Keep electric and magnetic forces signed at entry
Electric and magnetic forces must be added as vectors before a cancellation is claimed. The supplied entry velocity is along +x, electric field .6 N/C along +y and magnetic field .2 T along +z. A right-handed coordinate system gives v cross B along −y. Therefore the net entry force is q(.6−.2v) in the y direction, with charge in coulombs.
Learning outcome: Compare speeds2,3,4 m/s for both charge signs and neutral charge. Retain component forces before summing. Explain why a neutral undeflected record and a nonzero-charge balanced record support different inferences.
Starting recap
Use a right-handed coordinate system, oriented surface normal, signed charge/current and declared energy boundary. Separate an entry force from a trajectory, emf from current, and rms from signed average.
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At v=2 m/s and q=+1 µC, electric force is +.6 µN and magnetic force −.4 µN, leaving +.2 µN. At q=−1 µC both reverse and net force is −.2 µN. At v=3 m/s either nonzero sign balances.
Open the experimentPredict
A neutral particle with zero entry deflection establishes:
Experiment
Compare speeds2,3,4 m/s for both charge signs and neutral charge. Retain component forces before summing. Explain why a neutral undeflected record and a nonzero-charge balanced record support different inferences.
Keep electric and magnetic forces signed at entry
Worked teaching model · synthetic values · Uniform E=.6 N/C along +y and B=.2 T along +z at entry. Velocity has only +x component. Calculate entry forces, not finite-aperture transmission or a complete later trajectory. Fictional passive particle, virtual only.
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Signed entry force versus speed
Unlike graph quantities rescale independently; read units and values, not screen angles. 1: same fields and charge (solid). Joined curves show the supplied model, not additional observations.
Equivalent numerical coordinates for Signed entry force versus speed
same fields and charge
| entry speed (m/s) | net y force (N) |
|---|---|
| 1 | 4e-7 |
| 1.1 | 3.8e-7 |
| 1.2 | 3.6e-7 |
| 1.3 | 3.4e-7 |
| 1.4 | 3.2e-7 |
| 1.5 | 3e-7 |
| 1.6 | 2.8e-7 |
| 1.7 | 2.6e-7 |
| 1.8 | 2.4e-7 |
| 1.9 | 2.2e-7 |
| 2 | 2e-7 |
| 2.1 | 1.8e-7 |
| 2.2 | 1.6e-7 |
| 2.3 | 1.4e-7 |
| 2.4 | 1.2e-7 |
| 2.5 | 1e-7 |
| 2.6 | 8e-8 |
| 2.7 | 6e-8 |
| 2.8 | 4e-8 |
| 2.9 | 2e-8 |
| 3 | 0 |
| 3.1 | -2e-8 |
| 3.2 | -4e-8 |
| 3.3 | -6e-8 |
| 3.4 | -8e-8 |
| 3.5 | -1e-7 |
| 3.6 | -1.2e-7 |
| 3.7 | -1.4e-7 |
| 3.8 | -1.6e-7 |
| 3.9 | -1.8e-7 |
| 4 | -2e-7 |
| 4.1 | -2.2e-7 |
| 4.2 | -2.4e-7 |
| 4.3 | -2.6e-7 |
| 4.4 | -2.8e-7 |
| 4.5 | -3e-7 |
| 4.6 | -3.2e-7 |
| 4.7 | -3.4e-7 |
| 4.8 | -3.6e-7 |
| 4.9 | -3.8e-7 |
| 5 | -4e-7 |
At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.
| Quantity | Value | Unit |
|---|---|---|
| Electric force y | 6e-7 | N |
| Magnetic force y | -6e-7 | N |
| Net entry force y | 0 | N |
| Nonzero-charge balanced entry speed | 3 | m/s |
| Entry magnetic power | 0 | W |
| Entry electric power | 0 | W |
Charge sign reverses both entry forces; the nonzero-charge balance speed depends on the field ratio.
For velocity +x and field +z, v cross B points −y. Add qE and q(v cross B) as vectors. At balanced entry v=E/B=3 m/s, nonzero charges have zero net entry force. A neutral particle has zero force at every speed and therefore does not test that ratio. Both forces are transverse at this instant, but a later deflected velocity can allow electric work; this entry calculation does not certify finite-aperture transmission.
Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.
Charge sign reverses both entry forces; the nonzero-charge balance speed depends on the field ratio.
For velocity +x and field +z, v cross B points −y. Add qE and q(v cross B) as vectors. At balanced entry v=E/B=3 m/s, nonzero charges have zero net entry force. A neutral particle has zero force at every speed and therefore does not test that ratio. Both forces are transverse at this instant, but a later deflected velocity can allow electric work; this entry calculation does not certify finite-aperture transmission.
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