Force on a moving charge · 运动电荷受到的力
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| centripetal force/senˈtrɪpɪtl fɔːs/ | 向心力 | xiàng xīn lì |
| velocity selector/vəˈlɒsɪti sɪˈlektə/ | 速度选择器 | sù dù xuǎn zé qì |
| Hall voltage/hɔːl ˈvəʊltɪdʒ/ | 霍尔电压 | huò ěr diàn yā |
| Hall probe/hɔːl prəʊb/ | 霍尔探头 | huò ěr tàn tóu |
A force that can never speed anything up
- A magnetic force on a moving charge is always at right angles to the velocity. Always, without exception.
- A force perpendicular to the motion does no work, so it cannot change the particle's speed. It can only change its direction.
- A constant-magnitude force that only ever turns you is exactly the recipe for a circle, which is why every particle accelerator is round.
- This lesson is $F = BQv\sin\theta$, the circular motion it produces, and the two instruments built on it.
一个永远无法让东西加速的力
- 作用在运动电荷上的磁力总是垂直于速度。总是如此,没有例外。
- 垂直于运动的力不做功,所以它无法改变粒子的速率。它只能改变方向。
- 一个大小不变、永远只让你转弯的力,正是画圆的配方,而这就是每台粒子加速器都是圆形的原因。
- 这一课讲 $F = BQv\sin\theta$、它产生的圆周运动,以及基于它的两种仪器。
The force on a moving charge
- $\theta$ is the angle between the velocity and the field. The force is greatest at right angles and zero when the charge moves along the field.
- The direction comes from Fleming's left-hand rule, with the second finger along the motion of a positive charge. For a negative charge, such as an electron, the force is the opposite way.
- A charge at rest feels no force at all: $v = 0$ makes $F = 0$.
运动电荷所受的力
- $\theta$ 是速度与磁场之间的夹角。力在垂直时最大,当电荷沿着磁场运动时为零。
- 方向由弗莱明左手定则给出,中指沿正电荷的运动方向。对电子这样的负电荷,力的方向相反。
- 静止的电荷完全不受力:$v = 0$ 使 $F = 0$。
A $2.0\ \text{C}$ charge moves at $3.0\ \dfrac{\text{m}}{\text{s}}$ at right angles to a $0.50\ \text{T}$ field. What is the force? · 一个 $2.0\ \text{C}$ 的电荷以 $3.0\ \dfrac{\text{m}}{\text{s}}$ 运动,与 $0.50\ \text{T}$ 的场成直角。力是多少?
$F = BQv = 0.50 \times 2.0 \times 3.0 = 3.0\ \text{N}$. · $F = BQv = 0.50 \times 2.0 \times 3.0 = 3.0\ \text{N}$。
For motion at right angles, the magnetic force on a moving charge is $F = BQ$. · 对于成直角的运动,运动电荷受到的磁力是 $F = BQ$。
$F = BQv$ — proportional to the charge's speed. · $F = BQv$——与电荷的速率成正比。
Why the path is a circle
- The force is always perpendicular to $v$, so it does no work, and the speed is constant. Only the direction changes.
- A constant-magnitude force always pointing at right angles to the motion is a centripetal force 向心力, so the path is a circle:
- The radius is proportional to the momentum and inversely proportional to the charge and the field. A stronger field bends the path more tightly.
The force turns the velocity without ever adding to it
路径为什么是圆
- 力始终垂直于 $v$,所以它不做功,而速率恒定。变的只有方向。
- 一个大小不变、始终垂直于运动的力就是向心力(centripetal force),所以路径是一个圆:
- 半径与动量成正比,与电荷和磁场成反比。场越强,路径弯得越紧。
这个力扭转速度,却从不给它增添
Force on a moving charge · 运动电荷受力
F = BQv
The magnetic force on a charge is proportional to its speed (for a fixed field and charge). · 电荷受到的磁力与其速度成正比（对于固定的场和电荷）。
The magnetic force does no work on a charge moving in a circle. · 磁力对沿圆运动的电荷不做功。
The force is always perpendicular to the velocity, so it changes direction but not speed — no work done. · 力总是垂直于速度,所以它改变方向而不改变速率——不做功。
The radius of a charge's circular path in a magnetic field is: · 电荷在磁场中圆形路径的半径是:
Setting $BQv = \dfrac{mv^{2}}{r}$ gives $r = \dfrac{mv}{BQ}$ — bigger momentum, bigger circle. · 令 $BQv = \dfrac{mv^{2}}{r}$ 得到 $r = \dfrac{mv}{BQ}$——动量越大,圆越大。
The period is independent of speed
- Substituting $v = 2\pi r/T$ into $r = mv/(BQ)$ gives:
- The speed has cancelled. A faster particle travels a bigger circle in exactly the same time.
