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en/11.1.23.md
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| + | ### Statement | ||
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| + | $11.1.23.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + |  | ||
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| + | The angle between the rod and the vertical (shown in green) is $\omega t$ at time $t$ after the initial momment. 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, | ||
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| + | \[IBL\cdot\frac{L}{2}=mg\cdot\frac{L}{2}\sin(\omega t)\qquad\Rightarrow\qquad I=\frac{mg}{BL}\sin(\omega t).\] | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $11.1.23.$ [Insert the problem statement] | |||
| ### Solution | |||
|  | |||
| The angle between the rod and the vertical (shown in green) is $\omega t$ at time $t$ after the initial momment. 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, | |||
| \[IBL\cdot\frac{L}{2}=mg\cdot\frac{L}{2}\sin(\omega t)\qquad\Rightarrow\qquad I=\frac{mg}{BL}\sin(\omega t).\] | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||