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Physics · Moving Charges and Magnetic Field

JEE Advanced 2024 — Paper 2 — Question 18

A positive, singly ionized atom of mass number AMA_{M} is accelerated from rest by the voltage 192 V . Thereafter, it enters a rectangular region of width ww with magnetic field B→0=0.1k^\overrightarrow{\mathrm{B}}_{0}=0.1 \hat{\mathrm{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)×10−27 kg=(5 / 3) \times 10^{-27} \mathrm{~kg}, charge of the electron =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C}.] Which of the following option(s) is(are) correct?

Question figure
  1. Option A:

    The value of xx for H+\mathrm{H}^{+}ion is 4 cm

    Correct
  2. Option B:

    The value of xx for an ion with Am=144A_{m}=144 is 48 cm

    Correct
  3. Option C:

    For detecting ions with 1≤AM≤1961 \leq A_{M} \leq 196, the minimum height ( x1−x0x_{1}-x_{0} ) of the detector is 55 cm

  4. Option D:

    The minimum width ww of the region of the magnetic field for detecting ions with Am=196A_{m}=196 is 56 cm

Answer: A, B

Step-by-step solution

x=2R=2mvqB=8mVqB2x=2 R=\frac{2 m v}{q B}=\sqrt{\frac{8 m V}{q B^{2}}} x1−x0=14×4−4=52 cm\mathrm{x}_{1}-\mathrm{x}_{0}=14 \times 4-4=52 \mathrm{~cm} Minimum width w for Am=196⇒R=14×2=28 cmA_{m}=196 \Rightarrow R=14 \times 2=28 \mathrm{~cm}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 2
Subject
Physics
Chapter
Moving Charges and Magnetic Field
Topic
Motion of Charged Particles in magnetic Fields

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