Electromagnetic induction > Vortex electric field > Problem 11.2.20

Statement

$11.2.20.$

The circuit of an electric circuit includes a resistance R and an uncharged
capacitor of capacitance C.
a. Prove that the charge on the capacitor during the appearance and then
disappearance of the magnetic flux through the circuit does not exceed the
valueΦT
CR2, where T is the lifetime of this magnetic flux, Φ is its maximum
value.
b. To determine the direct current I flowing in the circuit during time T, if
there is an alternating current caused by electromagnetic induction in the
same time interval, measure the capacitance potential V after all the cur-
rents disappear, and then estimate the direct current by the formula I =CV
.
Determine the maximum error of such an estimate in the case when the AC
amplitude is k times greater than I.

Solution

firs analyse the Field in the capacitor at rest

Plate area: A.
Separation: d.
Capacitance: $C = \dfrac{\varepsilon_0 A}{d}$.
Charge:$ Q = CV$.
Surface density: $\sigma = \dfrac{Q}{A} = \dfrac{CV}{A}$.
· Electric field between plates (from the positive to the negative):
$ \mathbf{E}_0 = -\frac{V}{d}\, \hat{\mathbf{z}} \quad \text{(uniform)}$.

At rest, there is no magnetic field.

Now ,at a given instant, the capacitor moves with velocity v (variable due to acceleration). We assume$ v \ll c$ (non‑relativistic).

The moving charged plates constitute surface currents:

Upper plate $(charge +Q) at z = d/2: \mathbf{K}_+ = \sigma v\, \hat{\mathbf{x}}$.
· Lower plate $(charge -Q) at z = -d/2: \mathbf{K}_- = -\sigma v\, \hat{\mathbf{x}}$.

These currents generate a magnetic field between the plates. Using Ampère's law for each plane and superposing, we obtain a uniform field in the region between the plates:

$\mathbf{B} = \mu_0 \sigma v\, \hat{\mathbf{y}}$.

Outside the plates,

$\mathbf{B} = 0$.

We calculate the electromagnetic momentum

The momentum density of the electromagnetic field is

$\varepsilon_0(\mathbf{E} \times \mathbf{B})$.

Integrating over the volume between the plates $ (V = A d)$:

$\mathbf{P}_{\text{em}} = \varepsilon_0 \int \mathbf{E} \times \mathbf{B} \, dV$.

With$ \mathbf{E} = -\frac{V}{d}\hat{\mathbf{z}}, \mathbf{B} = \mu_0 \sigma v \, \hat{\mathbf{y}}:$

$\mathbf{E} \times \mathbf{B} = \left(-\frac{V}{d}\hat{\mathbf{z}}\right) \times (\mu_0 \sigma v \, \hat{\mathbf{y}}) = -\frac{V}{d}\mu_0 \sigma v \, (\hat{\mathbf{z}} \times \hat{\mathbf{y}}) = \frac{V}{d}\mu_0 \sigma v \, \hat{\mathbf{x}}$.

Substituting

$\sigma = CV/A $ and $

C = \varepsilon_0 A/d$:

$\mathbf{E} \times \mathbf{B} = \frac{V}{d}\mu_0 \frac{CV}{A} v \, \hat{\mathbf{x}} = \mu_0 \frac{C V^2}{A d} v \, \hat{\mathbf{x}}$.

$\mathbf{P}_{\text{em}} = \varepsilon_0 \left( \mu_0 \frac{C V^2}{A d} v \right) (A d) \, \hat{\mathbf{x}} = \varepsilon_0 \mu_0 C V^2 v \, \hat{\mathbf{x}} = \frac{C V^2}{c^2}\, v\, \hat{\mathbf{x}}$.

Now the calculate electromagnetic mass

The total momentum of the field is proportional to the instantaneous velocity. The electromagnetic mass is defined as the proportionality constant:

$m_{\text{em}} = \frac{P_{\text{em}}}{v} = \frac{C V^2}{c^2}$.

Answer

$\boxed{m_{\text{em}} = \frac{C V^2}{c^2}}$.