New solution

Valter edited
View solution History
revision #20038 newer →
en/11.1.7.md +51 −0
@@ -0,0 +1,51 @@
+### 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.$^{*)}$
+
+![For problem $11.1.7$|465x222, 45%](../../img/11.1.7/11.1.7.png)
+
+### 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:
+
+$$
+B=\frac{V}{S\omega}=\frac{V}{a^2\omega}
+$$
+
+#### Answer
+
+$B = V /(a^2\omega )$
+
+#### Answer
+
+[Insert a concise answer or boxed result]