How to Decide Whether a Magnetic Trajectory Is Circular or Helical
Learn how to distinguish circular from helical particle motion by resolving velocity into parallel and perpendicular components. Includes worked calculations, magnetic force direction, charge identification, and challenging HSC Physics practice questions.
A charged particle enters a uniform magnetic field at an angle. Its path curves, but does it travel in a circle or a helix?
The answer depends on one thing: whether the particle has a velocity component parallel to the magnetic field.
- If its velocity is entirely perpendicular to the field, it follows a circle.
- If its velocity has both perpendicular and parallel components, it follows a helix.
- If its velocity is entirely parallel to the field, it travels in a straight line.
This distinction also gives you a reliable way to interpret particle tracks and determine whether a particle is positively or negatively charged.
01Start by splitting the velocity
Imagine a charged particle moving through a uniform magnetic field. The magnetic force acts perpendicular to both the particle’s velocity and the magnetic field.
However, the force does not act equally on every part of the particle’s velocity.
Resolve the velocity into two components:
- \(v_\perp\): the component perpendicular to the magnetic field.
- \(v_\parallel\): the component parallel to the magnetic field.
The perpendicular component produces circular motion. The parallel component remains unchanged and carries the particle along the magnetic field.
Think of a spring. Moving around its circular cross-section corresponds to the perpendicular velocity, while progressing along its length corresponds to the parallel velocity.
The analogy helps you picture a helix, although the particle is not physically following a spring. Its sideways acceleration comes from the magnetic force.
Why only one component produces magnetic force
The magnetic force on a moving charge is described by
\[ F_B=|q|vB\sin\theta \]
where:
- \(F_B\) is the magnetic force in newtons (N),
- \(q\) is the particle’s charge in coulombs (C),
- \(v\) is its speed in metres per second,
- \(B\) is the magnetic field strength in tesla (T),
- \(\theta\) is the angle between velocity and magnetic field.
Since
\[ v_\perp=v\sin\theta \]
we can rewrite the force equation as
\[ F_B=|q|v_\perp B \]
Notice that \(v_\parallel\) does not appear.
This is because a velocity component parallel to the magnetic field produces no magnetic force. Only the perpendicular component is affected.
For the ideal HSC model, assume a uniform, constant magnetic field, negligible forces other than the magnetic force, and non-relativistic particle speeds.
The three possible trajectories
| Velocity condition | Particle trajectory | Explanation |
|---|---|---|
| \(v_\parallel=0\), \(v_\perp>0\) | Circle | All velocity is perpendicular to the field. |
| \(v_\parallel\neq0\), \(v_\perp>0\) | Helix | The particle circles while advancing along the field. |
| \(v_\perp=0\) | Straight line | No magnetic force acts on the particle. |
In terms of the angle:
- \(\theta=90^\circ\): circular motion.
- \(0^\circ<\theta<90^\circ\), or \(90^\circ<\theta<180^\circ\): helical motion.
- \(\theta=0^\circ\) or \(180^\circ\): straight-line motion.
The sign of the parallel component determines whether the particle advances along or against the magnetic field.
The important distinction is that experiencing a magnetic force does not automatically mean a particle follows a circle. It only follows a circle when there is no parallel velocity component.
02How experiments revealed the direction of particle curvature
Before interpreting a particle’s trajectory, consider a problem physicists faced when studying cosmic rays.
They could photograph curved particle tracks, but how could they determine whether a particle was positively or negatively charged?
A curved track alone was not enough. The direction of travel also had to be established.
The cloud chamber experiment
In 1932, physicist Carl Anderson used a cloud chamber surrounded by a large electromagnet to investigate cosmic-ray particles.
The apparatus worked as follows:
- The chamber contained supersaturated water and alcohol vapour.
- A magnetic field was established throughout the chamber using an electromagnet surrounding it.
- A charged particle travelled through the vapour, ionising molecules along its path.
- Tiny droplets condensed around the ions, making the particle’s path visible.
- A camera recorded the resulting track.
The particle itself was not visible. The droplets showed where it had travelled.
The magnetic field caused charged particles to curve. For a particle travelling perpendicular to the field, the radius of curvature was
\[ r=\frac{mv}{|q|B} \]
where \(m\) is mass and \(r\) is the radius of the circular path.
