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Consider an ideal solenoid that is infinitely long. As the current flowing through the solenoid varies over time, I am taught that an electric field will be induced and the field lines will be circles that are concentric with the axis of the solenoid.

Why does this have to be the case? Why can't there be a component of the electric field that is parallel to the solenoid?

It would be helpful for somebody to explain this directly from Maxwell's equations.

My own guess (if you know the answer please tell me if I am right or wrong): Focusing on Maxwell's equations in vacuum. The rate of change of B-field is equal to the negative of the curl of E-field. This implies that, when the B-field's strength is changed with it's direction being kept constant (which is the case here), the E-field created must be normal to the B-field's direction. We also know that, in vacuum, the divergence of the E-field is 0. The only possibility is for the E-field to be circles that are concentric with the axis of the solenoid.

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Inside the solenoid the magnetic field is uniform and parallel to the axis, so we may write$$\vec{B}=B\ \vec{\hat{z}}.$$ So, using the Faraday-Maxwell equation anywhere in the solenoid,$$\text{curl}| \vec{E}=-\frac{\partial{\vec{B}}}{dt}$$We now use the curl in cylindrical polars, to suit the symmetry of the set-up. Since $\vec{B}$ has no components in the $\vec {\hat{r}}$ or $\vec {\hat{\theta}}$ directions, we need only equate $-\frac{\partial{{B}}}{dt}$ to the $\vec {\hat{z}}$ component of curl in cylindrical polars, so $$-\frac{\partial{{B}}}{dt}=\frac{1}{r} \left(\frac{\partial (r E_{\theta})}{\partial r}\ -\frac{\partial E_{r}}{\partial \theta}\right).$$ Because the magnetic field is uniform, symmetry shows us that the second term in the bracket disappears. We can now integrate up wrt r, remembering that $-\frac{\partial{{B}}}{dt}$ is a constant from the point of view of this integration. So $$\int_{0}^{R}-\frac{\partial{B}}{dt} r\ dr =\int_{r=0}^R d(rE_\theta).$$ Thus $$R E_{\theta}=-\frac{R^2}{2} \frac{\partial{B}}{dt}.$$ Strictly we should add arbitrary functions of $\theta$ and $z$, since these would disappear in a partial differentiation wrt $r$, but symmetry considerations make it unnecessary to do so. Finally we note that the emf in a loop of radius $R$ is simply $E_{\theta}$ times the loop circumference, so,$$\mathscr{E}=(-)2\pi R \times \frac{R}{2} \frac{\partial{B}}{dt}=(-)\pi R^2 \frac{\partial{B}}{dt}.$$This result should be familiar!

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    $\begingroup$ I've now completed the formal derivation from the F-M equation! $\endgroup$ – Philip Wood Jan 2 '18 at 18:55

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