A positive, singly ionized atom of mass number A_M is accelerated from rest by the voltage 192V. The

Q: Why does the ion move in a semi-circle?

The magnetic force is always perpendicular to the velocity, providing a centripetal force. Since it enters perpendicular to the boundary, it completes exactly half a circle before hitting the same boundary again.

Q: Why is the distance $x$ equal to $2R$?

The ion enters at $x=0$ and curves back to the same vertical plane. The distance between the entry and exit points of a semi-circle is its diameter, which is $2R$.

Q: How did we get the mass of the ion?

The mass number $A_M$ represents the total number of nucleons (protons and neutrons). We multiply $A_M$ by the given mass of a single nucleon to get the total mass.

Q: Why is the minimum width $w$ equal to $R_{max}$?

If the width $w$ is less than the radius $R$, the ion will exit the magnetic field from the opposite side before completing its semi-circle. To detect all ions, $w$ must be at least the largest radius.

Q: Why is the detector height $x_{max} - x_{min}$?

The detector needs to cover the entire range of impact points. The lightest ion hits at $x_{min}$ and the heaviest at $x_{max}$. The distance between these two extreme points is the required height.

Question

A positive, singly ionized atom of mass number $A_M$ is accelerated from rest by the voltage $192	ext{V}$. Thereafter, it enters a rectangular region of width $w$ with magnetic field $\vec{B}_0 = 0.1\hat{k}	ext{ Tesla}$, as shown in the figure. The ion finally hits a detector at the distance $x$ below its starting trajectory.
[Given: Mass of neutron/proton = $(5/3) 	imes 10^{-27}	ext{kg}$, charge of the electron = $1.6 	imes 10^{-19}	ext{C}$.]
Which of the following option(s) is(are) correct?

Options:

(A) The value of $x$ for $	ext{H}^+$ ion is $4	ext{ cm}$.

(B) The value of $x$ for an ion with $A_M = 144$ is $48	ext{cm}$.

(C) For detecting ions with $1 \le A_M \le 196$, the minimum height $(x_1 - x_0)$ of the detector is $55	ext{cm}$.

(D) The minimum width $w$ of the region of the magnetic field for detecting ions with $A_M = 196$ is $56	ext{cm}$.

Sigma Percentile · Accepted Answer

### The Setup: Accelerating into the Unknown
Imagine you are standing at the starting line of a particle accelerator. Our protagonist is a positive, singly ionized atom. It starts from rest and is pushed forward by a voltage of $192	ext{V}$.

What happens here? The electrical potential energy is completely converted into kinetic energy. We can write this as:

$$K = qV \implies \frac{1}{2}mv^2 = qV$$

From this, we can easily find the velocity $v$ of the ion just as it enters the magnetic field:

$$v = \sqrt{\frac{2qV}{m}}$$

### The Magnetic Dance: Finding the Radius
As soon as the ion enters the rectangular region, it encounters a uniform magnetic field $\vec{B}_0$ pointing straight out of the page. Because the ion is moving perpendicular to this field, it experiences a magnetic force that acts as a centripetal force, forcing it into a circular path.

The radius $R$ of this circular path is a classic result in electromagnetism:

$$R = \frac{mv}{qB_0}$$

Let's substitute the velocity $v$ we found earlier into this equation. This gives us a direct formula for the radius in terms of the accelerating voltage:

$$R = \frac{m}{qB_0}\sqrt{\frac{2qV}{m}} = \frac{1}{B_0}\sqrt{\frac{2mV}{q}}$$

### Crunching the Numbers: The Master Formula
Now, let's look at the specific properties of our ion. It is singly ionized, meaning it has lost exactly one electron. Therefore, its charge $q$ is just the elementary charge, $1.6 	imes 10^{-19}	ext{ C}$.

