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Magnetism question

2024 · Shift 2 · Q41
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Magnetism question

2024 · Shift 2 · Q41

JEE AdvancedPhysicsMagnetismMultiple correct+4 / −2
A positive, singly ionized atom of mass number AMA_{\mathrm{M}}AM​ is accelerated from rest by the voltage 192 V192 \mathrm{~V}192 V. Thereafter, it enters a rectangular region of width www with magnetic field B⃗0=0.1k^\vec{B}_0=0.1 \hat{k}B0​=0.1k^ Tesla, as shown in the figure. The ion finally hits a detector at the distance xxx below its starting trajectory. [Given: Mass of neutron/proton =(5/3)×10−27 kg=(5 / 3) \times 10^{-27} \mathrm{~kg}=(5/3)×10−27 kg, charge of the electron =1.6×10−19C=1.6 \times 10^{-19} \mathrm{C}=1.6×10−19C.] JEE Advanced 2024 Paper 2 Online Physics - Magnetism Question 2 English Which of the following option(s) is(are) correct?
  1. A
    The value of xxx for H+H^{+}H+ ion is 4 cm4 \mathrm{~cm}4 cm.
  2. B
    The value of xxx for an ion with AM=144A_{\mathrm{M}}=144AM​=144 is 48 cm48 \mathrm{~cm}48 cm.
  3. C
    For detecting ions with 1≤AM≤1961 \leq A_{\mathrm{M}} \leq 1961≤AM​≤196, the minimum height (x1−x0)\left(x_1-x_0\right)(x1​−x0​) of the detector is 55 cm55 \mathrm{~cm}55 cm.
  4. D
    The minimum width www of the region of the magnetic field for detecting ions with AM=196A_{\mathrm{M}}=196AM​=196 is 56 cm56 \mathrm{~cm}56 cm.
View written solutionFree

Correct answer: A, B

  1. Speed of the ion after acceleration

A singly ionized positive atom has charge q=e=1.6×10−19 C.q=e=1.6\times 10^{-19}\,\text{C}.q=e=1.6×10−19C.

It is accelerated from rest through potential difference V=192 VV=192\,\text{V}V=192V, so qV=12mv2.qV=\frac12 mv^2.qV=21​mv2.

Hence, v=2qVm.v=\sqrt{\frac{2qV}{m}}.v=m2qV​​.

The mass of the ion is approximately m=AM(53×10−27) kg.m=A_M\left(\frac53\times 10^{-27}\right)\,\text{kg}.m=AM​(35​×10−27)kg.

So, v=2(1.6×10−19)(192)AM(5/3×10−27).v=\sqrt{\frac{2(1.6\times 10^{-19})(192)}{A_M(5/3\times 10^{-27})}}.v=AM​(5/3×10−27)2(1.6×10−19)(192)​​.

Now, 2(1.6×10−19)(192)=6.144×10−17.2(1.6\times 10^{-19})(192)=6.144\times 10^{-17}.2(1.6×10−19)(192)=6.144×10−17.

Thus,

\sqrt{\frac{3.6864\times 10^{10}}{A_M}}.$$ 2. **Radius of circular path in magnetic field** Inside the magnetic field, the ion moves in a circular arc of radius $$r=\frac{mv}{qB}.$$ Using $v=\sqrt{2qV/m}$, $$r=\sqrt{\frac{2mV}{qB^2}}.$$ Substitute $m=A_M\left(\frac53\times 10^{-27}\right)$, $V=192$, $q=1.6\times 10^{-19}$, $B=0.1$ T: $$r=\sqrt{\frac{2\cdot A_M\cdot (5/3\times 10^{-27})\cdot 192}{(1.6\times 10^{-19})(0.1)^2}}.$$ Compute the constant: $$2\cdot \frac53\cdot 192=640.$$ So, $$r=\sqrt{\frac{640\times 10^{-27}}{1.6\times 10^{-21}}A_M} =\sqrt{4\times 10^{-4}A_M}.$$ Therefore, $$r=2\times 10^{-2}\sqrt{A_M}\,\text{m}=2\sqrt{A_M}\,\text{cm}.$$ 3. **Deflection below original trajectory** From the figure/setup, the ion enters the magnetic region tangentially and exits after tracing a semicircle before hitting the detector below. Therefore the vertical displacement is the diameter: $$x=2r=4\sqrt{A_M}\,\text{cm}.$$ --- 4. **Check each option** ### Option A: For $H^+$, $A_M=1$ $$x=4\sqrt{1}=4\,\text{cm}.$$ So **A is correct**. ### Option B: For $A_M=144$ $$x=4\sqrt{144}=4\times 12=48\,\text{cm}.$$ So **B is correct**. ### Option C: For $1\le A_M\le 196$ Minimum deflection: $$x_0=4\sqrt{1}=4\,\text{cm}.$$ Maximum deflection: $$x_1=4\sqrt{196}=4\times 14=56\,\text{cm}.$$ Thus required detector height is $$(x_1-x_0)=56-4=52\,\text{cm}.$$ Not $55\,\text{cm}$. So **C is incorrect**. ### Option D: Minimum width of magnetic field region for $A_M=196$ For semicircular motion, the rectangular magnetic field region must at least contain the radius horizontally up to the turning point and in fact the width required is $$w_{\min}=r=2\sqrt{196}=28\,\text{cm},$$ not $56\,\text{cm}$. So **D is incorrect**. --- 5. **Final answer** The correct options are: $$\boxed{A,\ B}$$
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