Velocity Selector
In a velocity selector, charged particles pass undeflected through a region where a uniform electric field of 3×104V m−1 acts perpendicular to a uniform magnetic field of 0.6T. What speed must the selected particles have to traverse the region in a straight line?
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About This Question
- Subject
- physics
- Chapter
- magnetic effects of current and magnetism
- Topic
- velocity selector
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
5×104m s−1
A velocity selector exploits the fact that the electric force qE and the magnetic force qvB act in opposite directions on a moving charge when the fields are mutually perpendicular and oriented appropriately. Only particles for which these forces exactly cancel pass straight through, so the selection condition is qE=qvB, independent of charge and mass. Solving for speed gives v=BE=0.63×104=5×104m s−1. The option 1.8×104m s−1 wrongly multiplies E by B instead of dividing. The option 2×105m s−1 inverts the ratio, using B/E scaled wrongly. The option 5×105m s−1 misplaces a power of ten. This crossed-field configuration is the basis of the velocity selector that feeds mass spectrometers, as treated in NCERT and J. J. Thomson's experiment. A unit check confirms volt per metre divided by tesla yields metre per second, and the result is a physically reasonable beam speed. Because the selection condition is independent of charge and mass, the velocity selector delivers a beam in which every particle shares one common speed, regardless of how the particles were produced. This makes it the essential first stage before a mass spectrometer, where the cleanly monoenergetic beam then enters a pure magnetic field and separates by mass. Historically the same crossed-field arrangement let Thomson measure the charge-to-mass ratio of the electron.
This hard difficulty physics question is from the chapter magnetic effects of current and magnetism, covering the topic of velocity selector. It appeared in the 2025 exam.
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