| English | 意味 |
|---|---|
| centripetal force/senˈtrɪpɪtl fɔːs/ | 向心力 |
| velocity selector/vəˈlɒsɪti sɪˈlektə/ | 速度選択器 |
| Hall voltage/hɔːl ˈvəʊltɪdʒ/ | ホール電圧 |
| Hall probe/hɔːl prəʊb/ | ホールプローブ |
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.
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 $F = 0$.
$2.0\ \text{C}$ の電荷が $3.0\ \dfrac{\text{m}}{\text{s}}$ で $0.50\ \text{T}$ 磁束密度方向と直角に運動している。力はいくらか？
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$F = BQv = 0.50 \times 2.0 \times 3.0 = 3.0\ \text{N}$.
垂直な運動の場合、移動する電荷にかかる磁力は $F = BQ$____ である。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$F = BQv$ ——電荷の速度に比例する。
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
移動する電荷にかかる力
F = BQv
ある電荷への磁力はその速度に比例する（一定の磁界と電荷の場合）。
円軌道で動く電荷に対して磁力は仕事を行わない。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
力は常に速度に垂直なので、方向は変わるが速さは変わらない——したがって仕事は行われない。
磁場中で電荷が描く円軌道の半径は：
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$BQv = \dfrac{mv^{2}}{r}$ を設定することで $r = \dfrac{mv}{BQ}$ が得られる——運動量越大きいほど、円も大きい。
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.
一周分の周期は速度に依存しない。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$T = \dfrac{2\pi m}{BQ}$ ——速い粒子ほど大きな円を描くが、同じ時間で一周する。
磁場中で円運動をする帯電粒子について、一定の値を保つ物理量はどれか？すべて選択してください。
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
力が仕事を行わないため、速さや運動エネルギーは変わらず、方向だけが変化する。これが軌道を円形にする原因である。
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.
電子が 2.0e7 m/s の速さで、1.5 mT の磁場に対して垂直に入射しました。軌道の半径はいくらくらいメートルですか？（質量 m = 9.11e-31 kg、電荷 e = 1.60e-19 C）
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
r = mv/(BQ) = (9.11e-31 × 2.0e7)/(1.5e-3 × 1.60e-19) = 0.076 m。速度を倍にすると r も倍になるが、周期は変わらない。
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
速度選択器（直交するEおよびBの電界・磁界）は、特定の速度を持つ粒子のみ通過させる。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
力の釣り合い（$qE = qvB$）が成り立つのは $v = \dfrac{E}{B}$ の時だけで、他は偏折される。
速度選択器において、速度 v = ____ の粒子に対して、電気力と磁力が釣り合う。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
QE = BQv であり、電荷項が相殺されるため、粒子の電荷や質量に関わらず、一つの速度のみが直進できる。
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.
ホールプローブが磁束密度を測定できる理由は何か？
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
電荷が一方の面に押し付けられ、形成された電界が磁力と釣り合うことで、Bに比例する定常的な電位差が残る。
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.
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