Question

The (angular) frequency, ω, of the alternating p.d applied across the dees, and the strength, B, of the (normal) magnetic field, used, in a given cyclotron, have been adjusted to values (say ω₀ and B₀), that would enable this cyclotron to accelerate protons.

If the same cyclotron were to be used for accelerating alpha particles, say, the new value of (i) ω, when B is left unchanged at its earlier value (B₀) and (ii) B, when ω is left unchanged at its earlier value (ω₀), would equal, respectively

(1) \(\frac{ω_0}{2}\) and \(\frac{B_0}{2}\)

(2) \(\frac{ω_0}{2}\) and 2B₀

(3) 2B0 and \(\frac{ω_0}{2}\)

(4) 2ω₀ and 2B₀

Answer 1

(2) \(\frac{ω_0}{2}\) and 2B₀

The radius of the circular path, of the charged particle, in a magnetic field, is given by mv²/r = qvB

Hence r = mv/qB

The time period, for one cycle of the motion of the charged particle, in this circular path, is given by

The corresponding angular frequency, ω, equals 2π/T = qB/m

In the adjusted cyclotron, the (angular) frequency of the applied alternating potential equals this (cyclotron) frequency.

Hence for the protons, we have ω₀ = eB₀/m

For the α-particle, the required (angular) frequency, say, ω_α , is given by

(i) When there is no change in the value of B, we would have ω_α = eB/m x 1/2 = ω₀ /2

(ii) When there is no change in the value of ω₀ , we would have ω₀ = eB_α / 2m