WAEC - Physics (2023 - No. 8)
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
\(\frac{qBr}{m}\)
\(\frac{mr}{qB}\)
\(\frac{qm}{Br}\)
\(\frac{mq}{r}\)
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 = \(\frac{qBr}{m}\)
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