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Tổ chức Baccalaureate Quốc tế · IB Diploma · Physics · HL Chủ đề 18 5:00 Giọng đọc tiếng Anh · phụ đề tiếng Anh + 中文 được ghi trực tiếp
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A charged particle curves in a uniform magnetic field.
带电粒子在均匀磁场中发生偏转。
Its direction changes although the magnetic force 磁力 alone does not increase its kinetic energy.
其运动方向发生改变，但仅靠磁场力不会增加其动能。
The magnetic force magnitude is F equals the absolute value of q, times v, times B, times sine of theta.
磁力大小为F等于q的绝对值乘以v乘以B乘以sinθ。
It is perpendicular to velocity and magnetic field, with direction determined by charge sign.
它垂直于速度和磁场，方向由电荷符号决定。
For perpendicular entry into a uniform field, it can provide the centripetal force for circular motion.
当垂直进入均匀磁场时，它可以提供圆周运动的向心力。
Magnetic force is defined as the force due to a magnetic field on a moving charge or current.
磁力：磁场对运动电荷或电流的作用力。
Momentum is defined as mass multiplied by velocity in the classical model.
动量：经典模型中质量与速度的乘积。
Equate magnetic force with m v squared over r to obtain r equals m v divided by the absolute value of q times B for perpendicular motion.
将磁力与 m v²/r 等同，可得垂直运动时 r = mv/(|q|B)。
A stronger field makes a smaller radius at fixed momentum.
在动量固定时，更强的磁场会产生更小的半径。
Parallel entry gives zero magnetic force in this model.
在此模型中，平行入射产生的磁力为零。
In the lab: Draw velocity, field and force as separate arrows using stated into/out-of-page conventions.
在实验室中，使用规定的进入或离开纸面的约定，将速度、场和力分别画成箭头。
Use supplied beam data or a simulation.
使用提供的束流数据或模拟结果。
Do not treat a current-carrying wire direction as identical to electron-motion direction.
不要将载流导线的方向等同于电子运动的方向。
Known: a particle has m equal to 2.0 times 10 to the minus 27 kg, v equal to 3.0 times 10 to the 6 metres per second, the magnitude of q equal to 1.0 times 10 to the minus 19 C, and B equal to 0.20 T.
已知：粒子质量 m 为 2.0×10⁻²⁷ kg，速度 v 为 3.0×10⁶ m/s，电荷量大小 |q| 为 1.0×10⁻¹⁹ C，磁感应强度 B 为 0.20 T。
r equals m v divided by the product of the magnitude of q and B, which is 0.30 m.
r 等于 mv 除以 |q| 与 B 的乘积，即 0.30 m。
The calculation assumes non-relativistic motion and perpendicular entry.
计算假设为非相对论运动且垂直射入。
Your turn.
轮到你了。
Use r equals m v over the size of q times B.
使用公式 r 等于 mv 除以 q 的绝对值与 B 的乘积。
For m equal to 2 times 10 to the minus 27 kilograms, v equal to 2 times 10 to the 6 metres per second, q of size 1 times 10 to the minus 19 coulombs and B equal to 0.20 tesla, find r.
已知 m 等于 2 乘以 10 的负 27 次方千克，v 等于 2 乘以 10 的 6 次方米每秒，q 的绝对值等于 1 乘以 10 的负 19 次方库仑，B 等于 0.20 特斯拉，求 r。
A magnetic force can change momentum direction without doing work.
磁场力可以改变动量方向而不做功。
Adding an electric field can change the energy, so conclusions about a magnetic field alone do not cover every combined-field apparatus.
加入电场后能量会发生变化，因此仅针对磁场的结论无法涵盖所有复合场装置。
Not true that a magnetic force alone continuously increases the kinetic energy of a charged particle.
这种说法不正确。 仅靠磁场力不会持续增加带电粒子的动能。
A force perpendicular to the velocity does no work, because work is force times distance along the force, so a magnetic field alone cannot change a particle's kinetic energy, only its direction.
垂直于速度的力不做功，因为功等于力乘以沿力方向的位移。 因此，仅靠磁场无法改变粒子的动能，只能改变其运动方向。
Speeding up needs a field component along the motion, which only an electric field provides.
加速需要一个沿运动方向的场分量，而这只有电场才能提供。
The magnetic force magnitude is F equals the absolute value of q, times v, times B, times sine of theta.
磁力大小为F等于|q|vBsinθ。
Equate magnetic force with m v squared over r to obtain r equals m v divided by the absolute value of q, times B, for perpendicular motion.
将磁力与mv²/r等同，可得垂直运动时r = mv/(|q|B)。
Watch for: A magnetic force can change momentum direction without doing work.
注意：磁力可以改变动量方向而不做功。
Use r equals m v divided by the product of charge magnitude and magnetic field.
使用公式 r = mv / (qB)。
Take m equals 2 times 10 to the minus 27 kg, v equals 2 times 10 to the 6 metres per second, charge magnitude 1 times 10 to the minus 19 coulomb, and B equals 0.10 T.
取 m = 2 × 10⁻²⁷ kg，v = 2 × 10⁶ m/s，电荷量大小为 1 × 10⁻¹⁹ C，B = 0.10 T。
A magnetic force perpendicular to the velocity does no work.
垂直于速度的磁场力不做功。
The result is an orbital radius of 0.40 metres.
结果是轨道半径为0.40米。
Entering perpendicular to the field, the charge feels a force at right angles to its velocity.
垂直进入磁场时，电荷受到与速度垂直的力。
That force is centripetal: it changes direction but not speed, so kinetic energy stays the same and the radius is m v over q B.
这个力提供向心力：只改变方向而不改变速率，所以动能不变，半径等于m v除以q B。
At fixed perpendicular speed and field, reverse charge sign with mass and charge magnitude fixed.
在垂直速度和磁场固定的条件下，保持质量和电荷量大小不变，反转电荷符号。
Compare radius, curvature direction and magnetic work.
比较半径、曲率方向和磁场做功。
Radius is unchanged; curvature reverses because q changes sign.
半径保持不变；曲率反转，因为电荷q改变了符号。
Magnetic force is perpendicular to velocity and does no work.
磁场力垂直于速度方向，不做功。
Oblique entry requires the perpendicular velocity component.
斜向进入需要垂直于磁场的速度分量。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。