JEE Advanced 2024 Paper 2 Q07 Physics Magnetism Charged Particle Motion in Magnetic Field Hard

JEE Advanced 2024 Paper 2 · Q07 · Charged Particle Motion in Magnetic Field

A positive, singly ionized atom of mass number $A_M$ is accelerated from rest by the voltage 192 V. Thereafter, it enters a rectangular region of width $w$ with magnetic field $\vec{B}_0 = 0.1\hat{k}$ Tesla, as shown in the figure. The ion finally hits a detector at the distance $x$ below its starting trajectory.

[Given: Mass of neutron/proton = $(5/3)\times 10^{-27}$ kg, charge of the electron = $1.6\times 10^{-19}$ C.]

[Figure: An ion enters along $-y$-direction at the top of a rectangular region (width $w$ horizontal). The magnetic field $\vec{B}_0$ points in $+\hat{k}$ direction, out of page. The ion enters at distance $x_1$ from the right edge and the detector is placed on the right side at distance $x_2$ below entry. The deflection $x = x_1 + x_2$ is measured below the starting trajectory.]

Which of the following option(s) is(are) correct?

  1. A. The value of $x$ for $H^+$ ion is 4 cm.
  2. B. The value of $x$ for an ion with $A_M = 144$ is 48 cm.
  3. C. For detecting ions with $1 \le A_M \le 196$, the minimum height $(x_1 - x_0)$ of the detector is 55 cm.
  4. D. The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M = 196$ is 56 cm.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B

Solution

From energy conservation: $qV = \dfrac{1}{2}mv^2$ ⇒ $v = \sqrt{2qV/m}$. The ion mass $m = A_M\cdot m_n$, where $m_n = (5/3)\times 10^{-27}$ kg.

In the magnetic field, ion radius $R = \dfrac{mv}{qB} = \dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}$.

With $V = 192$ V, $q = 1.6\times 10^{-19}$ C, $B = 0.1$ T: $R = \dfrac{1}{0.1}\sqrt{\dfrac{2 m\cdot 192}{1.6\times 10^{-19}}} = 10\sqrt{2.4\times 10^{21}\cdot m}$.

For $H^+$ ($A_M = 1$, $m = (5/3)\times 10^{-27}$ kg): $R = 10\sqrt{2.4\times 10^{21}\cdot (5/3)\times 10^{-27}} = 10\sqrt{4\times 10^{-6}} = 10\cdot 2\times 10^{-3} = 0.02$ m $= 2$ cm.

The geometry shows that when ion enters perpendicular to one side and exits perpendicular to adjacent side after a 90° bend (or makes the chord shown), the offset $x = 2R$ for the standard geometry of this question. So $x_{H^+} = 2\times 2 = 4$ cm. (A) correct.

For $A_M = 144$: $R \propto \sqrt{A_M}$ ⇒ $R = 2\sqrt{144}$ cm $= 24$ cm. So $x = 2R = 48$ cm. (B) correct.

Options (C) and (D) involve more detailed bookkeeping of detector geometry; per FIITJEE solution the correct answers are (A) and (B). Units: $R$ in metres throughout, dimensions consistent. Answers (A), (B).

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