| @@ -1,44 +1,37 @@ |
| | | ### Statement |
| | | |
| | − | $10.2.2.$ [Insert problem description here] |
| | | |
| | − | __Example Statement__: |
| | − | $1.1.1.$ Determine the coordinate $x(t)$ of a body as a function of time $t$, given that its acceleration is defined as $a(t) = bt$, where $b$ is a constant. |
| | + | $10.2.2.$ Estimate the electron drift velocity across an inhomogeneous magnetic field, |
| | + | the induction components of which are $B_{x}=0 $,$B_{y}=0 $, |
| | + | $B_{z}=B_{0}(1+\alpha x)$.Electron velocity :$v <<\frac{e B_{0}}{\alpha m_{e}}$. |
| | | |
| | | |
| | | ### Solution |
| | | |
| | − | [Your solution should be placed here] |
| | | |
| | + | |
| | | __Example Solution__: |
| | − | The acceleration of the body defined by |
| | + | The equation of motion for electron in magnetic field is : |
| | | |
| | − | $$a(t) = bt$$ |
| | + | $$m\frac{d\vec{v}}{dt} = -e(\vec{v} \times\vec{B})$$ |
| | | |
| | − | We know that acceleration is the time derivative of velocity: |
| | + | Calculating the cross product yields two equations : |
| | | |
| | − | $$a(t) = \frac{d v(t)}{d t}$$ |
| | + | $$ O\hat{i} : m\frac{d{v_{x}}}{dt}=ev_{y}B_{0}(1+\alpha x) $$ |
| | + | $$ O\hat{j} : m\frac{d{v_{y}}}{dt}=-ev_{x}B_{0}(1+\alpha x) $$ |
| | + | The drift velocity should be perpendicular to the direction in which the field varies . Using the fact that $\alpha$ is small , and using the solutions for the case in which $\alpha=0$ : |
| | + | $$x^{(0)}(t)=r_{L}cos(\omega t)$$ |
| | + | $$y^{(0)}(t)=r_{L}sin(\omega t)$$ |
| | + | , $r_{L} = \frac{mv_{0}}{eB_{0}}$, by perturbation theory $v \approx v^{0}+\delta v$, |
| | | |
| | − | To find the velocity $v(t)$, we integrate $a(t)$ with respect to time: |
| | + | $$ m\frac{d{v^{0}_{x}}}{dt}=e(v^{0}_{y}+\delta v)B_{0}(1+\alpha x) $$ |
| | + | $$ m\frac{d{(v^{0}_{y}+\delta v)}}{dt}=-ev^{0}_{x}B_{0}(1+\alpha x) $$ |
| | + | $=>$ , Averaging over time yields yields: |
| | + | left part equal to zero and : |
| | + | $e<v^{0}_{y}x>B_{0}\alpha+eB_{0}<v^{0}_{y}>=0$, time average over one period of $<v^{0}_{y}x> = \frac{1}{T}\int_{0}^{T} \omega r_{L}^{2} cos^{2}\omega t \,dt = \frac{\omega^{2} T }{2\pi} r_{L}^{2} \frac{1+sin2\omega T}{4} = \frac{v_{0}r_{L}}{4} $ , $v_{D} = \alpha \frac{mv_{0}^{2}}{4eB_{0}}$ |
| | | |
| | − | $$v(t) = \int a(t) \, dt = \int b t \, dt$$ |
| | − | |
| | − | If the initial velocity is $v(0) = 0$, then the velocity becomes: |
| | − | |
| | − | $$v(t) = \frac{b t^2}{2}$$ |
| | − | |
| | − | Likewise, integrate $v(t)$ with respect to time: |
| | − | |
| | − | $$x(t)= \int v(t) \, dt = \frac{b}{2} \int t^2 \, dt$$ |
| | − | |
| | − | From where the coordinate from time, considering the initial conditions: |
| | − | |
| | − | $$\boxed{x(t)=\frac{bt^3}{6}}$$ |
| | − | |
| | | #### Answer |
| | + | $v_{D} = \alpha \frac{mv_{0}^{2}}{4eB_{0}}$ |
| | | |
| | − | [Insert a concise answer or boxed result, like this:] |
| | | |
| | | |
| | − | __Example Answer__: |
| | − | $$ x(t)=\frac{bt^3}{6} $$ |