The radius of the circular path of a charged particle of mass $m$, charge $q$, and kinetic energy $K$ moving perpendicular to a magnetic field $B$ is:

$$r = \frac{m v}{q B} = \frac{\sqrt{2 m K}}{q B}$$

For a proton ($p$) and an $\alpha$-particle ($\alpha$):

• $m_p = m$, $q_p = e$
• $m_\alpha = 4m$, $q_\alpha = 2e$

Taking the ratio of their radii for identical kinetic energy $K$ and field $B$:

$$\frac{r_p}{r_\alpha} = \frac{\sqrt{m_p}}{q_p} \times \frac{q_\alpha}{\sqrt{m_\alpha}} = \frac{\sqrt{m}}{e} \times \frac{2e}{\sqrt{4m}} = 1 \times \frac{2}{2} = 1$$

The magnetic force on a charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ is given by:

$$\vec{F} = q(\vec{v} \times \vec{B}) \implies F = |q| v B \sin\theta$$

Here, the electron is projected along the direction of the magnetic field, so $\theta = 0^\circ$:

$$\sin(0^\circ) = 0 \implies F = 0$$

The wire consists of straight segments and a semicircular arc of radius $R$ carrying current $I$:

• The straight wire segments lie along lines passing through the centre $O$, so the angle between $d\vec{l}$ and $\vec{r}$ is $0^\circ$ or $180^\circ$, giving $B_{\text{straight}} = 0$.
• For a circular arc subtending angle $\theta$ at the centre, the magnetic field is:

$$B = \frac{\mu_0 I}{4\pi R} \times \theta$$

For a semicircular arc, $\theta = \pi$ radians:

$$B = \frac{\mu_0 I}{4\pi R} \times \pi = \frac{\mu_0 I}{4R}$$

The centripetal force for circular motion in a magnetic field is provided by the magnetic Lorentz force:

$$\frac{m v^2}{r} = q v B \implies r = \frac{m v}{q B}$$

The period of revolution is:

$$T = \frac{2\pi r}{v} = \frac{2\pi m}{q B}$$

The frequency of revolution (cyclotron frequency) is:

$$f = \frac{1}{T} = \frac{q B}{2\pi m}$$

Notice that the frequency depends only on charge $q$, magnetic field $B$, and mass $m$. It is completely independent of the speed $v$ and radius $r$. Hence, option 1 is the correct answer.