From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Thu, 20 Oct 2022 18:25:31 +0200 Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex' --- source/know/concept/lorentz-force/index.md | 58 +++++++++++++++--------------- 1 file changed, 29 insertions(+), 29 deletions(-) (limited to 'source/know/concept/lorentz-force') diff --git a/source/know/concept/lorentz-force/index.md b/source/know/concept/lorentz-force/index.md index d1b216d..03178f6 100644 --- a/source/know/concept/lorentz-force/index.md +++ b/source/know/concept/lorentz-force/index.md @@ -10,10 +10,10 @@ layout: "concept" --- 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: +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: $$\begin{aligned} \boxed{ @@ -25,9 +25,9 @@ $$\begin{aligned} ## Uniform electric field -Consider the simple case of an electric field $\vb{E}$ +Consider the simple case of an electric field $$\vb{E}$$ that is uniform in all of space. -In the absence of a magnetic field $\vb{B} = 0$ +In the absence of a magnetic field $$\vb{B} = 0$$ and any other forces, Newton's second law states: @@ -38,16 +38,16 @@ $$\begin{aligned} \end{aligned}$$ This is straightforward to integrate in time, -for a given initial velocity vector $\vb{u}_0$: +for a given initial velocity vector $$\vb{u}_0$$: $$\begin{aligned} \vb{u}(t) = \frac{q}{m} \vb{E} t + \vb{u}_0 \end{aligned}$$ -And then the particle's position $\vb{x}(t)$ +And then the particle's position $$\vb{x}(t)$$ is found be integrating once more, -with $\vb{x}(0) = \vb{x}_0$: +with $$\vb{x}(0) = \vb{x}_0$$: $$\begin{aligned} \boxed{ @@ -56,16 +56,16 @@ $$\begin{aligned} } \end{aligned}$$ -In summary, unsurprisingly, a uniform electric field $\vb{E}$ -accelerates the particle with a constant force $\vb{F} = q \vb{E}$. -Note that the direction depends on the sign of $q$. +In summary, unsurprisingly, a uniform electric field $$\vb{E}$$ +accelerates the particle with a constant force $$\vb{F} = q \vb{E}$$. +Note that the direction depends on the sign of $$q$$. ## Uniform magnetic field Consider the simple case of a uniform magnetic field -$\vb{B} = (0, 0, B)$ in the $z$-direction, -without an electric field $\vb{E} = 0$. +$$\vb{B} = (0, 0, B)$$ in the $$z$$-direction, +without an electric field $$\vb{E} = 0$$. If there are no other forces, Newton's second law states: @@ -76,7 +76,7 @@ $$\begin{aligned} \end{aligned}$$ Evaluating the cross product yields -three coupled equations for the components of $\vb{u}$: +three coupled equations for the components of $$\vb{u}$$: $$\begin{aligned} \dv{u_x}{t} @@ -89,15 +89,15 @@ $$\begin{aligned} = 0 \end{aligned}$$ -Differentiating the first equation with respect to $t$, -and substituting $\idv{u_y}{t}$ from the second, +Differentiating the first equation with respect to $$t$$, +and substituting $$\idv{u_y}{t}$$ from the second, we arrive at the following harmonic oscillator: $$\begin{aligned} \dvn{2}{u_x}{t} = - \omega_c^2 u_x \end{aligned}$$ -Where we have defined the **cyclotron frequency** $\omega_c$ as follows, +Where we have defined the **cyclotron frequency** $$\omega_c$$ as follows, which may be negative: $$\begin{aligned} @@ -108,15 +108,15 @@ $$\begin{aligned} \end{aligned}$$ Suppose we choose our initial conditions so that -the solution for $u_x(t)$ is given by: +the solution for $$u_x(t)$$ is given by: $$\begin{aligned} u_x(t) = u_\perp \cos(\omega_c t) \end{aligned}$$ -Where $u_\perp \equiv \sqrt{u_x^2 + u_y^2}$ is the constant total transverse velocity. -Then $u_y(t)$ is found to be: +Where $$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) @@ -126,10 +126,10 @@ $$\begin{aligned} \end{aligned}$$ This means that the particle moves in a circle, -in a direction determined by the sign of $\omega_c$. +in a direction determined by the sign of $$\omega_c$$. Integrating the velocity yields the position, -where we refer to the integration constants $x_{gc}$ and $y_{gc}$ +where we refer to the integration constants $$x_{gc}$$ and $$y_{gc}$$ as the **guiding center**, around which the particle orbits or **gyrates**: $$\begin{aligned} @@ -140,7 +140,7 @@ $$\begin{aligned} = \frac{u_\perp}{\omega_c} \cos(\omega_c t) + y_{gc} \end{aligned}$$ -The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_L$, given by: +The radius of this orbit is known as the **Larmor radius** or **gyroradius** $$r_L$$, given by: $$\begin{aligned} \boxed{ @@ -151,7 +151,7 @@ $$\begin{aligned} \end{aligned}$$ Finally, it is easy to integrate the equation -for the $z$-axis velocity $u_z$, which is conserved: +for the $$z$$-axis velocity $$u_z$$, which is conserved: $$\begin{aligned} z(t) @@ -159,9 +159,9 @@ $$\begin{aligned} = u_z t + z_0 \end{aligned}$$ -In conclusion, the particle's motion parallel to $\vb{B}$ +In conclusion, the particle's motion parallel to $$\vb{B}$$ is not affected by the magnetic field, -while its motion perpendicular to $\vb{B}$ +while its motion perpendicular to $$\vb{B}$$ is circular around an imaginary guiding center. The end result is that particles follow a helical path when moving through a uniform magnetic field: @@ -177,9 +177,9 @@ $$\begin{aligned} } \end{aligned}$$ -Where $\vb{x}_{gc}(t) \equiv (x_{gc}, y_{gc}, z_{gc})$ +Where $$\vb{x}_{gc}(t) \equiv (x_{gc}, y_{gc}, z_{gc})$$ is the position of the guiding center. -For a detailed look at how $\vb{B}$ and $\vb{E}$ +For a detailed look at how $$\vb{B}$$ and $$\vb{E}$$ can affect the guiding center's motion, see [guiding center theory](/know/concept/guiding-center-theory/). -- cgit v1.2.3