# Pre-Quantum Electrodynamics
Author: Jean-Sébastien Caux

#### The Induced Electric Field
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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: \[ \oint_{\cal P} {\bf B} \cdot d{\bf l} = \mu_0 I_{enc} \hspace{3cm} \oint_{\cal P} {\bf E} \cdot d{\bf l} = -\frac{d}{dt} \Phi \]

{\bf Example 7.7:} \({\bf B}(t)\) points up in circular region of radius \(R\). What is the induced \({\bf E}(t)\) ? \paragraph{Solution:} amperian loop of radius \(s\), apply Faraday: \[ \oint {\bf E} \cdot d{\bf l} = E (2\pi s) = -\frac{d\Phi}{dt} = -\pi s^2 \frac{dB}{dt} \Rightarrow {\bf E} = -\frac{s}{2} \frac{dB}{dt} \hat{\boldsymbol \phi}. \] Increasing \({\bf B}\): clockwise (viewed from above) \({\bf E}\) from Lenz.

{\bf Example 7.8:} wheel or radius \(b\) with line charge \(\lambda\) on the rim. Uniform magnetic field \({\bf B}_0\) in central region up to \(a < b\), pointing up. Field turned off. What happens ? \paragraph{Solution:} the wheel starts spinning to compensate the reduction of field. Faraday: \[ \oint {\bf E} \cdot d{\bf l} = -\frac{d\Phi}{dt} = - \pi a^2 \frac{dB}{dt} \Rightarrow {\bf E} = -\frac{a^2}{2b} \frac{dB}{dt} \hat{\boldsymbol \phi}. \] 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 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.

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

{\bf Example 7.9:} infinitely long straight wire carries \(I(t)\). Find induced \({\bf E}\) field as a function of distance \(s\) from wire. \paragraph{Solution:} quasistatic: magnetic field is \(B = \frac{\mu_0 I}{2\pi s}\) and circles the wire. Like \({\bf B}\) field of solenoid, \({\bf E}\) runs parallel to wire. Amperian loop with sides at distances \(s_0\) and \(s\): \[ \oint {\bf E} \cdot d{\bf l} = E(s_0)l - E(s)l = -\frac{d}{dt} \int {\bf B} \cdot d{\bf a} = -\frac{\mu_0 l}{2\pi} \frac{dI}{dt} \int_{s_0}^s \frac{ds'}{s'} = -\frac{\mu_0 l}{2\pi} \frac{dI}{dt} \ln(s/s_0). \] So: \[ {\bf E} (s) = \left[ \frac{\mu_0}{2\pi} \frac{dI}{dt} \ln s + K \right] \hat{\bf x} \label{Gr(7.19)} \] where \(K\) is a constant (depends on the history of \(I(t)\)).

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

Author: Jean-Sébastien Caux

Created: 2022-02-08 Tue 06:55

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