Anderson could use the curvature to investigate the particle’s charge, but there was a problem.
A negatively charged particle travelling in one direction could produce the same curved track as a positively charged particle travelling in the opposite direction.
He needed to establish which way the particle was moving.
The modification: introducing a lead plate
Anderson placed a thin lead plate across the chamber.
The prediction was straightforward: a particle passing through the plate would lose kinetic energy as it interacted with the material. Its momentum and speed would decrease.
Since
\[ r=\frac{mv}{|q|B} \]
a decrease in speed would produce a smaller radius, provided the particle’s direction remained perpendicular to the field.
The observation supported this prediction. The track became more tightly curved after passing through the plate.
This established the direction of travel: the particle had moved from the larger-radius section towards the smaller-radius section.
With the direction known, Anderson could apply the magnetic force rule to establish the charge sign.
His famous 1932 observation identified a positively charged particle with the mass of an electron: the positron. The lead plate was crucial in resolving the ambiguity about the direction of travel. 【turn684321search0】【turn684321search16】
The lesson for interpreting magnetic trajectories is important:
You need to know both the direction of the magnetic field and the direction of particle motion before you can reliably determine the charge sign from curvature.
03How to identify positive and negative charges from curvature
The direction of the magnetic force is given by
\[ \vec F_B=q(\vec v\times\vec B) \]
The cross product determines the force direction for a positive charge. For a negative charge, reverse that direction.
You can use the right-hand rule to determine the force on a positive charge.
A useful reference arrangement
Suppose:
- The particle initially moves to the right.
- The magnetic field points into the page.
- The particle is moving perpendicular to the field.
For a positive charge, the magnetic force points upwards.
For a negative charge, the force points downwards.
This gives you a useful decision rule:
| Initial curvature | Charge sign |
|---|---|
| Upwards | Positive |
| Downwards | Negative |
This table applies only to the stated arrangement: velocity to the right and magnetic field into the page.
If you reverse either the magnetic field or the direction of travel, the conclusions reverse.
What if the particle follows a helix?
The same rule still works.
You simply use the perpendicular velocity component when determining the magnetic force.
The parallel component does not contribute to the cross product, so it does not change the direction of the force.
For a helical trajectory, the magnetic force continuously points towards the centre of the circular motion around the field axis.
One common mistake is to look at the overall direction of the helix and use that as the direction of the magnetic force. Don’t do this. The magnetic force is determined by the instantaneous velocity perpendicular to the field.
04Don’t confuse a helix with a spiral
These two shapes can look similar in a particle detector, but they describe different features of the motion.
An ideal helix has:
- A constant radius.
- A constant axial advance per revolution, called its pitch.
- A fixed axis parallel to the magnetic field.
A spiral has a changing radius.
Why might a real particle produce a spiral-shaped track?
In a cloud or bubble chamber, a particle can lose kinetic energy through interactions with the surrounding material. Its momentum decreases, which reduces its radius of curvature.
A particle may therefore follow helical motion while its radius gradually shrinks.
This is different from an ideal helix with constant velocity components. Real detector tracks can include energy loss and scattering, so the simple constant-radius model becomes an approximation. 【turn375736search3】
A second complication: your viewing angle
A helix is a three-dimensional trajectory. A photograph is usually two-dimensional.
This creates another possible misconception.
Imagine looking directly along the magnetic field axis. The particle is advancing towards or away from you while circling around the axis.
From this viewpoint, the helix can appear to be a perfect circle.
Now imagine viewing the same particle from the side. You can see its progress along the field, and its projected trajectory looks wave-like.
This means a circular-looking photograph does not automatically prove that the particle followed a circular trajectory.
You need to consider the orientation of the magnetic field and the viewing direction.
If a second view or other measurements reveal that the particle advances along the field during each revolution, its actual trajectory is helical.
05The equations for radius and pitch
Once you have established that a particle follows a helix, you can calculate its radius and pitch.
These quantities describe different parts of the motion.
