Edit to “Statement”
en/11.1.23.md
+1 −1
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $11.1.23.$ A metal ring is placed vertically in the gravity field. A metal rod of length $L$ and mass $m$ is pivotally fixed in the center of the ring and touches it with the other end. A uniform magnetic field of induction $B$ is perpendicular to the ring plane. By what law is it necessary to change the current in the rod so that the rod rotates evenly with an angular velocity $\omega$, if at the initial moment the rod was in the upper position? Ignore the friction. | ||
| + | $11.1.23^*.$ A metal ring is placed vertically in the gravity field. A metal rod of length $L$ and mass $m$ is pivotally fixed in the center of the ring and touches it with the other end. A uniform magnetic field of induction $B$ is perpendicular to the ring plane. By what law is it necessary to change the current in the rod so that the rod rotates evenly with an angular velocity $\omega$, if at the initial moment the rod was in the upper position? Ignore the friction. | ||
| ### Solution | |||
|  | |||
| The angle between the rod and the vertical (shown in green) is $\omega t$ at time $t$ after the initial moment. If there is current $I$ flowing radially outward through the rod, then the magnetic force on the rod is $F=IBL$. The angular speed is constant when counterclockwise moment due to $F$ equals clockwise moment due to gravity $mg$. Thus, | |||
| \[F\cdot\frac{L}{2}=mg\cdot\frac{L}{2}\cdot\sin(\omega t)\qquad\Rightarrow\qquad I=\frac{mg}{BL}\cdot\sin(\omega t).\] | |||
| #### Answer | |||
| $I=\frac{mg}{BL}\cdot\sin(\omega t)$ | |||
| unchanged lines 11 | |||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $11.1.23.$ A metal ring is placed vertically in the gravity field. A metal rod of length $L$ and mass $m$ is pivotally fixed in the center of the ring and touches it with the other end. A uniform magnetic field of induction $B$ is perpendicular to the ring plane. By what law is it necessary to change the current in the rod so that the rod rotates evenly with an angular velocity $\omega$, if at the initial moment the rod was in the upper position? Ignore the friction. | $11.1.23^*.$ A metal ring is placed vertically in the gravity field. A metal rod of length $L$ and mass $m$ is pivotally fixed in the center of the ring and touches it with the other end. A uniform magnetic field of induction $B$ is perpendicular to the ring plane. By what law is it necessary to change the current in the rod so that the rod rotates evenly with an angular velocity $\omega$, if at the initial moment the rod was in the upper position? Ignore the friction. | ||
| ### Solution | ### Solution | ||
|  |  | ||
| The angle between the rod and the vertical (shown in green) is $\omega t$ at time $t$ after the initial moment. If there is current $I$ flowing radially outward through the rod, then the magnetic force on the rod is $F=IBL$. The angular speed is constant when counterclockwise moment due to $F$ equals clockwise moment due to gravity $mg$. Thus, | The angle between the rod and the vertical (shown in green) is $\omega t$ at time $t$ after the initial moment. If there is current $I$ flowing radially outward through the rod, then the magnetic force on the rod is $F=IBL$. The angular speed is constant when counterclockwise moment due to $F$ equals clockwise moment due to gravity $mg$. Thus, | ||
| \[F\cdot\frac{L}{2}=mg\cdot\frac{L}{2}\cdot\sin(\omega t)\qquad\Rightarrow\qquad I=\frac{mg}{BL}\cdot\sin(\omega t).\] | \[F\cdot\frac{L}{2}=mg\cdot\frac{L}{2}\cdot\sin(\omega t)\qquad\Rightarrow\qquad I=\frac{mg}{BL}\cdot\sin(\omega t).\] | ||
| #### Answer | #### Answer | ||
| $I=\frac{mg}{BL}\cdot\sin(\omega t)$ | $I=\frac{mg}{BL}\cdot\sin(\omega t)$ | ||
| unchanged lines 11 | |||