Edits to “Statement”, “Answer”
en/11.1.7.md
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| @@ -1,11 +1,5 @@ | |||
| ### Statement | |||
| − | $11.1.7.$ [Insert the problem statement] | ||
| − | |||
| − | ### Solution | ||
| − | |||
| − | ### Statement | ||
| − | |||
| $11.1.7.$ The induction of a constant magnetic field is measured using a square frame with dimensions $a \times a$, rotating with an angular velocity $\omega$. Its axis of rotation is perpendicular to the direction of the magnetic field. The amplitude of the electric voltage taken from the frame is $V$. Find the magnetic field induction.$^{*)}$ | |||
|  | |||
| ### Solution | |||
| The magnetic flux passing through the square frame, rotated by an angle $\alpha=\omega t$, is: | |||
| $$ | |||
| \Phi = BS \cdot \cos\alpha = BS \cdot \cos \omega t | |||
| $$ | |||
| The induced EMF arising in the rotating square frame is: | |||
| $$ | |||
| \varepsilon_i = - \frac{d\Phi}{dt} = - {BS \cdot \frac{d}{dt}\left(\cos \omega t\right)} | |||
| $$ | |||
| From this, we find the induced EMF: | |||
| $$ | |||
| \varepsilon_i = BS\omega \cdot \sin \omega t | |||
| $$ | |||
| Thus, the amplitude of the induced EMF is: | |||
| $$ | |||
| V=BS\omega | |||
| $$ | |||
| Expressing the magnetic field induction: | |||
| $$ | |||
| B=\frac{V}{S\omega}=\frac{V}{a^2\omega} | |||
| $$ | |||
| @@ -45,7 +39,3 @@Solution | |||
| #### Answer | |||
| $B = V /(a^2\omega )$ | |||
| − | |||
| − | #### Answer | ||
| − | |||
| − | [Insert a concise answer or boxed result] | ||
| @@ -1,11 +1,5 @@ | |||
| ### Statement | ### Statement | ||
| $11.1.7.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $11.1.7.$ The induction of a constant magnetic field is measured using a square frame with dimensions $a \times a$, rotating with an angular velocity $\omega$. Its axis of rotation is perpendicular to the direction of the magnetic field. The amplitude of the electric voltage taken from the frame is $V$. Find the magnetic field induction.$^{*)}$ | $11.1.7.$ The induction of a constant magnetic field is measured using a square frame with dimensions $a \times a$, rotating with an angular velocity $\omega$. Its axis of rotation is perpendicular to the direction of the magnetic field. The amplitude of the electric voltage taken from the frame is $V$. Find the magnetic field induction.$^{*)}$ | ||
|  |  | ||
| ### Solution | ### Solution | ||
| The magnetic flux passing through the square frame, rotated by an angle $\alpha=\omega t$, is: | The magnetic flux passing through the square frame, rotated by an angle $\alpha=\omega t$, is: | ||
| $$ | $$ | ||
| \Phi = BS \cdot \cos\alpha = BS \cdot \cos \omega t | \Phi = BS \cdot \cos\alpha = BS \cdot \cos \omega t | ||
| $$ | $$ | ||
| The induced EMF arising in the rotating square frame is: | The induced EMF arising in the rotating square frame is: | ||
| $$ | $$ | ||
| \varepsilon_i = - \frac{d\Phi}{dt} = - {BS \cdot \frac{d}{dt}\left(\cos \omega t\right)} | \varepsilon_i = - \frac{d\Phi}{dt} = - {BS \cdot \frac{d}{dt}\left(\cos \omega t\right)} | ||
| $$ | $$ | ||
| From this, we find the induced EMF: | From this, we find the induced EMF: | ||
| $$ | $$ | ||
| \varepsilon_i = BS\omega \cdot \sin \omega t | \varepsilon_i = BS\omega \cdot \sin \omega t | ||
| $$ | $$ | ||
| Thus, the amplitude of the induced EMF is: | Thus, the amplitude of the induced EMF is: | ||
| $$ | $$ | ||
| V=BS\omega | V=BS\omega | ||
| $$ | $$ | ||
| Expressing the magnetic field induction: | Expressing the magnetic field induction: | ||
| $$ | $$ | ||
| B=\frac{V}{S\omega}=\frac{V}{a^2\omega} | B=\frac{V}{S\omega}=\frac{V}{a^2\omega} | ||
| $$ | $$ | ||
| @@ -45,7 +39,3 @@Solution | |||
| #### Answer | #### Answer | ||
| $B = V /(a^2\omega )$ | $B = V /(a^2\omega )$ | ||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||