New solution
en/11.2.15.md
+33 −0
| @@ -0,0 +1,33 @@ | |||
| + | ### Statement | ||
| + | |||
| + | $11.2.15.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + |  | ||
| + | |||
| + | The magnetic force $evB$ provides centripetal acceleration to the electron, so | ||
| + | |||
| + | \[evB=\frac{mv^2}{r}\qquad\Rightarrow\qquad v=\frac{eBr}{m},\] | ||
| + | |||
| + | where $-e$, $m$, and $v$ are the charge, mass, and speed of the electron around the circle, respectively. Since $B=\alpha t$, the tangential acceleration of the electron | ||
| + | |||
| + | \[a=\frac{dv}{dt}=\frac{d}{dt}\left(\frac{eBr}{m}\right)=\frac{e\alpha r}{m}\] | ||
| + | |||
| + | It is caused by induced electric field $E$ as a result of changing magnetic flux. Thus, | ||
| + | |||
| + | \[a=\frac{eE}{m}\qquad\Rightarrow\qquad E=\alpha r.\] | ||
| + | |||
| + | Let $B'$ be the magnetic field inside the cylinder in the same direction as $B$. By the law of induction, | ||
| + | |||
| + | \[E\cdot2\pi r=\frac{d}{dt}(B'\cdot\pi r_0^2+B\cdot\pi(r^2-r_0^2)).\] | ||
| + | |||
| + | As a result, | ||
| + | |||
| + | \[\frac{dB'}{dt}=\alpha\left[\left(\frac{r}{r_0}\right)^2+1\right].\] | ||
| + | |||
| + | Since $B'$ is initially zero, we have $B'=\alpha t[(r/r_0)^2+1]$. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
| @@ -0,0 +1,33 @@ | |||
| ### Statement | |||
| $11.2.15.$ [Insert the problem statement] | |||
| ### Solution | |||
|  | |||
| The magnetic force $evB$ provides centripetal acceleration to the electron, so | |||
| \[evB=\frac{mv^2}{r}\qquad\Rightarrow\qquad v=\frac{eBr}{m},\] | |||
| where $-e$, $m$, and $v$ are the charge, mass, and speed of the electron around the circle, respectively. Since $B=\alpha t$, the tangential acceleration of the electron | |||
| \[a=\frac{dv}{dt}=\frac{d}{dt}\left(\frac{eBr}{m}\right)=\frac{e\alpha r}{m}\] | |||
| It is caused by induced electric field $E$ as a result of changing magnetic flux. Thus, | |||
| \[a=\frac{eE}{m}\qquad\Rightarrow\qquad E=\alpha r.\] | |||
| Let $B'$ be the magnetic field inside the cylinder in the same direction as $B$. By the law of induction, | |||
| \[E\cdot2\pi r=\frac{d}{dt}(B'\cdot\pi r_0^2+B\cdot\pi(r^2-r_0^2)).\] | |||
| As a result, | |||
| \[\frac{dB'}{dt}=\alpha\left[\left(\frac{r}{r_0}\right)^2+1\right].\] | |||
| Since $B'$ is initially zero, we have $B'=\alpha t[(r/r_0)^2+1]$. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||