Radius of the circular component
The magnetic force supplies the centripetal force associated with the perpendicular motion:
\[ |q|v_\perp B=\frac{mv_\perp^2}{r} \]
Rearranging gives
\[ \boxed{r=\frac{mv_\perp}{|q|B}} \]
The radius depends on:
- Particle mass.
- Perpendicular velocity.
- Charge magnitude.
- Magnetic field strength.
Notice that the total speed does not appear.
A particle can have a large total speed but a relatively small circular radius if most of its velocity is parallel to the magnetic field.
Period of the circular component
The time for one complete revolution is
\[ T=\frac{2\pi r}{v_\perp} \]
Substituting the radius equation:
\[ \boxed{T=\frac{2\pi m}{|q|B}} \]
For a given particle in a uniform field, the period is independent of its velocity in this non-relativistic model.
Pitch of the helix
The pitch is the distance the particle advances along the magnetic field during one complete revolution.
Since the parallel velocity remains constant,
\[ \boxed{h=|v_\parallel|T} \]
where \(h\) is the pitch in metres.
Combining the equations gives
\[ h=\frac{2\pi m|v_\parallel|}{|q|B} \]
This is particularly useful when interpreting experimental tracks.
- The radius tells you about the perpendicular velocity.
- The pitch tells you about the parallel velocity.
Together, they allow you to reconstruct the particle’s full velocity.
For a more detailed derivation of the circular-motion equations, see Circular Motion of a Charge in a Magnetic Field: HSC Physics.
06Worked examples
Worked example: Determine the trajectory and charge sign
A particle with mass \(1.67\times10^{-27}\text{ kg}\) and charge magnitude \(1.60\times10^{-19}\text{ C}\) enters a uniform magnetic field of strength \(0.25\text{ T}\).
Its speed is \(5.00\times10^6\text{ m s}^{-1}\), and its velocity makes an angle of \(60^\circ\) with the magnetic field.
The perpendicular velocity initially points to the right, while the parallel component points into the page. The observed track initially curves upwards.
Determine:
(a) Whether the particle follows a circular or helical trajectory.
(b) The sign of its charge.
(c) The radius and pitch of its trajectory.
Step 1
The perpendicular component is
\[ \begin{aligned} v_\perp&=v\sin60^\circ\\ &=5.00\times10^6(0.866)\\ &=4.33\times10^6\text{ m s}^{-1} \end{aligned} \]
The parallel component is
\[ \begin{aligned} v_\parallel&=v\cos60^\circ\\ &=5.00\times10^6(0.5)\\ &=2.50\times10^6\text{ m s}^{-1} \end{aligned} \]
Step 2
Both velocity components are non-zero.
The perpendicular component produces circular motion, while the parallel component carries the particle into the page.
Therefore, the particle follows a helical trajectory.
Step 3
The perpendicular velocity points right, and the magnetic field points into the page.
Using the right-hand rule, a positive charge would experience an upwards magnetic force.
The observed curvature is upwards.
Therefore, the particle has a positive charge.
Step 4
Use
\[ r=\frac{mv_\perp}{|q|B} \]
Substituting:
\[ \begin{aligned} r&=\frac{(1.67\times10^{-27})(4.33\times10^6)} {(1.60\times10^{-19})(0.25)}\\ &=0.181\text{ m} \end{aligned} \]
The radius of the circular component is \(0.181\text{ m}\).
Step 5
First, calculate the period:
\[ \begin{aligned} T&=\frac{2\pi m}{|q|B}\\ &=\frac{2\pi(1.67\times10^{-27})} {(1.60\times10^{-19})(0.25)}\\ &=2.62\times10^{-7}\text{ s} \end{aligned} \]
Now use the parallel velocity:
\[ \begin{aligned} h&=v_\parallel T\\ &=(2.50\times10^6)(2.62\times10^{-7})\\ &=0.656\text{ m} \end{aligned} \]
The particle advances approximately \(0.656\text{ m}\) along the magnetic field during each revolution.
Final result: The particle is positively charged and follows a helix with a radius of \(0.181\text{ m}\) and a pitch of \(0.656\text{ m}\).
Worked example: Reconstruct a particle’s velocity from its track
A particle has mass \(1.67\times10^{-27}\text{ kg}\) and charge magnitude \(1.60\times10^{-19}\text{ C}\).