- That surprising fact is what makes the cyclotron possible: one fixed accelerating frequency works for the particle throughout, however fast it becomes.
周期与速率无关
- 把 $v = 2\pi r/T$ 代入 $r = mv/(BQ)$,得到:
- 速率被约掉了。更快的粒子走更大的圆,用的时间恰好相同。
- 这个出人意料的事实让回旋加速器成为可能:一个固定的加速频率对该粒子全程有效,不管它变得多快。
The time for one circular orbit does not depend on the speed. · 转一个圆所用的时间不取决于速率。
$T = \dfrac{2\pi m}{BQ}$ — a faster particle just goes round a bigger circle in the same time. · $T = \dfrac{2\pi m}{BQ}$——更快的粒子只是在相同的时间里绕一个更大的圆。
A charged particle moves in a circle in a magnetic field. Which quantities stay constant? Select all · 所有 that apply. · 一个带电粒子在磁场中做圆周运动。哪些量保持不变?选出所有适用的。
The force does no work, so speed and kinetic energy cannot change; only the direction does, which is what makes the path circular. · 这个力不做功,所以速率和动能不可能改变;变的只有方向,而这正是路径成为圆的原因。
Worked example: bending an electron beam
- An electron of speed $2.0 \times 10^7\ \text{m/s}$ enters a field of $1.5\ \text{mT}$ at right angles. Find the radius of its path. ($m_{\text{e}} = 9.11 \times 10^{-31}\ \text{kg}$, $e = 1.60 \times 10^{-19}\ \text{C}$.)
- What if the speed doubles? The radius doubles, but the period does not change, since $T = 2\pi m/(BQ)$ contains no $v$.
- And the kinetic energy? Unchanged by the magnetic force, which does no work. Only an electric field can change the speed.
例题:弯折电子束
- 一个速率 $2.0 \times 10^7\ \text{m/s}$ 的电子垂直进入 $1.5\ \text{mT}$ 的磁场。求它路径的半径。($m_{\text{e}} = 9.11 \times 10^{-31}\ \text{kg}$,$e = 1.60 \times 10^{-19}\ \text{C}$。)
- *如果速率加倍呢?*半径加倍,但周期不变,因为 $T = 2\pi m/(BQ)$ 里没有 $v$。
- *那动能呢?*不受磁力影响,因为它不做功。只有电场才能改变速率。
An electron at 2.0e7 m/s enters a 1.5 mT field at right angles. What is the radius of its path, in metres? (m = 9.11e-31 kg, e = 1.60e-19 C) · 一个 2.0e7 m/s 的电子垂直进入 1.5 mT 的磁场。它路径的半径是多少米?(m = 9.11e-31 kg,e = 1.60e-19 C)
r = mv/(BQ) = (9.11e-31 x 2.0e7)/(1.5e-3 x 1.60e-19) = 0.076 m. Doubling the speed doubles r but leaves the period unchanged. · r = mv/(BQ) = (9.11e-31 x 2.0e7)/(1.5e-3 x 1.60e-19) = 0.076 m。速率加倍使 r 加倍,而周期不变。
The velocity selector
- Cross an electric field with a magnetic field so their forces on a moving charge are opposite.
- The electric force $QE$ does not depend on speed; the magnetic force $BQv$ does. So they balance for exactly one speed:
- Particles at that speed pass straight through the slit. Faster ones are pushed one way and slower ones the other, whatever their charge or mass. That is a velocity selector 速度选择器.
