Magnetic forces: turn velocity without increasing speed
Введено| English | Значение |
|---|---|
| magnetic force/mæɡˈnetɪk fɔːs/ | магнитная сила |
| momentum/məʊˈmentəm/ | импульс |
What would explain this observation?
- A charged particle curves in a uniform magnetic field. Its direction changes although the magnetic force 磁力 alone does not increase its kinetic energy.
Build the model
- The magnetic force magnitude is F = |q|vB sinθ.
- It is perpendicular to velocity and magnetic field, with direction determined by charge sign.
- For perpendicular entry into a uniform field, it can provide the centripetal force for circular motion.
- magnetic force: Force due to a magnetic field on a moving charge or current.
- momentum 动量: Mass multiplied by velocity in the classical model.
What changes under a magnetic force alone in perpendicular circular motion?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
A magnetic force acts perpendicular to the velocity, so it changes the velocity's DIRECTION only, producing circular motion. A perpendicular force does no work, so kinetic energy cannot increase, and the charge magnitude is intrinsic.
Match each technical term to its precise meaning.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Use the definitions to distinguish related quantities and processes.
Choose evidence that can test it
- Equate magnetic force with mv²/r to obtain r = mv/(|q|B) for perpendicular motion.
- A stronger field makes a smaller radius at fixed momentum.
- Parallel entry gives zero magnetic force in this model.
In the lab: Draw velocity, field and force as separate arrows using stated into/out-of-page conventions. Use supplied beam data or a simulation. Do not treat a current-carrying wire direction as identical to electron-motion direction.
Which two habits make the investigation or model in this case more defensible?
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
Stating the conditions and assumptions lets others repeat the work exactly, and recording the supporting evidence lets them check the conclusion. Hiding disagreeing results removes the evidence that could challenge a wrong conclusion, and one measurement can never establish a wider claim.
Work from known quantities
- Known: a particle has m = 2.0×10⁻²⁷ kg, v = 3.0×10⁶ m/s, |q| = 1.0×10⁻¹⁹ C and B = 0.20 T. r = mv/(|q|B) = 0.30 m.
- The calculation assumes non-relativistic motion and perpendicular entry.
Your turn: Use r = mv/(|q|B). For m=2×10⁻²⁷ kg, v=2×10⁶ m/s, |q|=1×10⁻¹⁹ C and B=0.20 T, find r.
Use r = mv/(|q|B). For m=2×10⁻²⁷ kg, v=2×10⁶ m/s, |q|=1×10⁻¹⁹ C and B=0.20 T, find r.
Сложность 2/5 · 140 монет + небольшой бонус за первое выполнение сегодня
The result is 0.2 m. Known: m = 2x10^-27 kg; v = 2x10^6 m/s; |q| = 1x10^-19 C; B = 0.20 T, perpendicular entry. Use r = mv/(|q|B). r = (2x10^-27 x 2x10^6)/(1x10^-19 x 0.20) = 0.2 m. The path is a circle of this radius under the stated model.
Check the conclusion and its limits
- A magnetic force can change momentum direction without doing work.
- Adding an electric field can change the energy, so conclusions about a magnetic field alone do not cover every combined-field apparatus.
Not true: A magnetic force alone continuously increases the kinetic energy of a charged particle. A force perpendicular to the velocity does no work (work = force x distance ALONG the force), so a magnetic field alone cannot change a particle's kinetic energy - only its direction. Speeding up needs a field component along the motion, which only an electric field provides.
- The magnetic force magnitude is F = |q|vB sinθ.
- Equate magnetic force with mv²/r to obtain r = mv/(|q|B) for perpendicular motion.
- Watch for: A magnetic force can change momentum direction without doing work.
Write from an orbital radius of 0.50 m
- Use r = mv divided by the product of charge magnitude and magnetic field.
- Take m = 2 times 10 to the minus 27 kg, v = 2 times 10 to the 6 metres per second, charge magnitude 1 times 10 to the minus 19 coulomb and B = 0.08 T. A magnetic force perpendicular to the velocity does no work.
A force perpendicular to the velocity does no work (work = force x distance ALONG the force), so a magnetic field alone cannot change a particle's kinetic energy - only its direction.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
True. A force perpendicular to the velocity does no work (work = force x distance ALONG the force), so a magnetic field alone cannot change a particle's kinetic energy - only its direction. Speeding up needs a field component along the motion, which only an electric field provides.
Force due to a magnetic field on a moving charge or current: write the technical term.
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
magnetic force means Force due to a magnetic field on a moving charge or current.
A charged particle has mass 2 times 10 to the minus 27 kg, speed 2 times 10 to the 6 metres per second and charge magnitude 1 times 10 to the minus 19 coulomb, in a 0.08 T magnetic field perpendicular to its velocity. What is the radius of its path, and why does this force not increase the kinetic energy?
The magnetic radius uses momentum divided by charge magnitude times field. Work needs a force component along the displacement.
What is the radius of the path?
Сложность 1/5 · 100 монет + небольшой бонус за первое выполнение сегодня
4 divided by 8 is 0.50, so the radius is 0.50 m. Other options — 0.20 m (the lesson path in a 0.20 T field); 0.08 T (the magnetic field).