Velocity Selector
For mutually perpendicular E, B, and v arranged so qE and q(v cross B) oppose, undeflected motion requires E = vB, hence v = E/B. The selected speed is independent of charge sign when geometry is reversed consistently.
Why this shows up in the exam
Selecting a narrow beam speed · Finding E from an undeflected charged particle · Feeding known-speed ions into a mass spectrometer
Learn the idea
Crossed electric and magnetic forces cancel only for particles with the selected speed E/B and the correct direction. The electric force pushes along E, while magnetic force can oppose it. Only one speed makes the two pushes equal, so an undeflected beam emerges with a known velocity.
🧠 Memory hook: No deflection means electric push equals magnetic turn.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- q E = |q| v B — force-balance magnitude for undeflected passage
- v = E/B — selected speed for mutually perpendicular fields and velocity
How to approach it
- 1Draw both force directions
- 2Verify they oppose
- 3Set magnitudes equal and solve v = E/B
Common slip-ups that cost marks
- •Using v = B/E
- •Balancing magnitudes without checking opposite directions
- •Applying E/B when fields or velocity are not mutually perpendicular
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A charge of 2 microC moves perpendicular to a 3 T magnetic field at 4 x 10^5 m/s. Find the magnetic force.
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