Electromagnetic induction > Vortex electric field > Problem 11.2.6

Statement

$11.2.6.$ A sinusoidal current $I = I \sin{(2\pi\nu t)}$ flows through a solenoid of length $\ell_0$ = 20 cm and radius $r$ = 2 cm, where $I_0$ = 10 A, $\nu$ = 50 Hz. Number of turns in the solenoid $n_0$ = 200. Find the distribution of the eddy electric field strength inside the solenoid. What is the amplitude of the voltage produced by this field in a coil of length $l$ = 5 cm and radius $r$ = 1 cm placed inside the solenoid along its axis? The number of turns in this coil is $n$ = 100.

Solution

The Eddy electric field is an induced electric field inside a conductive material when it is subjected to a time-varying magnetic field.
Supposing a ring of radius $y$ whose center is in the coil axis:
$\varepsilon = -\frac{d\Phi_B}{dt} = -\frac{d(\vec{B}\cdot\vec{S})}{dt} = -\pi y^2 \frac{dB}{dt}$
but $\varepsilon = -\vec{E}\cdot\vec{l} = -2\pi y E$ (where $\vec{l}$ is the vector in direction of induced electric current on the ring that follows the lenght of the ring)
$2E = y\frac{dB}{dt}$ (1)
Moreover,
$B(t) = \mu_0 I(t) \frac{n_0}{\ell_0}$, (2)
Putting (2) into (1) and taking in account that $I(t) = I_0 \sin{(2\pi\nu t)}$
$E(y,t) = \frac{\mu_0 n_0 y}{\ell_0}I_0\pi\nu\cos{(2\pi\nu t)}$

The EFM for the coil is given by,ç
$\varepsilon = - n \pi r^2 \frac{dB}{dt}$ (3)
From (2) and (3),
$\varepsilon = -\frac{2\pi^2 r^2 n n_0 \mu_0 I_0\nu}{\ell_0}\cos{(2\pi\nu t)}$
The maximum value is reached when $\cos{(2\pi\nu t)} = -1$, so
$\varepsilon = \frac{2\pi^2 r^2 n n_0 \mu_0 I_0\nu}{\ell_0}$
Calculating:
$\varepsilon \simeq 0.12\; \rm{V}$

Answer

$E(y,t) = \frac{\mu_0 n_0 y}{\ell_0}I_0\pi\nu\cos{(2\pi\nu t)}$
$\varepsilon \simeq 0.12\; \rm{V}$