A deuteron and a proton moving with equal kinetic energy enter into a uniform magnetic field at right angle to the field. If rd and rp are the radii of their circular paths respectively, then the ratio r_d \over r_p will be \sqrtx : 1 where x is \underline\hspace2cm .

Magnetics PYQ — JEE Main 2022 · 27 Jun · Shift 2 · Q70 with Solution

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Correct answer: 2

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

r = \frac{mv}{qB}

Since kinetic energy is given equal, it is better to express $r$ in terms of kinetic energy.

For a particle,

K = \frac{1}{2}mv^2

So,

v = \sqrt{\frac{2K}{m}}

Substitute into the radius formula:

r = \frac{m}{qB}\sqrt{\frac{2K}{m}}

r = \frac{\sqrt{2mK}}{qB}

Thus,

r \propto \frac{\sqrt{m}}{q}

when $K$ and $B$ are same.

Therefore,

\frac{r_d}{r_p} = \frac{\sqrt{m_d}/e}{\sqrt{m_p}/e} = \sqrt{\frac{m_d}{m_p}} = \sqrt{2}

So,

\frac{r_d}{r_p} = \sqrt{2} : 1

Comparing with $\sqrt{x} : 1$ , we get

\sqrt{x} = \sqrt{2}

Hence,

x = 2

Stored correct answer = 2

Our derived answer = 2

So the answer agrees with the stored correct answer.