#### The Induced Electric Field
emd.Fl.ief

Two sources of electric fields: electric charges, and changing magnetic fields.

Electric fields induced by a changing magnetic field are determined in an exactly parallel way as magnetostatic fields from the current: exploit parallel between Ampère and Faraday \[ {\boldsymbol \nabla} \times {\bf B} = \mu_0 {\bf J} \hspace{3cm} {\boldsymbol \nabla} \times {\bf E} = -\frac{\partial {\bf B}}{\partial t}. \] For electric field induced by changing magnetic field: use tricks of Ampère's law in integral form: \[ \oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 I_{enc} \hspace{3cm} \oint_{\cal P} {\bf E} \cdot d{\bf l} = -\frac{d}{dt} \Phi \]

Example: loop with time-dependent flux

Consider a time-dependent magnetic field \({\bf B}(t)\) directed vertically through a horizontal circular region of radius \(R\).

Task: find the induced \({\bf E}(t)\).

Solution: amperian loop of radius \(r\), apply Faraday: \[ \oint {\bf E} \cdot d{\bf l} = E (2\pi r) = -\frac{d\Phi}{dt} = -\pi r^2 \frac{dB}{dt} \Rightarrow {\bf E} = -\frac{r}{2} \frac{dB}{dt} \hat{\boldsymbol \varphi}. \] Increasing \({\bf B}\): clockwise (viewed from above) \({\bf E}\) from Lenz.

Example: wheel with charged rim traversed by flux

Consider a wheel of radius \(b\) with line charge \(\lambda\) on the rim. A uniform magnetic field \({\bf B}_0\) pointing up is traversing the central region up to radius \(a < b\). The field is then turned off. What happens?

Solution: the wheel starts spinning to compensate the reduction of field. Faraday: \[ \oint {\bf E} \cdot d{\bf l} = -\frac{d\Phi}{dt} = - \pi a^2 \frac{dB}{dt} \Rightarrow {\bf E} = -\frac{a^2}{2b} \frac{dB}{dt} \hat{\boldsymbol \varphi}. \] Torque on segment \(d{\bf l}\): \(|{\bf r} \times {\bf F}| = b \lambda E dl\). Total torque: \[ N = b\lambda \oint E dl = -b \lambda \pi a^2 \frac{dB}{dt} \] so total angular momentum imparted to the wheel is \[ \int N dt = -\lambda \pi a^2 b \int_{B_0}^0 dB = \lambda \pi a^2 b B_0. \]

The precise way the field is turned off doesn't matter. Only electric field does work.

N.B.: we use magnetostatic formulas for changing fields. This is called the quasistatic approximation, and works provided we deal with slow enough phenomena.

Example: field from wire with time-dependent current

Consider an infinitely long straight wire which carries current \(I(t)\).

Task: find the induced \({\bf E}\) field as a function of distance \(r\) from wire.

Solution: assuming we can use the quasistatic approximation, the magnetic field is \(B = \frac{\mu_0 I}{2\pi r}\) and circles the wire. Like \({\bf B}\) field of solenoid, \({\bf E}\) runs parallel to wire. Amperian loop with sides at distances \(r_0\) and \(r\): \[ \oint {\bf E} \cdot d{\bf l} = E(r_0)l - E(r)l = -\frac{d}{dt} \int {\bf B} \cdot d{\bf a} = -\frac{\mu_0 l}{2\pi} \frac{dI}{dt} \int_{r_0}^s \frac{dr'}{r'} = -\frac{\mu_0 l}{2\pi} \frac{dI}{dt} \ln(r/r_0). \] So: \[ {\bf E} (r) = \left[ \frac{\mu_0}{2\pi} \frac{dI}{dt} \ln r + K \right] \hat{\bf x} \label{Gr(7.19)} \] where \(K\) is a constant (depends on the history of \(I(t)\)).

N.B.: this can't be true always, since it blows up as \(r \rightarrow \infty\). Reason: in this case, we've overstepped the quasistatic limit. We need \(r \ll c\tau\) where \(\tau\) is a typical time scale for change of \(I(t)\).

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Author: Jean-Sébastien Caux

Created: 2022-03-24 Thu 08:42