It moves through a uniform magnetic field of \(0.50\text{ T}\). Measurements of its helical trajectory give:
- Radius: \(0.104\text{ m}\).
- Pitch: \(0.262\text{ m}\).
An end-on observation shows the magnetic field pointing out of the page, the perpendicular velocity initially pointing right, and clockwise curvature.
Calculate the perpendicular velocity, parallel velocity, total speed, and angle between the velocity and magnetic field. Determine the charge sign.
Step 1
Use
\[ T=\frac{2\pi m}{|q|B} \]
Substituting:
\[ \begin{aligned} T&=\frac{2\pi(1.67\times10^{-27})} {(1.60\times10^{-19})(0.50)}\\ &=1.31\times10^{-7}\text{ s} \end{aligned} \]
Step 2
Rearrange the radius equation:
\[ v_\perp=\frac{r|q|B}{m} \]
Substituting:
\[ \begin{aligned} v_\perp&=\frac{(0.104)(1.60\times10^{-19})(0.50)} {1.67\times10^{-27}}\\ &=4.98\times10^6\text{ m s}^{-1} \end{aligned} \]
Step 3
Use the pitch equation:
\[ h=|v_\parallel|T \]
Therefore,
\[ \begin{aligned} |v_\parallel|&=\frac{h}{T}\\ &=\frac{0.262}{1.31\times10^{-7}}\\ &=2.00\times10^6\text{ m s}^{-1} \end{aligned} \]
Step 4
The perpendicular and parallel components are at right angles, so use Pythagoras:
\[ v=\sqrt{v_\perp^2+v_\parallel^2} \]
\[ \begin{aligned} v&=\sqrt{(4.98\times10^6)^2+(2.00\times10^6)^2}\\ &=5.37\times10^6\text{ m s}^{-1} \end{aligned} \]
Step 5
Using
\[ \tan\theta=\frac{v_\perp}{|v_\parallel|} \]
we obtain
\[ \theta=\tan^{-1}\left(\frac{4.98}{2.00}\right) \]
\[ \theta=68.1^\circ \]
Both components are non-zero, confirming that the particle follows a helix.
Step 6
The magnetic field points out of the page, and the perpendicular velocity points right.
For a positive charge, the magnetic force points downwards.
This produces clockwise curvature when viewed from the stated direction.
The observed curvature is clockwise, so the particle is positively charged.
Final result: The particle has a positive charge, a perpendicular velocity of \(4.98\times10^6\text{ m s}^{-1}\), a parallel velocity of \(2.00\times10^6\text{ m s}^{-1}\), and a total speed of \(5.37\times10^6\text{ m s}^{-1}\). Its trajectory is helical.
07Questions and solutions
Question 1
An electron enters a uniform magnetic field of strength \(0.30\text{ T}\).
Its velocity is \(4.0\times10^6\text{ m s}^{-1}\), directed horizontally to the right and perpendicular to the magnetic field.
The magnetic field points into the page.
Determine:
(a) Whether the trajectory is circular or helical.
(b) The initial direction of the magnetic force.
(c) Whether the particle curves clockwise or anticlockwise.
Solution 1
The electron follows a circular trajectory, initially curving downwards in a clockwise direction.
(a) Trajectory
The velocity is entirely perpendicular to the magnetic field.
Therefore,
\[ v_\parallel=0 \]
and the particle follows a circle.
There is no movement along the magnetic field axis.
(b) Magnetic force
For a positive charge moving right with a magnetic field into the page, the right-hand rule gives an upwards force.
An electron is negatively charged, so its force acts in the opposite direction.
The magnetic force is therefore downwards.
(c) Direction of curvature
The magnetic force acts towards the centre of the circular path.
Since the particle initially moves right and experiences a downward force, it begins curving clockwise.
The electron’s speed remains constant because the magnetic force is perpendicular to its instantaneous velocity.
Question 2
Two identical positively charged particles, A and B, move through the same uniform magnetic field.
Both have the same mass and charge magnitude. Their circular radii are equal, and their periods are both \(2.0\times10^{-7}\text{ s}\).
Both particles advance along the magnetic field.
The measured pitches are:
- Particle A: \(0.20\text{ m}\).
