JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Magnetic Effects of Current

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 \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 1 0^{- 27} kg

, charge of the electron

= 1.6 \times 1 0^{- 19} C

.]

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

(A)The value of x for $H^{+}$ ion is 4 cm .✓ Correct
(B)The value of x for an ion with $A_{M} = 144$ is 48 cm.✓ Correct
(C)For detecting ions with $1 \leq A_{M} \leq 196$ , the minimum height $(x_{1} - x_{0})$ of the detector is 55 cm.
(D)The minimum width $w$ of the region of the magnetic field for detecting ions with $A_{M} = 196$ is 56 cm.

Answer: (A), (B)

Solution

After acceleration through voltage

V

, the ion's speed is

v = \frac{2 q V}{m}

.

In the magnetic field, the radius of circular path is:

R = \frac{m v}{q B} = \frac{2 mV / q}{B} = \frac{1}{B} \frac{2 mV}{q}

The vertical displacement is

x = 2 R = \frac{2}{B} \frac{2 mV}{q}

.

(A) For

H^{+}

m = \frac{5}{3} \times 1 0^{- 27}

x = \frac{2}{0.1} \frac{2 \times \frac{5}{3} \times 1 0 ^{- 27} \times 192}{1.6 \times 1 0 ^{- 19}} = 4 cm

(B) For

A_{M} = 144

m = 144 \times \frac{5}{3} \times 1 0^{- 27}

x = 4144 = 48 cm

(C) For

A_{M} = 1

x_{0} = 4

cm; for

A_{M} = 196

x_{1} = 56

x_{1} - x_{0} = 52 cm \neq = 55 cm

(D) Minimum width

w = R

for

A_{M} = 196

w = \frac{x}{2} = 28 cm \neq = 56 cm

Options (A) and (B) are correct.

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · One or more correct options