Q.An electron is moving with a speed of $3.2\times10^7$ m/s in a magnetic field of $6.00\times10^{-4}$ T perpendicular to its path. What will be the radius of the path? What will be the frequency and the kinetic energy in keV? [Given: mass of electron $=9.1\times10^{-31}$ kg, charge $e=1.6\times10^{-19}$ C, $1\ \text{eV}=1.6\times10^{-19}$ J] [Ans (as printed in the book): 3.0 cm, 18.7 MHz, 2.53 keV]

The cyclotron period T = 2πm/(qB) is independent of the speed V, so none of the offered proportionalities holds.

The cyclotron frequency and period come from equating the magnetic force to the centripetal force:

$qvB = \dfrac{mv^2}{r} \Rightarrow r = \dfrac{mv}{qB}$

The period is

$T = \dfrac{2\pi r}{v} = \dfrac{2\pi m}{qB}$

The speed v cancels out, so T depends only on the mass, charge, and magnetic field — NOT on the speed (or radius) of the particle. This velocity-independence is the very principle that makes the cyclotron work. Therefore none of the options T ∝ V, T ∝ V², T ∝ 1/V, or T ∝ 1/V² is physically valid; the honest answer is that T is independent of V.

The device that accelerates charged particles or ions to high kinetic energies using crossed static magnetic and oscillating electric fields is the cyclotron.

A cyclotron consists of two hollow D-shaped electrodes ("dees") placed in a strong uniform magnetic field, with a high-frequency alternating voltage applied between them. A charged particle injected near the centre is accelerated each time it crosses the gap between the dees, and the magnetic field bends it into a circular path of increasing radius as its speed grows. Because the time for one half-revolution is independent of speed (as long as the particle is non-relativistic), the applied AC frequency can be kept fixed (the cyclotron frequency), and the particle gains energy on every crossing until it is extracted at high energy.

The magnetic field only bends the path (does no work); the oscillating electric field across the dee-gap does the actual accelerating, once every half-cycle, in resonance with the constant cyclotron frequency.

A cyclotron has two hollow, semicircular metal electrodes called dees ($D_1$, $D_2$), placed in a strong, uniform magnetic field $B$ perpendicular to their plane, with a narrow gap between them connected to a high-frequency oscillating voltage source.

Role of the magnetic field: Inside a dee (a field-free, hollow region electrically), the particle experiences only the magnetic force, which makes it move in a semicircular arc of radius

$r = \frac{mv}{qB}$

Since $\vec F = q\vec v\times\vec B$ is always perpendicular to $\vec v$, the magnetic field changes only the direction of motion, doing no work and not changing the speed.

Role of the electric field: Each time the particle crosses the narrow gap between the two dees, it passes through the oscillating electric field. If the field's direction is correctly synchronised, the particle is given a push and gains kinetic energy $qV$ (where $V$ is the gap's instantaneous potential difference) every single crossing.

Why this works repeatedly (resonance condition): The time for the particle to complete a semicircle in a dee is

$t = \frac{\pi m}{qB}$

which is independent of the particle's speed and radius (since larger $v$ exactly gives a proportionally larger $r$, keeping the transit time constant). So if the oscillator frequency is fixed at the cyclotron frequency

$u_c = \frac{qB}{2\pi m}$

the electric field reverses its direction exactly in time with the particle's arrival at the gap on every crossing — so the particle is accelerated (never decelerated) at each gap crossing, again and again.

Net effect: With every crossing the particle's speed $v$ increases, so its radius $r = mv/qB$ increases too — it traces a growing outward spiral, gaining kinetic energy in small increments at each gap crossing, until it reaches the outer edge of the dees where it is extracted as a high-energy beam.

Since the cyclotron period depends on mass, charge and magnetic field but not speed, equal periods for a proton and an alpha particle require the magnetic field ratio $B_p:B_\alpha = 1:2$.

Inside a cyclotron, a charged particle moves in a circular path because the magnetic force provides the centripetal force: $qvB=\dfrac{mv^2}{r}$, giving radius $r=\dfrac{mv}{qB}$. The period of one revolution is $T=\dfrac{2\pi r}{v}=\dfrac{2\pi m}{qB}$, which is independent of the particle's speed $v$.

For the proton: $T_p = \dfrac{2\pi m_p}{q_p B_p}$. For the alpha particle: $T_\alpha = \dfrac{2\pi m_\alpha}{q_\alpha B_\alpha}$.

The alpha particle has mass $m_\alpha = 4m_p$ (it is a helium nucleus: 2 protons + 2 neutrons, each roughly the proton mass) and charge $q_\alpha = 2q_p$ (it carries two units of elementary charge).

Setting $T_p = T_\alpha$: $\dfrac{m_p}{q_p B_p} = \dfrac{m_\alpha}{q_\alpha B_\alpha} = \dfrac{4m_p}{2q_p B_\alpha}$

This simplifies to $\dfrac{1}{B_p} = \dfrac{2}{B_\alpha} \implies \dfrac{B_p}{B_\alpha} = \dfrac{1}{2}$

So the magnetic field needed for the proton must be half that needed for the alpha particle for both to have the same cyclotron period, i.e. $B_p : B_\alpha = 1:2$.

The period of revolution of a charged particle inside a cyclotron is independent of its velocity.

Inside a cyclotron, the magnetic force provides the centripetal force for the particle's circular path: $qvB = \dfrac{mv^2}{r}$, which gives an orbit radius $r = \dfrac{mv}{qB}$.

The period of one revolution is:

$T = \dfrac{2\pi r}{v} = \dfrac{2\pi}{v}\cdot\dfrac{mv}{qB} = \dfrac{2\pi m}{qB}$

The velocity $v$ cancels out of this expression completely — as the particle speeds up during acceleration, its orbit radius grows correspondingly, but the time taken for one full revolution stays exactly the same. $T$ does, however, depend on the mass $m$ of the particle (heavier particles take longer per revolution), the charge $q$ (options c, d change $T$), and the magnetic induction $B$ (a stronger field shortens the period). This velocity-independence of $T$ is precisely what allows a cyclotron to use a single fixed-frequency oscillator (matched to $T$) to keep re-accelerating the particle every half-cycle as it spirals outward.

A cyclotron is used to accelerate charged particles to very high kinetic energies for nuclear research and for producing medical radioisotopes.

Concept: In a cyclotron a perpendicular magnetic field keeps the charged particle moving in circular arcs inside two dees, while an alternating voltage across the gap gives it an energy "kick" each half-cycle, spiralling it outward to high speed.

Applications: bombarding target nuclei to study nuclear reactions, producing radioactive isotopes (e.g. for PET scans and cancer therapy), and generating high-energy beams for research.