A proton and an \alpha -particle, having kinetic energies Kp and K \alpha respectively, enter into a magnetic field at right angles. The ratio of the radii of trajectory of proton to that of \alpha -particle is 2 : 1. The ratio of Kp : K \alpha is :

Magnetics PYQ — JEE Main 2021 · 18 Mar · Shift 2 · Q58 with Solution

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Correct answer: B

When a charged particle enters a magnetic field perpendicular to it, it moves in a circular path of radius

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

Using kinetic energy,

$K = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2K}{m}}$

Substitute into the radius formula:

$r = \frac{m}{qB}\sqrt{\frac{2K}{m}} = \frac{1}{qB}\sqrt{2mK}$

So,

$r \propto \frac{\sqrt{mK}}{q}$

For proton:

Thus,

$r_p \propto \frac{\sqrt{m_p K_p}}{e}$

For $\alpha$ -particle:

Thus,

$r_\alpha \propto \frac{\sqrt{4m_p K_\alpha}}{2e}$

Now simplify:

$r_\alpha \propto \frac{2\sqrt{m_p K_\alpha}}{2e} = \frac{\sqrt{m_p K_\alpha}}{e}$

Given:

$r_p : r_\alpha = 2:1$

So,

$\frac{r_p}{r_\alpha} = \frac{\frac{\sqrt{m_p K_p}}{e}}{\frac{\sqrt{m_p K_\alpha}}{e}} = \sqrt{\frac{K_p}{K_\alpha}}$

Hence,

$\sqrt{\frac{K_p}{K_\alpha}} = 2$

Squaring both sides,

$\frac{K_p}{K_\alpha} = 4$

Therefore,

$K_p : K_\alpha = 4:1$

So the correct answer is B.