Physics · Moving Charges And Magnetism · NEET
The radius is r = mv/(qB), so a faster particle moves in a bigger circle. Time period T = distance/speed = 2πr/v = 2π(mv/qB)/v. The v in the top and bottom cancel, leaving T = 2πm/(qB). So even though a fast particle travels a longer path, it does so at higher speed, and the two effects cancel exactly. Every particle of the same q and m completes one loop in the same time.
They measure the same rotation but in different units. Frequency f = qB/(2πm) counts loops per second (unit Hz). Angular frequency ω = qB/m counts radians per second. They are linked by ω = 2πf. In NEET numericals, read the question: if it asks 'frequency of revolution' use f; if it asks 'angular frequency' use ω = qB/m.
Yes. T = 2πm/(qB), so B is in the denominator. If you double B, the time period becomes half and the frequency doubles. B is the one thing (besides q and m of the particle) that changes T. Speed and radius do not.
They come from the same physics. The magnetic force qvB supplies the centripetal force mv²/r, giving r = mv/(qB). Frequency then follows because the particle covers 2πr per loop at speed v: f = v/(2πr) = v/(2π·mv/qB) = qB/(2πm). So the radius formula and the frequency formula are two views of the same circular motion.
No. Frequency f = qB/(2πm) depends on the charge-to-mass ratio q/m. An electron and proton have the same charge magnitude but very different masses (the proton is about 1836 times heavier). So the electron has a much higher frequency and a much shorter time period than the proton in the same field.
An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 × 10⁻² T. If the value of e/m is 1.76 × 10¹¹ C/kg, the frequency of revolution of the electron is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
f = qB/(2πm), where q is charge, B is magnetic field strength, and m is mass. The matching time period is T = 2πm/(qB) and the angular frequency is ω = qB/m.
Yes. f = qB/(2πm) contains no v term. A faster particle simply moves in a larger circle (r = mv/qB) but keeps the same frequency and time period.
Only the particle's charge q, its mass m, and the magnetic field B. It is T = 2πm/(qB). It does NOT depend on speed, kinetic energy, or radius.
Angular frequency ω = qB/m is measured in radians per second (rad/s). Ordinary frequency f = ω/2π is in hertz (Hz).
NEET tests it two ways: direct plug-in numericals (like NEET 2016) and conceptual traps asking what happens to T when speed or radius changes. Knowing T is speed-independent lets you answer both quickly.