Its mass $m$ depends on its mass number $A_M$. We are given the mass of a single nucleon (proton or neutron) as $\frac{5}{3} 	imes 10^{-27}	ext{ kg}$. So, the total mass is:

$$m = A_M 	imes \left(\frac{5}{3} 	imes 10^{-27}ight)	ext{ kg}$$

Let's plug all these numbers, along with $V = 192	ext{V}$ and $B_0 = 0.1	ext{ T}$, into our radius formula. I know this looks like a terrifying calculation, but let's take a breath and simplify it step by step:

$$R = \frac{1}{0.1}\sqrt{\frac{2 \cdot A_M \left(\frac{5}{3} 	imes 10^{-27}ight) \cdot 192}{1.6 	imes 10^{-19}}}$$

After carefully canceling out the powers of ten and simplifying the constants, we arrive at a beautifully elegant relation:

$$R = 0.02\sqrt{A_M}	ext{ meters} = 2\sqrt{A_M}	ext{ cm}$$

Because the ion enters perpendicular to the boundary and the magnetic field is uniform, it will complete exactly a semi-circle before hitting the detector on the same boundary. The distance $x$ from the entry point is simply the diameter of this semi-circle:

$$x = 2R = 4\sqrt{A_M}	ext{ cm}$$

### The Final Verdict: Testing the Options
Now we have our master key: $x = 4\sqrt{A_M}$. Let's unlock the options one by one.

**Option (A):** For a Hydrogen ion ($	ext{H}^+$), the mass number $A_M = 1$. 
$$x = 4\sqrt{1} = 4	ext{ cm}$$
This matches perfectly. **Option (A) is correct.**

**Option (B):** For an ion with $A_M = 144$, we substitute it into our formula:
$$x = 4\sqrt{144} = 4 	imes 12 = 48	ext{ cm}$$
This also matches perfectly. **Option (B) is correct.**

**Option (C):** We need to find the minimum height of the detector to catch all ions from $A_M = 1$ to $196$. 
The lightest ion hits at $x_{\min} = 4\sqrt{1} = 4	ext{ cm}$.
The heaviest ion hits at $x_{\max} = 4\sqrt{196} = 4 	imes 14 = 56	ext{ cm}$.
The detector must span this entire range, so its minimum height is:
$$	ext{Height} = x_{\max} - x_{\min} = 56 - 4 = 52	ext{ cm}$$
The option claims $55	ext{ cm}$, so **Option (C) is incorrect.**

**Option (D):** What is the minimum width $w$ of the magnetic field region? To ensure that even the largest ion completes its semi-circle without exiting the right side, the width $w$ must be at least equal to the maximum radius $R_{\max}$.
$$w_{\min} = R_{\max} = \frac{x_{\max}}{2} = \frac{56}{2} = 28	ext{ cm}$$
The option claims $56	ext{ cm}$, so **Option (D) is incorrect.**

Select Answer:

Visualized Solution

$A na l y z in g t h e S e t u p$

$K in e t i c E n er g y f r o m A cce l er a t in g V o l t a g e$

$R a d i u s o f C i r c u l a r P a t h$

$S u b s t i t u t in g G i v e n V a l u es$

$C a l c u l a t in g R a d i u s R$

$D i s t an ce o f I m p a c t x$

$C h ec k in g O pt i o n s (A) an d (B)$

$C h ec k in g O pt i o n (C)$

$C h ec k in g O pt i o n (D)$

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Setup

The Magnetic Dance

Crunching the Numbers

The Final Verdict

Similar Questions

In an experiment, electrons are accelerated, from rest by applying a voltage of $500 V$ . Calculate the radius of the path, if a magnetic field $100 mT$ is then applied. (Take, charge of the electron $= 1.6 \times 1 0^{- 19} C$ and mass of the electron $= 9.1 \times 1 0^{- 31} kg$ )

The region between $y = 0$ and $y = d$ contains a magnetic field $B = B \hat{k}$ . A particle of mass $m$ and charge $q$ enters the region with a velocity $v = v \hat{i}$ . If $d = \frac{m v}{2 q B}$ , then the acceleration of the charged particle at the point of its emergence at the other side is

A particle of mass $m$ and charge $q$ , moving with velocity $v$ enters Region II normal to the boundary as shown in the figure. Region II has a uniform magnetic field $B$ perpendicular to the plane of the paper. The length of the Region II is $l$ . Choose the correct choice (s).

A particle of charge $q$ and mass $m$ is moving with a velocity $- v \hat{i} (v \neq = 0)$ towards a large screen placed in the $y z$ -plane at a distance $d$ . If there is a magnetic field $B = B_{0} \hat{k}$ , the minimum value of $v$ for which the particle will not hit the screen is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kinetic Energy from Accelerating Voltage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Radius of Circular Path

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Given Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Radius R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance of Impact x

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Options (A) and (B)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (C)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Checking Option (D)