Two ions of masses 4 amu and 16 amu have charges +2e and +3e, respectively. These ions pass through

Q: Why does the magnetic field cause a circular path?

The magnetic force is always perpendicular to the velocity vector ($\vec{F} = q(\vec{v} \times \vec{B})$). A force that is always perpendicular to motion acts as a centripetal force, changing the direction but not the speed, resulting in circular motion.

Q: How did we get $v = \sqrt{\frac{2K}{m}}$?

Kinetic energy is defined as $K = \frac{1}{2}mv^2$. Multiplying both sides by 2 gives $2K = mv^2$. Dividing by $m$ gives $v^2 = \frac{2K}{m}$, and taking the square root yields the result.

Q: Why is $\sin\theta = \frac{d}{r}$?

If you draw a line from the center of the circular path to the point where the particle enters and another to where it exits the field of width $d$, the angle between these two radii is $\theta$. Dropping a perpendicular from the exit point to the initial radius forms a right triangle where the opposite side is $d$ and the hypotenuse is $r$.

Q: What if the magnetic field region was infinitely wide?

If the region was infinitely wide, the particles would complete a full semicircle and exit from the same side they entered, effectively being deflected by $180^\circ$, assuming they don't hit any other boundaries.

Q: Does the charge sign matter for the deflection magnitude?

The sign of the charge determines the *direction* of the deflection (up vs. down, or left vs. right) according to the right-hand rule, but the *magnitude* of the deflection angle depends only on the magnitude of the charge, mass, and kinetic energy.

Question

Two ions of masses 4 amu and 16 amu have charges +2e and +3e, respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then,

Options:

(A) lighter ion will be deflected less than heavier ion

(B) lighter ion will be deflected more than heavier ion

(C) both ions will be deflected equally

(D) no ion will be deflected

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine two ions, one light and one heavy, entering a region with a uniform magnetic field pointing into the screen.

As they enter, the magnetic force acts perpendicular to their velocity. This perpendicular force acts as a centripetal force, bending their paths into circular arcs.

To understand how much they bend, we need to determine the radius of this circular path.

### The Master Equation

The formula for the radius of a charged particle moving perpendicularly through a magnetic field is:
$$r = \frac{mv}{qB}$$

However, the problem gives us their **kinetic energies**, not their velocities. We know that kinetic energy is related to velocity by:
$$K = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2K}{m}}$$

Substituting this expression for velocity into our radius formula, we get:
$$r = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB}$$

### Comparing the Radii

Look closely at this equation. The problem states that both ions have the exact same kinetic energy $K$, and they are in the same magnetic field $B$.

So, the radius $r$ is directly proportional to the square root of mass $m$, and inversely proportional to the charge $q$:
$$r \propto \frac{\sqrt{m}}{q}$$

Let's set up the ratio of their radii, $\frac{r_1}{r_2}$. Plugging in our proportionalities, this equals:
$$\frac{r_1}{r_2} = \sqrt{\frac{m_1}{m_2}} 	imes \frac{q_2}{q_1}$$

Now, we substitute the given values. The masses are $4	ext{ amu}$ and $16	ext{ amu}$, and the charges are $+2e$ and $+3e$:
$$\frac{r_1}{r_2} = \sqrt{\frac{4}{16}} 	imes \frac{3e}{2e}$$

Calculating this out:
$$\frac{r_1}{r_2} = \frac{1}{2} 	imes \frac{3}{2} = \frac{3}{4}$$

This means $r_1$ is less than $r_2$. The lighter ion traces a tighter circle!

### The Deflection Angle

Now, how does the radius relate to the deflection angle $	heta$?

Look at the geometry of the path. If the magnetic field region has a width $d$, the sine of the deflection angle $	heta$ is exactly equal to $d$ divided by the radius $r$:
$$\sin	heta = \frac{d}{r}$$

Since the width $d$ is constant for both ions, $\sin	heta$ is inversely proportional to the radius $r$.

A smaller radius means a larger $\sin	heta$, which in turn means a larger deflection angle.

### Final Conclusion

Because the lighter ion has a smaller radius ($r_1 < r_2$), it will experience a larger deflection ($	heta_1 > 	heta_2$).

**Therefore, the lighter ion will be deflected more than the heavier ion.**

Animated Solution for Physics - Magnetic Effects of Current: Two ions of masses 4 amu and 16 amu have charges +2e and +3e, respectively. These ions pass through the region of constant perpendicular magnetic field. The kinetic energy of both ions is same. Then,

Select Answer:

Visualized Solution

Visualizing the Setup

Radius of Curvature $r$

Proportionality Relation

Substituting Values

Calculating Ratio $\frac{r _{1}}{r _{2}}$

Deflection Angle $θ$

Comparing Deflections

The Way Forward

The Sigma Insight: Motion of a Charge in Magnetic Fields

Analyzing the Setup

The Master Equation

Comparing the Radii

The Deflection Angle

Final Conclusion

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