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| ### Statement |
| ### Statement |
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| $11.2.6.$ 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. |
| $11.2.6.$ 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. |
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| ### Solution |
| ### Solution |
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| 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:\ |
| Supposing a ring of radius $y$ whose center is in the coil axis:\ |
| $\varepsilon = -\frac{d\Phi_B}{dt} = -\frac{d(\vec{B}\cdot\vec{S})}{dt} = -\pi y^2 \frac{dB}{dt}$\ |
| $\varepsilon = -\frac{d\Phi_B}{dt} = -\frac{d(\vec{B}\cdot\vec{S})}{dt} = -\pi y^2 \frac{dB}{dt}$\ |
| but $\varepsilon = -\vec{E}\cdot\vec{l} = -2\pi y E$ (where $l$ is the vector in direction of induced electric current on the ring that follows the lenght of the ring)\ |
| but $\varepsilon = -\vec{E}\cdot\vec{l} = -2\pi y E$ (where $l$ 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)\ | | $2E = y\frac{dB}{dt}$ (1)\ |
| Moreover,\ | | Moreover,\ |
| $B(t) = \mu_0 I(t) \frac{n_0}{\ell_0}$, (2)\ | | $B(t) = \mu_0 I(t) \frac{n_0}{\ell_0}$, (2)\ |
| Putting (2) into (1) and taking in account that $I(t) = I_0 \sin{(2\pi\nu t)}$\ | | Putting (2) into (1) and taking in account that $I(t) = I_0 \sin{(2\pi\nu t)}$\ |
| $E(y,t) = \frac{\mu_0 n_0 y}{\ell_0}I_0\pi\nu\cos{(2\pi\nu t)}$\ | | $E(y,t) = \frac{\mu_0 n_0 y}{\ell_0}I_0\pi\nu\cos{(2\pi\nu t)}$\ |
| \ | | \ |
| The EFM for the coil is given by,ç | | The EFM for the coil is given by,ç |
| $\varepsilon = - n \pi r^2 \frac{dB}{dt}$ (3)\ | | $\varepsilon = - n \pi r^2 \frac{dB}{dt}$ (3)\ |
| From (2) and (3),\ | | From (2) and (3),\ |
| $\varepsilon = -\frac{2\pi^2 r^2 n n_0 \mu_0 I_0\nu}{\ell_0}\cos{(2\pi\nu t)}$ \ | | $\varepsilon = -\frac{2\pi^2 r^2 n n_0 \mu_0 I_0\nu}{\ell_0}\cos{(2\pi\nu t)}$ \ |
| The maximum value is reached when $\cos{2\pi\nu t} = -1$, so\ | | The maximum value is reached when $\cos{2\pi\nu t} = -1$, so\ |
| $\varepsilon = \frac{2\pi^2 r^2 n n_0 \mu_0 I_0\nu}{\ell_0}$\ | | $\varepsilon = \frac{2\pi^2 r^2 n n_0 \mu_0 I_0\nu}{\ell_0}$\ |
| Calculating:\ | | Calculating:\ |
| $\varepsilon \simeq 0.12\;\rm{V}$ | | $\varepsilon \simeq 0.12\;\rm{V}$ |
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| #### Answer | | #### Answer |
| $E(y,t) = \frac{\mu_0 n_0 y}{\ell_0}I_0\pi\nu\cos{(2\pi\nu t)}$\ | | $E(y,t) = \frac{\mu_0 n_0 y}{\ell_0}I_0\pi\nu\cos{(2\pi\nu t)}$\ |
| $\varepsilon \simeq 0.12\;\rm{V}$ | | $\varepsilon \simeq 0.12\;\rm{V}$ |
