is uniform in space but it is changing in time with a certain law $B(t)=\alpha t$, and it is directed pointing outwards from screen. (for example one can consider an ideal solenoid with current varying in time, which would give $B(t)=\mu_0 n i(t)$).

Eq $(1)$ says that an electric field is generated. Since we have cilindrycal symmetry I can use Stokes theorem and, considering a surface $S$ that has as border the circle of radius $r$ perpendicular to the direction of $B$:

$$\int_{S} ∇×E \,\,\, dS=-\int_S ∂B/∂t \,\, dS=\oint E \cdot dl$$

$$E(r) 2\pi r=-\pi r^2 \frac{dB}{dt}$$

$$E(r)=-r \alpha \frac{1}{2}$$

Electric field is perpendicular to $B$ but the question is: what rule should I use to predict the orientation of E with respect to B?

I tried with the classic right-hand grip rule (or coffee-mug rule or the corkscrew-rule) (the same as Ampere's law) but it looks wrong in this case, infact, if $B$ is pointing outwards from the screen, then $E$ is directed (tangentially) clockwise! While it should not by the right-hand grip rule.

So how can I predict the direction of $E$ knowing the direction of $B$ in this case?

The same question holds for eq $2$ (in that case, I would like to predict the direction of $B$ generated by changing $E$). In particular, does the minus sign in the first equation affect the rule to use in the two case to find the direction of the generated field?

  • $\begingroup$ Your confusion might stem from the fact that you're missing the vector notation of vector quantities. Faraday's law speaks about an oriented closed curve and a corresponding surface whose normal vector is oriented with respect to the orientation of the closed curve according to the right-hand rule. If you start from there, the sign of the electric field should be meaningful with respect to the orientation of the closed curve. $\endgroup$ – Andras Deak Oct 31 '16 at 0:47

The answer is very simple. Stokes theorem describes the relation between an oriented surface integral and an oriented line integral along the boundary of this surface. The mathematical convention is the following. When you have chosen the side of the surface where the normal surface element vector $d\vec S $ protrudes, and you look from there to the line integral path around the surface, then the line element $d\vec r$ in the line integral is pointing in counterclockwise direction.

Therefore, for the first equation with $\vec B$ and $d\vec S$ pointing out of the screen your line integral is taken anti-clockwise. Because of the minus sign of the surface integral of the positive $∂\vec B/∂t$ this means that $\vec E$ is oriented clockwise (opposite to $d\vec r$).

Analogously, you will find that in equation (2) and positive $∂ \vec E/∂t$, $\vec B$ will be oriented counter-clockwise (same as $d\vec r$).

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