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JEE Advanced 2024 · Paper 2 · one or more correct

JEE Advanced 2024 Physics question on Magnetism and Moving Charges

A positive, singly ionized atom of mass number AMA_{\text{M}} is accelerated from rest by the voltage 192 V. Thereafter, it enters a rectangular region of width ww with magnetic field B⃗0=0.1k^\vec{B}_0 = 0.1\hat{k} Tesla, as shown in the figure. The ion finally hits a detector at the distance xx below its starting trajectory.

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

Which of the following option(s) is(are) correct?
  1. A.
    The value of xx for H+H^+ ion is 4 cm.
  2. B.
    The value of xx for an ion with AM=144A_{\text{M}} = 144 is 48 cm.
  3. C.
    For detecting ions with 1≤AM≤1961 \leq A_{\text{M}} \leq 196, the minimum height (x1−x0)(x_1 - x_0) of the detector is 55 cm.
  4. D.
    The minimum width ww of the region of the magnetic field for detecting ions with AM=196A_{\text{M}} = 196 is 56 cm.
Show answer

Correct options: A, B

[JEE Advanced 2024 · Paper 2]

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