From 942035bfe0c19be78efe1452d88b85490f035aab Mon Sep 17 00:00:00 2001 From: Prefetch Date: Sun, 12 Sep 2021 19:22:22 +0200 Subject: Expand knowledge base --- content/know/concept/lorentz-force/index.pdc | 248 +++++++++++++++++++++++++++ 1 file changed, 248 insertions(+) create mode 100644 content/know/concept/lorentz-force/index.pdc (limited to 'content/know/concept/lorentz-force/index.pdc') diff --git a/content/know/concept/lorentz-force/index.pdc b/content/know/concept/lorentz-force/index.pdc new file mode 100644 index 0000000..f0e9850 --- /dev/null +++ b/content/know/concept/lorentz-force/index.pdc @@ -0,0 +1,248 @@ +--- +title: "Lorentz force" +firstLetter: "L" +publishDate: 2021-09-08 +categories: +- Physics +- Electromagnetism + +date: 2021-09-08T17:00:32+02:00 +draft: false +markup: pandoc +--- + +# Lorentz force + +The **Lorentz force** is an empirical force used to define +the [electric field](/know/concept/electric-field/) $\vb{E}$ +and [magnetic field](/know/concept/magnetic-field/) $\vb{B}$. +For a particle with charge $q$ moving with velocity $\vb{u}$, +the Lorentz force $\vb{F}$ is given by: + +$$\begin{aligned} + \boxed{ + \vb{F} + = q (\vb{E} + \vb{u} \cross \vb{B}) + } +\end{aligned}$$ + + +## Uniform magnetic field + +Consider the simple case of a uniform magnetic field +$\vb{B} = (0, 0, B)$ in the $z$-direction, +without an electric field $\vb{E} = 0$. +If there are no other forces, +Newton's second law states: + +$$\begin{aligned} + \vb{F} + = m \dv{\vb{u}}{t} + = q \vb{u} \cross \vb{B} +\end{aligned}$$ + +Evaluating the cross product yields +three coupled equations for the components of $\vb{u}$: + +$$\begin{aligned} + \dv{u_x}{t} + = \frac{q B}{m} u_y + \qquad \quad + \dv{u_y}{t} + = - \frac{q B}{m} u_x + \qquad \quad + \dv{u_z}{t} + = 0 +\end{aligned}$$ + +Differentiating the first equation with respect to $t$, +and substituting $\dv*{u_y}{t}$ from the second, +we arrive at the following harmonic oscillator: + +$$\begin{aligned} + \dv[2]{u_x}{t} = - \omega_c^2 u_x +\end{aligned}$$ + +Where we have defined the **cyclotron frequency** $\omega_c$ as follows, +which is always positive: + +$$\begin{aligned} + \boxed{ + \omega_c + \equiv \frac{|q| B}{m} + } +\end{aligned}$$ + +Suppose we choose our initial conditions so that +the solution for $u_x(t)$ is given by: + +$$\begin{aligned} + u_x(t) + = - u_\perp \sin\!(\omega_c t) +\end{aligned}$$ + +Where $u_\perp \equiv \sqrt{u_x^2 + u_y^2}$ is the constant total transverse velocity. +Then $u_y(t)$ is found to be: + +$$\begin{aligned} + u_y(t) + = \frac{m}{q B} \dv{u_x}{t} + = - \frac{m \omega_c}{q B} u_\perp \cos\!(\omega_c t) + = - \mathrm{sgn}(q) \: u_\perp \cos\!(\omega_c t) +\end{aligned}$$ + +Where $\mathrm{sgn}$ is the signum function. +This tells us that the particle moves in a circular orbit, +and that the direction of rotation is determined by $q$. + +Integrating the velocity yields the position, +where we refer to the integration constants $x_{gc}$ and $y_{gc}$ +as the **guiding center**, around which the particle orbits or **gyrates**: + +$$\begin{aligned} + x(t) + = \frac{u_\perp}{\omega_c} \cos\!(\omega_c t) + x_{gc} + \qquad \quad + y(t) + = - \mathrm{sgn}(q) \: \frac{u_\perp}{\omega_c} \sin\!