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JEE Advanced Physics Magnetic Effects of Current 2024 JEE Advanced 2024 (Paper 2)

JEE Advanced Physics Question (2024) — Solution

Question

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 B _0=0.1 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 , charge of the electron =1.6 10^ -19 C .] Which of the following option(s) is(are) correct?

Options

  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 A_ M 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 .

Answer

B. The value of x for an ion with A_ M =144 is 48 ~cm .

Step-by-step solution

aligned & x =2 R \\ & x =2 P qB x = 2 2 mqV qB x = 2 ~B 2 mV q aligned Option A For H ^ + m = 5 3 10^ -27 ~kg x= 2 0.1 2 5 3 10^ -27 192 1.6 10^ -19 =4 ~cm Option B For A _ m =144 x= 2 0.1 2 144 5 3 10^ -27 192 1.6 10^ -19 =48 ~cm Option C for A _m=1 x =4 ~cm \& for A _ m =196 x =56 ~cm . so x _0=4 ~cm \& x _1=56 ~cm x _1- x _0=52 ~cm . Option D Minimum width =R for A _ M =196 aligned & R = P qB = 2 mqV qB \\ & R = 1 ~B 2 mV q \\ & w _ = R = 1 0.1 2 196 5 3 10^ -27 192 1.6 10^ -19 =28 ~cm aligned

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