One speed goes straight; everything else is deflected
速度选择器
- 让电场与磁场交叉,使它们对运动电荷的作用力方向相反。
- 电场力 $QE$ 与速率无关;磁力 $BQv$ 与速率有关。所以它们恰好对一个速率平衡:
- 具有这个速率的粒子径直穿过狭缝。更快的被推向一边,更慢的被推向另一边,不论它们的电荷或质量如何。这就是速度选择器(velocity selector)。
一个速率直着走;其余的都被偏折
A velocity selector (crossed E and B fields) lets through only particles with speed: · 速度选择器(交叉的 E 和 B 场)只让具有什么速率的粒子通过:
The forces balance ($qE = qvB$) only at $v = \dfrac{E}{B}$; others are deflected. · 只有在 $v = \dfrac{E}{B}$ 时力才平衡($qE = qvB$);其他的都被偏转。
In a velocity selector the electric and magnetic forces balance for particles of speed v = ____. · 在速度选择器中,电场力和磁力对速率 v = ____ 的粒子相互平衡。
QE = BQv, and the charge cancels, so only one speed passes straight through whatever the particle's charge or mass. · QE = BQv,电荷约掉,所以不论粒子的电荷或质量如何,只有一个速率能直着通过。
The Hall effect
- Pass a current through a flat slab in a perpendicular field. The magnetic force pushes the moving charges to one face, which builds up until the electric field it creates just balances the magnetic force.
- The steady p.d. across the slab is the Hall voltage 霍尔电压:
- Since everything except $B$ is fixed for a given slab, $V_{\text{H}} \propto B$, which is how a Hall probe 霍尔探头 measures a magnetic field.
霍尔效应
- 让电流通过垂直磁场中的一块扁平薄片。磁力把运动电荷推向一个面,电荷堆积直到它产生的电场恰好平衡磁力。
- 薄片两面之间稳定的电压就是霍尔电压(Hall effect):
- 对给定的薄片,除 $B$ 之外一切都固定,所以 $V_{\text{H}} \propto B$,这就是霍尔探头(Hall probe)测量磁场的方式。
Why can a Hall probe be used to measure magnetic flux density? · 霍尔探头为什么能用来测量磁通密度?
Charges are pushed to one face until the electric field they build up balances the magnetic force, leaving a steady p.d. proportional to B. · 电荷被推向一个面,直到它们建立的电场平衡磁力,留下一个正比于 B 的稳定电压。
Marks that slip away
- The magnetic force does no work, so the speed and kinetic energy are unchanged. Only the direction changes.
- The period is independent of speed; the radius is not. Check which the question asks for.
- $\theta$ is between the velocity and the field, and a charge at rest feels no force.
- For a negative charge the force is opposite to the left-hand-rule direction for the conventional current. Say so when you use it.
容易丢掉的分
- 磁力不做功,所以速率和动能不变。变的只有方向。
- 周期与速率无关;半径则不然。看清题目问的是哪一个。
- $\theta$ 是速度与磁场之间的夹角,而静止电荷不受力。
- 对负电荷,力的方向与按常规电流用左手定则得到的方向相反。用的时候要说明。
You've got it
- $F = BQv\sin\theta$: greatest at right angles, zero along the field, zero for a charge at rest, and perpendicular to both $v$ and $B$
- because it does no work, the speed is constant and the path is a circle with $r = \dfrac{mv}{BQ}$
- the period $T = \dfrac{2\pi m}{BQ}$ is independent of speed, which is what makes a cyclotron work
- a velocity selector passes only $v = E/B$; a Hall probe uses $V_{\text{H}} = BI/(ntq)$, proportional to $B$, to measure a field
你掌握了
- $F = BQv\sin\theta$:垂直时最大,沿场方向为零,静止电荷为零,且同时垂直于 $v$ 和 $B$
- 因为它不做功,速率恒定,路径是半径 $r = \dfrac{mv}{BQ}$ 的圆
- 周期 $T = \dfrac{2\pi m}{BQ}$ 与速率无关,这正是回旋加速器能工作的原因
- 速度选择器只放行 $v = E/B$;霍尔探头用正比于 $B$ 的 $V_{\text{H}} = BI/(ntq)$ 测量磁场