Physics · Moving Charges And Magnetism · NEET
No. The cyclotron frequency f = qB/2pi.m depends only on the charge q, mass m and magnetic field B. As the particle speeds up, its radius grows (r = mv/qB), so it travels a bigger circle at higher speed, and the time for one circle stays the same. This constant time is exactly why the same alternating voltage keeps pushing the particle correctly every half-circle.
The ELECTRIC field in the gap between the two dees accelerates it. The magnetic force is always perpendicular to velocity, so it does zero work and cannot change speed, it only bends the path into a circle. Inside the hollow dees there is no electric field, so there the particle just moves in a semicircle at constant speed.
Electrons are very light, so even at modest energy their speed becomes close to the speed of light. Then their mass increases (relativistic effect), which changes m in f = qB/2pi.m. The frequency is no longer constant, the electric push falls out of step, and the particle stops gaining energy. Cyclotrons are used for heavier particles like protons, deuterons and alpha particles.
Resonance means the frequency of the applied alternating voltage equals the cyclotron frequency of the particle: f(applied) = qB/2pi.m. When this holds, the electric field reverses at the exact moment the particle crosses the gap, so it always gets a forward push. If it is not matched, the particle sometimes gets slowed down instead of sped up.
The radius increases in steps. Each time the particle crosses the gap it gains speed, so from r = mv/qB its radius is a little larger on the next semicircle. The path is a spiral of growing semicircles. The particle finally leaves the cyclotron at the outer edge (maximum radius R), with maximum speed and energy.
An electron is moving in a circular path under the influence of a transverse magnetic field of 3.57 x 10^-2 T. If the value of e/m is 1.76 x 10^11 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/(2 pi m), where q is charge, B is magnetic field and m is mass. The time period is T = 2 pi m / qB. Both are independent of the particle's speed.
K.E. max = q^2 B^2 R^2 / (2m), where R is the maximum radius (the dee radius). It comes from putting v = qBR/m into K.E. = (1/2) m v^2. So a bigger dee or a stronger field gives higher energy.
It accelerates charged particles to high energy for use in nuclear physics research, to produce radioactive isotopes for medicine (like PET scan tracers), for cancer treatment (proton beams), and to study nuclear reactions.
A dee is one of two hollow, D-shaped metal chambers. The particle moves in a semicircle inside each dee at constant speed, and gets accelerated by the electric field only in the gap between the two dees.
The magnetic field bends the particle into circles, sending it back to the gap again and again so the electric field can keep pushing it. Without bending, the particle would fly out after a single push. The field controls the geometry and timing, the electric field supplies the energy.
