An alpha-particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest.

Q: Why do we use $K.E. = qV$?

When a charged particle moves through a potential difference $V$, the electric field does work on it. By the work-energy theorem, this work ($W = qV$) is converted entirely into the particle's kinetic energy, assuming it starts from rest.

Q: Why does the particle move in a circular path in the magnetic field?

The magnetic force $\vec{F} = q(\vec{v} \times \vec{B})$ is always perpendicular to the velocity vector. A force that is always perpendicular to motion does no work and only changes the direction of the velocity, acting as a centripetal force, which results in uniform circular motion.

Q: How do we know the charge of the alpha particle and the sulfur ion?

An alpha particle is a helium nucleus, which consists of 2 protons and 2 neutrons, so its charge is $+2e$. The problem states the sulfur ion is 'singly charged', meaning it has lost or gained one electron, giving it a charge magnitude of $1e$.

Q: Why did we substitute momentum $p$ instead of velocity $v$ in the radius formula?

While $r = \frac{mv}{qB}$ is correct, we don't directly know the velocity. We know the kinetic energy ($qV$). It's mathematically cleaner to relate kinetic energy to momentum ($p = \sqrt{2mK}$) and use $r = \frac{p}{qB}$, rather than finding $v = \sqrt{\frac{2qV}{m}}$ and substituting it back, though both methods yield the exact same result.

Q: What if the particles were accelerated to the same momentum instead of the same potential?

If they had the same momentum $p$, the radius formula $r = \frac{p}{qB}$ shows that the radius would only be inversely proportional to their charge $q$. The mass would not affect the radius in that specific scenario!

Question

An $\alpha$-particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential V and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_\alpha$ and $r_s$, respectively. The ratio ($r_s/r_\alpha$) is______.

Sigma Percentile · Accepted Answer

Welcome to a beautiful intersection of electrostatics and magnetism! This problem is a classic JEE Advanced masterpiece because it doesn't just test your memory of formulas; it tests your ability to track a particle's journey across two completely different physical regimes.

### Phase 1: The Electric Accelerator

Imagine our two contenders: a hefty sulfur ion and a compact alpha particle. They are both sitting at rest, minding their own business, until we flip a switch and subject them to a massive electrical potential difference, $V$.

What does this potential do? It acts like a slingshot. The electric field does work on the charges, converting electrical potential energy into pure kinetic energy. By the work-energy theorem, the kinetic energy gained is simply the charge multiplied by the potential:

$$K.E. = qV$$

But kinetic energy isn't the most useful quantity when we are about to enter a magnetic field. We need momentum. Recall the beautiful relationship between momentum $p$ and kinetic energy $K.E.$:

$$p = \sqrt{2m(K.E.)}$$

Substituting our kinetic energy, we get the momentum of the particles right as they exit the accelerator:

$$p = \sqrt{2mqV}$$

Notice how the momentum depends on both the mass and the charge. The sulfur ion is much heavier, but the alpha particle has double the charge. It's a tug-of-war of variables!

### Phase 2: The Magnetic Deflector

Now, our particles fly into a region with a uniform magnetic field $\vec{B}$ pointing directly into the page. The moment they enter, they experience the Lorentz force. Because their velocity is perfectly perpendicular to the magnetic field, this force acts purely as a centripetal force, bending their straight-line paths into perfect circles.

Equating the magnetic force to the centripetal force:

$$qvB = \frac{mv^2}{r}$$

Solving for the radius $r$, we get the famous cyclotron radius formula:

$$r = \frac{mv}{qB} = \frac{p}{qB}$$

### Phase 3: The Master Equation

This is where the magic happens. We have the radius in terms of momentum, and we have the momentum in terms of the accelerating potential. Let's fuse them together:

$$r = \frac{\sqrt{2mqV}}{qB}$$

To make this elegant, let's bring that $q$ from the denominator inside the square root. It becomes $q^2$, and one $q$ cancels out:

$$r = \frac{1}{B}\sqrt{\frac{2mV}{q}}$$

Take a deep breath and look at this master equation. The problem states that both particles are accelerated through the **same** potential $V$ and enter the **same** magnetic field $B$. Therefore, $2$, $V$, and $B$ are identical for both particles. They are constants in our comparison.

This leaves us with a incredibly clean proportionality:

$$r \propto \sqrt{\frac{m}{q}}$$

### Phase 4: The Final Showdown

We are asked to find the ratio of the radius of the sulfur ion to the radius of the alpha particle, $\frac{r_s}{r_\alpha}$.

Using our proportionality, we can set up the ratio:

$$\frac{r_s}{r_\alpha} = \sqrt{\frac{m_s}{q_s} 	imes \frac{q_\alpha}{m_\alpha}}$$

Now, let's introduce the stats of our contenders:

- **Sulfur Ion ($S^+$):** Mass $m_s = 32 	ext{ amu}$, Charge $q_s = 1e$ (since it's singly charged).
- **Alpha Particle ($\alpha$):** Mass $m_\alpha = 4 	ext{ amu}$, Charge $q_\alpha = 2e$ (it's a helium nucleus).

Let's plug these numbers into our arena:

$$\frac{r_s}{r_\alpha} = \sqrt{\frac{32}{1} 	imes \frac{2}{4}}$$

Simplify the fraction inside the square root:

$$\frac{r_s}{r_\alpha} = \sqrt{\frac{64}{4}} = \sqrt{16}$$

And the square root of 16 is a perfect, satisfying integer:

**$$\frac{r_s}{r_\alpha} = 4$$**

The sulfur ion, being massive and only singly charged, is much harder for the magnetic field to bend, resulting in a circular path that is exactly four times larger than that of the highly charged, lighter alpha particle. Physics is beautiful when it all cancels out!

Enter Numerical Value:

Visualized Solution

Initial Setup: $v ⊥ B$

Kinetic Energy & Momentum

Radius of Circular Path

Radius in terms of V

Proportionality Relation

Setting up the Ratio

Substituting Values

Final Calculation

Conclusion & Variations

The Sigma Insight: Motion of a Charge in Magnetic Fields

Phase 1

Phase 2

Phase 3

Phase 4

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