(\omega_c t) + y_{gc} +\end{aligned}$$ + +The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_L$, given by: + +$$\begin{aligned} + \boxed{ + r_L + \equiv \frac{u_\perp}{\omega_c} + = \frac{m u_\perp}{|q| B} + } +\end{aligned}$$ + +Finally, it is trivial to integrate the equation for the $z$-direction velocity $u_z$: + +$$\begin{aligned} + z(t) + = u_z t + z_{gc} +\end{aligned}$$ + +In conclusion, the particle's motion parallel to $\vb{B}$ +is not affected by the magnetic field, +while its motion perpendicular to $\vb{B}$ +is circular around an imaginary guiding center. +The end result is that particles follow a helical path +when moving through a uniform magnetic field. + + +## Uniform electric and magnetic field + +Let us now consider a more general case, +with constant uniform electric and magnetic fields $\vb{E}$ and $\vb{B}$, +which may or may not be perpendicular. +The equation of motion is then: + +$$\begin{aligned} + \vb{F} + = m \dv{\vb{u}}{t} + = q (\vb{E} + \vb{u} \cross \vb{B}) +\end{aligned}$$ + +If we take the dot product with the unit vector $\vu{B}$, +the cross product vanishes, leaving: + +$$\begin{aligned} + \dv{\vb{u}_\parallel}{t} + = \frac{q}{m} \vb{E}_\parallel +\end{aligned}$$ + +Where $\vb{u}_\parallel$ and $\vb{E}_\parallel$ are +the components of $\vb{u}$ and $\vb{E}$ +that are parallel to $\vb{B}$. +This equation is easy to integrate: +the guiding center accelerates according to $(q/m) \vb{E}_\parallel$. + +Next, let us define the perpendicular component $\vb{u}_\perp$ +such that $\vb{u} = \vb{u}_\parallel \vu{B} + \vb{u}_\perp$. +Its equation of motion is found by +subtracting $\vb{u}_\parallel$'s equation from the original: + +$$\begin{aligned} + m \dv{\vb{u}_\perp}{t} + = q (\vb{E} + \vb{u} \cross \vb{B}) - q \vb{E}_\parallel + = q (\vb{E}_\perp + \vb{u}_\perp \cross \vb{B}) +\end{aligned}$$ + +To solve this, we go to a moving coordinate system +by defining $\vb{u}_\perp = \vb{v}_\perp + \vb{w}_\perp$, +where $\vb{v}_\perp$ is a constant of our choice. +The equation is now as follows: + +$$\begin{aligned} + m \dv{t} (\vb{v}_\perp + \vb{w}_\perp) + = m \dv{\vb{w}_\perp}{t} + = q (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B} + \vb{w}_\perp \cross \vb{B}) +\end{aligned}$$ + +We want to choose $\vb{v}_\perp$ such that the first two terms vanish, +or in other words: + +$$\begin{aligned} + 0 + = \vb{E}_\perp + \vb{v}_\perp \cross \vb{B} +\end{aligned}$$ + +To find $\vb{v}_\perp$, we take the cross product with $\vb{B}$, +and use the fact that $\vb{B} \cross \vb{E}_\perp = \vb{B} \cross \vb{E}$: + +$$\begin{aligned} + 0 + = \vb{B} \cross (\vb{E}_\perp + \vb{v}_\perp \cross \vb{B}) + = \vb{B} \cross \vb{E} + \vb{v}_\perp B^2 + \quad \implies \quad + \boxed{ + \vb{v}_\perp + = \frac{\vb{E} \cross \vb{B}}{B^2} + } +\end{aligned}$$ + +When $\vb{v}_\perp$ is chosen like this, +the perpendicular equation of motion is reduced to: + +$$\begin{aligned} + m \dv{\vb{w}_\perp}{t} + = q \vb{w}_\perp \cross \vb{B} +\end{aligned}$$ + +Which is simply the case we treated previously with $\vb{E} = 0$, +with a known solution +(assuming $\vb{B}$ still points in the positive $z$-direction): + +$$\begin{aligned} + w_x(t) + = - w_\perp \sin\!(\omega_c t) + \qquad + w_y(t) + = - \mathrm{sgn}(q) \: w_\perp \cos\!(\omega_c t) +\end{aligned}$$ + +However, this result is shifted by a constant $\vb{v}_\perp$, +often called the **drift velocity** $\vb{v}_d$, +at which the guiding center moves transversely. +Curiously, $\vb{v}_d$ is independent of $q$. + +Such a drift is not specific to an electric field. +In the equations above, $\vb{E}$ can be replaced +by a general force $\vb{F}/q$ (e.g. gravity) without issues. +In that case, $\vb{v}_d$ does depend on $q$. + +$$\begin{aligned} + \boxed{ + \vb{v}_d + = \frac{\vb{F} \cross \vb{B}}{q B^2} + } +\end{aligned}$$ + + + +## References +1. F.F. Chen, + *Introduction to plasma physics and controlled fusion*, + 3rd edition, Springer. -- cgit v1.2.3