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.

Correct answer: A. The value of x for H^ + ion is 4 ~cm . aligned & x=2 R & x=2 P qB x= 2 2 mqV qB x= 2 ~B 2 mV q aligned Option A For H^ + arrow 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 =144x=(2)/(0.1) 2 × 144 × (5)/(3) × 10^ -27 × 192 1.6 × 10^ -19 =48 ~cm Option C for A_m=1x=4 ~cm \& for A_ m =196x=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

JEE Advanced 2024Paper 2

Exam: JEE Advanced
Year: 2024
Paper: Paper 2
Subject: Physics
Chapter: Magnetic Effects of Current

JEE Advanced 2024 · Paper 2 · Physics · Magnetic Effects of Current

A positive, singly ionized atom of mass number A_ M is accelerated...

A positive, singly ionized atom of mass number

A_{\mathrm{M}}

is accelerated from rest by the voltage

192 \mathrm{~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} \mathrm{~kg}

, charge of the electron

=1.6 \times 10^{-19} \mathrm{C}

A positive, singly ionized atom of mass number A_ M is accelerated from rest by the voltage 192 ~V . Thereafter, it

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

Options

AThe value of $x$ for $H^{+}$ ion is $4 \mathrm{~cm}$ .
BThe value of $x$ for an ion with $A_{\mathrm{M}}=144$ is $48 \mathrm{~cm}$ .
CFor detecting ions with $1 \leq A_{\mathrm{M}} \leq 196$ , the minimum height $\left(x_1-x_0\right)$ of the detector is $55 \mathrm{~cm}$ .
DThe minimum width $w$ of the region of the magnetic field for detecting ions with $A_{\mathrm{M}}=196$ is $56 \mathrm{~cm}$ .

Correct answer

A. The value of $x$ for $H^{+}$ ion is $4 \mathrm{~cm}$ .

Step-by-step solution

\begin{aligned}& \mathrm{x}=2 \mathrm{R}\\ & \Rightarrow \mathrm{x}=2 \frac{\mathrm{P}}{\mathrm{qB}} \Rightarrow \mathrm{x}=\frac{2 \sqrt{2 \mathrm{mqV}}}{\mathrm{qB}} \Rightarrow \mathrm{x}=\frac{2}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{q}}} \end{aligned}

Option A For

\mathrm{H}^{+} \rightarrow \mathrm{m}=\frac{5}{3} \times 10^{-27} \mathrm{~kg}\therefore x=\frac{2}{0.1} \sqrt{\frac{2 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=4 \mathrm{~cm}

Option B For

\mathrm{A}_{\mathrm{m}}=144x=\frac{2}{0.1} \sqrt{\frac{2 \times 144 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=48 \mathrm{~cm}

Option C for

\mathrm{A}_m=1\mathrm{x}=4 \mathrm{~cm} \&

for

\mathrm{A}_{\mathrm{m}}=196\mathrm{x}=56 \mathrm{~cm}

. so

\mathrm{x}_0=4 \mathrm{~cm} \& \mathrm{x}_1=56 \mathrm{~cm}\therefore \mathrm{x}_1-\mathrm{x}_0=52 \mathrm{~cm}

. Option D Minimum width

=R

for

\mathrm{A}_{\mathrm{M}}=196

\begin{aligned} & \mathrm{R}=\frac{\mathrm{P}}{\mathrm{qB}}=\frac{\sqrt{2 \mathrm{mqV}}}{\mathrm{qB}} \\ & \mathrm{R}=\frac{1}{\mathrm{~B}} \sqrt{\frac{2 \mathrm{mV}}{\mathrm{q}}} \\ & \mathrm{w}_{\min }=\mathrm{R}=\frac{1}{0.1} \sqrt{\frac{2 \times 196 \times \frac{5}{3} \times 10^{-27} \times 192}{1.6 \times 10^{-19}}}=28 \mathrm{~cm} \end{aligned}

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