A proton against an alpha particle.

A proton and an alpha particle with the same kinetic energy enter the same uniform magnetic field at right angles to it. The ratio of the radii of their circular paths (proton : alpha) is:

Answer: 1 : 1.

The radius of the circle is r = mv/qB, and the momentum can be written in terms of the kinetic energy K as mv = √(2mK). The alpha particle has 4 times the proton's mass and 2 times its charge.

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

\Rightarrow \frac{r_{p}}{r_{α}} = \frac{\sqrt{m}}{e} \cdot \frac{2 e}{\sqrt{4 m}} = 1

The extra mass and the extra charge cancel exactly, so the two circles have the same radius. Comparing only the masses gives 1 : 2, and comparing only the charges gives 2 : 1; both are traps.

Original practice question · JEE Physics Decoded

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