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If a stationary charge is placed in a changing uniform magnetic field to measure the induced E field at some point, what would be the direction of E?

I think E must equal zero normal to B to maintain the symmetry of the problem, leaving E entirely along B.

Is this correct?

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What is the symmetry of the problem which you try to preserve (note that you have a time-dependent problem) ... – Fabian Apr 8 '11 at 18:20
@fabian you can rotate the axis of the coordinates around B, as well as moving the origin. – Larry Harson Apr 9 '11 at 12:57
up vote 4 down vote accepted

Faraday's law of induction - one of Maxwell's equations - says $$\nabla \times E = - \frac{\partial B}{\partial t}.$$ So the time derivative of $B$ induces an electric field. However, what you incorrectly assumed was that the time derivative of $B$ will directly be proportional to the induced electric field. In reality, it is the "curl" of the electric field that is equal to the time derivative of $B$.

In other words, if $B$ and its time derivative is uniform in space, the field $E$ will not be uniform. For example, if $-\partial B / \partial t = (0,0,\beta)$ in the $z$-direction, the most natural $E$ will depend linearly on spatial coordinates, and may be e.g. $$ E = (-y\beta/2, x\beta/2, 0) $$ This electric field is rotating around the $z$-axis - around the direction of the magnetic field (its change in time).

Note that it is constructed so that its curl is equal to $(0,0,\beta)$. However, this configuration is also nicely rotational symmetric because up to the extra factor of $\beta$, the vector $E$ is nothing else than the vector $r$ projected to the $xy$-plane and rotated by 90 degrees around the $z$-axis - and this definition doesn't depend on the choice of axes in the $xy$-plane, so the configuration is rotationally symmetric.

It is not translationally symmetric in the $x$ and $y$ directions, however. There exists a specific axis - in my case, the $z$-axis - where the induced $E$ vanishes. The solution to Maxwell's equations is not unique. I may add any other uniform $E$ field (or, more generally, any electromagnetic wave solution of Maxwell's equations with no right hand side) without spoiling the validity of the Maxwell's equations - this is a kind of trivial artifact of the linearity of these equations.

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I carelessly assumed that the induced E would be the same through out the region without bothering to check this with the Maxwell equation you quoted. Thanks for your answer. – Larry Harson Apr 9 '11 at 13:04
This is a nice illustration that symmetric problems need not have a symmetric solution. Also possible is a family of asymmetric solutions that are related by the symmetry. – wnoise May 8 '11 at 4:02

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