Правка разделов «Statement», «Solution», «Answer»
en/11.2.6.md
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| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $11.2.6.$ [Insert the problem 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 | |||
| 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 $l$ is the vector in direction of induced electric current on the ring that follows the lenght of the ring)\ | |||
| @@ -10,16 +10,18 @@Solution | |||
| $2E = y\frac{dB}{dt}$ (1)\ | |||
| Moreover,\ | |||
| $B(t) = \mu_0 I(t) \frac{n_0}{\ell_0}$, (2)\ | |||
| − | Putting (2) into (1) and taking in account that $I(t) = I_0 \sin{2\pi\nu t}$\ | ||
| − | $E(y,t) = \frac{\mu_0 n_0 y}{\ell_0}I_0\pi\nu\cos{2\pi\nu t}$\ | ||
| + | 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 = | ||
| + | $\varepsilon = -\frac{2\pi^2 r^2 n n_0 \mu_0 I_0\nu}{\ell_0}\cos{(2\pi\nu t)}$ \ | ||
| + | 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 | |||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $E(y,t) = \frac{\mu_0 n_0 y}{\ell_0}I_0\pi\nu\cos{(2\pi\nu t)}$\ | ||
| + | $\varepsilon \simeq 0.12\;\rm{V}$ | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $11.2.6.$ [Insert the problem 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 | ### Solution | ||
| 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)\ | ||
| @@ -10,16 +10,18 @@Solution | |||
| $2E = y\frac{dB}{dt}$ (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 = |
$\varepsilon = -\frac{2\pi^2 r^2 n n_0 \mu_0 I_0\nu}{\ell_0}\cos{(2\pi\nu t)}$ \ | ||
| 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:\ | Calculating:\ | ||
| $\varepsilon \simeq 0.12\;\rm{V}$ | $\varepsilon \simeq 0.12\;\rm{V}$ | ||
| #### Answer | #### Answer | ||
| $E(y,t) = \frac{\mu_0 n_0 y}{\ell_0}I_0\pi\nu\cos{(2\pi\nu t)}$\ | |||
| [Insert a concise answer or boxed result] | $\varepsilon \simeq 0.12\;\rm{V}$ | ||