Q.Give an application of cyclotron.

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

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

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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 …

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

$u_c = \frac{qB}{2\pi m}$ …

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).

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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:

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

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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.

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