A proton and a deuteron (q = +e, m = 2.0u) having same kinetic energies enter a region of uniform magnetic field B, moving perpendicular to B. The ratio of the radius rd of deuteron path to the radius rp of the proton path is: (a) 1: 2 (b) 1: 1 (c) √(2): 1 (d) 1: √(2)

Answer: (c). r = (√(2mK))/(qB); for equal K, B and q: r ∝ √(m). (rd)/(rp) = √(2/1) = √(2) So the answer is (c) √(2): 1.

A proton and a deuteron (q = +e, m = 2.0u) having same kinetic… | JEE Main 2024 (9 Apr, Shift 2)

Q37JEE Main 2024 · 9 APR · SHIFT 2Medium

Moving Charge in a Magnetic Field

A proton and a deuteron $(q = +e, m = 2.0\text{u})$ having same kinetic energies enter a region of uniform magnetic field $\vec{B}$ , moving perpendicular to $\vec{B}$ . The ratio of the radius $r_d$ of deuteron path to the radius $r_p$ of the proton path is:

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Answer: (c)

$r = \dfrac{\sqrt{2mK}}{qB}$ ; for equal $K$ , $B$ and $q$ : $r \propto \sqrt{m}$ .
$\dfrac{r_d}{r_p} = \sqrt{\dfrac{2}{1}} = \sqrt{2}$
So the answer is (c) $\sqrt{2} : 1$ .

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