A deuteron and an alpha-particle having equal kinetic energy enter perpendicular into a magnetic fie

Q: Why do we use kinetic energy instead of velocity in the radius formula?

The problem explicitly states that both particles have the *same kinetic energy*. If we used the velocity formula, we would have two unknown variables ($v_d$ and $v_\alpha$) which are not equal. Converting to kinetic energy allows us to cancel out the constant $K$.

Q: What exactly is a deuteron?

A deuteron is the nucleus of deuterium, an isotope of hydrogen. It contains one proton and one neutron. Therefore, its mass is approximately twice that of a proton ($2m_p$), and its charge is just $+e$ (from the single proton).

Q: Why is the mass of an alpha particle $4m_p$?

An alpha particle is a helium nucleus, consisting of two protons and two neutrons. Since protons and neutrons have roughly the same mass, the total mass is about $4m_p$. Its charge comes from the two protons, so it is $+2e$.

Q: Why does the magnetic field not change the kinetic energy of the particles?

The magnetic force is always perpendicular to the velocity of the particle ($\vec{F} = q(\vec{v} \times \vec{B})$). Because the force is perpendicular to the displacement, the work done by the magnetic field is zero. Hence, the kinetic energy and speed remain constant.

Q: What if the particles had the same momentum instead of kinetic energy?

If they had the same momentum $p$, the radius formula would be $r = \frac{p}{qB}$. In that case, the radius would only be inversely proportional to the charge $q$. The ratio $r_d / r_\alpha$ would simply be $q_\alpha / q_d = 2e / e = 2$.

Question

A deuteron and an $\alpha$-particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_d$ and $r_\alpha$ be their respective radii of circular path. The value of $\frac{r_d}{r_\alpha}$ is equal to

Options:

(A) $\frac{1}{\sqrt{2}}$

(B) $\sqrt{2}$

(C) $1$

(D) $2$

Sigma Percentile · Accepted Answer

### The Magnetic Dance

Imagine a charged particle entering a uniform magnetic field. If it enters perpendicularly, the magnetic field exerts a Lorentz force that acts exactly like a centripetal force.

This force forces the particle into a perfect circular path. The equation governing this dance is:

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

Rearranging this, we find the radius of the path:

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

### The Kinetic Energy Twist

Usually, we compare particles based on their velocities. But here, the problem throws a curveball: both the deuteron and the $\alpha$-particle have the **same kinetic energy** ($K$).

We need to express the momentum ($p = mv$) in terms of kinetic energy. We know that:

$$K = \frac{p^2}{2m}$$

Solving for momentum, we get:

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

Now, we substitute this back into our radius formula:

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

### Meet the Contenders

Let's look at the two particles entering the arena.

First, the **deuteron**. It is the nucleus of deuterium, containing one proton and one neutron. 
Mass of deuteron, $m_d = 2m_p$
Charge of deuteron, $q_d = e$

Next, the **$\alpha$-particle**. It is a helium nucleus, containing two protons and two neutrons.
Mass of $\alpha$-particle, $m_\alpha = 4m_p$
Charge of $\alpha$-particle, $q_\alpha = 2e$

### The Final Showdown

Since both particles have the same kinetic energy ($K$) and enter the same magnetic field ($B$), these terms are constant. The radius is only proportional to the mass and charge:

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

Let's find the ratio of their radii, $\frac{r_d}{r_\alpha}$:

$$\frac{r_d}{r_\alpha} = \frac{\sqrt{m_d}}{q_d} 	imes \frac{q_\alpha}{\sqrt{m_\alpha}}$$

Substitute the values we established:

$$\frac{r_d}{r_\alpha} = \frac{\sqrt{2m_p}}{e} 	imes \frac{2e}{\sqrt{4m_p}}$$

Simplify the expression:

$$\frac{r_d}{r_\alpha} = \sqrt{\frac{2}{4}} 	imes 2$$

$$\frac{r_d}{r_\alpha} = \frac{1}{\sqrt{2}} 	imes 2 = \sqrt{2}$$

The deuteron takes a wider turn! Its radius is $\sqrt{2}$ times larger than that of the $\alpha$-particle.

Animated Solution for Physics - Magnetic Effects of Current: A deuteron and an $α$ -particle having equal kinetic energy enter perpendicular into a magnetic field. Let $r_{d}$ and $r_{α}$ be their respective radii of circular path. The value of $\frac{r _{d}}{r _{α}}$ is equal to

Select Answer:

Visualized Solution

Visualizing the Paths

Radius of Circular Path

Momentum and Kinetic Energy

Radius in terms of Kinetic Energy

Proportionality Relation

Ratio of Radii

Substituting Particle Properties

Final Calculation

The Sigma Insight: Motion of a Charge in Magnetic Fields

The Magnetic Dance

The Kinetic Energy Twist

Meet the Contenders

The Final Showdown

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