In a mass spectrometer, an ion of charge, q, and mass, m, moving in a path of radius, r, in a field of flux density, B, has a speed of
-
\(\frac{qBr}{m}\)
- \(\frac{mr}{qB}\)
- \(\frac{qm}{Br}\)
- \(\frac{mq}{r}\)
The correct answer is: A
Explanation
F = qvB ( force on moving charge in a magnetic field) F = \(\frac{mv^2}{r}\)
equating F
qvB = \(\frac{mv^2}{r}\)
qB = \(\frac{mv}{r}\)
qBr = mv
Therefore, v = \(\frac{qBr}{m}\)