JEE Advanced 2024 · Paper 2 · Physics Q7

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PhysicsMultiple correct+4 / -2Question 24 of 51

A positive, singly ionized atom of mass number $A_M$ 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 $= (5/3)\times10^{-27}\,\mathrm{kg}$ , charge of the electron $= 1.6\times10^{-19}\,\mathrm{C}$ .]

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

Axes: x points down, y to the right, z out of the page. An ion moves in the +y direction along a horizontal line and enters, at 90° to its boundary, a vertical rectangular region of width w on the right (dotted shading) containing magnetic field B₀ pointing out of the page. The ion turns back and exits the field region lower down, moving in −y, and hits a vertical detector plate placed just left of the field boundary. Distances measured downward from the incoming trajectory: x₀ to the top of the detector, x to the point where the ion hits, x₁ to the bottom of the detector.

The value of

x

for

H^+

ion is

4\,\mathrm{cm}

The value of

x

for an ion with

A_M = 144

48\,\mathrm{cm}

For detecting ions with

1 \le A_M \le 196

, the minimum height

(x_1 - x_0)

of the detector is

55\,\mathrm{cm}

The minimum width

w

of the region of the magnetic field for detecting ions with

A_M = 196

56\,\mathrm{cm}

Answer and worked solution

Official answer: A, B

Radius. After acceleration through $V$ , $\frac12mv^2=eV$ . In the field the ion moves in a circle of radius
$R=\frac{mv}{eB}=\frac1B\sqrt{\frac{2mV}{e}}.$
It turns through a semicircle, so it lands a distance $x=2R$ below its entry line.

Hydrogen. For $H^+$ , $m=\frac53\times10^{-27}$ kg:
$\frac{2mV}{e}=\frac{2\cdot\frac53\times10^{-27}\cdot192}{1.6\times10^{-19}}=4\times10^{-6},$
so $R=\frac{2\times10^{-3}}{0.1}=2$ cm. In general $R=2\sqrt{A_M}$ cm.

(A) $x=2R=4$ cm. ✓
(B) $A_M=144$ : $R=24$ cm, so $x=48$ cm. ✓
(C) $A_M=196$ : $x=56$ cm. $A_M=1$ : $x=4$ cm. The detector must span $52$ cm, not $55$ . ✗
(D) The semicircle reaches a depth $R$ into the field, so for $A_M=196$ the minimum width is $28$ cm, not $56$ . ✗

Topics:Motion in magnetic and electric fields Force on a moving charge

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