- Particle B: \(0.60\text{ m}\).
The perpendicular velocity component of each particle is \(3.0\times10^6\text{ m s}^{-1}\).
Calculate the parallel velocity and total speed of each particle. Explain why the two particles have the same circular radius despite having different total speeds.
Solution 2
Particle B has a greater total speed, even though both particles have the same circular radius.
Both particles have the same perpendicular velocity, but different parallel velocities.
Step 1: Calculate the parallel velocity of particle A.
Use
\[ h=v_\parallel T \]
Rearranging:
\[ v_\parallel=\frac{h}{T} \]
For particle A:
\[ \begin{aligned} v_{\parallel,A} &=\frac{0.20}{2.0\times10^{-7}}\\ &=1.0\times10^6\text{ m s}^{-1} \end{aligned} \]
Step 2: Calculate the parallel velocity of particle B.
\[ \begin{aligned} v_{\parallel,B} &=\frac{0.60}{2.0\times10^{-7}}\\ &=3.0\times10^6\text{ m s}^{-1} \end{aligned} \]
Step 3: Calculate the total speed of each particle.
Use
\[ v=\sqrt{v_\perp^2+v_\parallel^2} \]
For particle A:
\[ \begin{aligned} v_A&=\sqrt{(3.0\times10^6)^2+(1.0\times10^6)^2}\\ &=3.16\times10^6\text{ m s}^{-1} \end{aligned} \]
For particle B:
\[ \begin{aligned} v_B&=\sqrt{(3.0\times10^6)^2+(3.0\times10^6)^2}\\ &=4.24\times10^6\text{ m s}^{-1} \end{aligned} \]
Step 4: Compare their trajectories.
The radius equation is
\[ r=\frac{mv_\perp}{|q|B} \]
Both particles have the same mass, charge magnitude, magnetic field strength, and perpendicular velocity.
Therefore, they have the same circular radius.
However, particle B has a greater parallel velocity, so it advances further along the magnetic field during each revolution.
Both particles follow helices, but particle B has the greater pitch and total speed.
The tempting mistake is to assume that equal circular radii mean equal total speeds. The radius only tells you about the perpendicular velocity component.
Question 3
A particle travels through a cloud chamber containing a uniform magnetic field directed into the page.
Its trajectory lies in the plane perpendicular to the magnetic field.
A thin lead plate is placed horizontally across the chamber.
The measured track has:
- A radius of \(0.120\text{ m}\) below the plate.
- A radius of \(0.040\text{ m}\) above the plate.
At the plate, the centre of curvature of the track lies to the left.
Assume that the particle loses kinetic energy when passing through the plate and experiences negligible scattering.
Determine:
(a) The direction of travel through the chamber.
(b) The ratio of the particle’s momentum after passing through the plate to its momentum before passing through.
(c) Whether the particle is positively or negatively charged.
Solution 3
The particle travels upwards through the chamber and has a positive charge. Its momentum after passing through the plate is one-third of its initial momentum.
(a) Direction of travel
The particle loses kinetic energy when passing through the lead plate.
Its speed decreases, so its radius of curvature becomes smaller.
The radius is smaller above the plate:
\[ 0.040\text{ m}<0.120\text{ m} \]
Therefore, the particle must have travelled from the lower section towards the upper section.
Its direction of travel is upwards.
(b) Momentum ratio
For motion perpendicular to a uniform magnetic field,
\[ r=\frac{mv}{|q|B} \]
Since the particle’s mass, charge, and magnetic field remain unchanged,
\[ r\propto p \]
where \(p=mv\) is its momentum.
Therefore,
\[ \frac{p_{\text{after}}}{p_{\text{before}}} = \frac{r_{\text{after}}}{r_{\text{before}}} \]
Substituting:
\[ \begin{aligned} \frac{p_{\text{after}}}{p_{\text{before}}} &=\frac{0.040}{0.120}\\ &=\frac13 \end{aligned} \]
The particle retains one-third of its original momentum after passing through the plate.
(c) Charge sign
The particle moves upwards, and the magnetic field points into the page.
Using the right-hand rule, a positive charge experiences a force to the left.
The observed curvature is towards the left.
Therefore, the particle is positively charged.
This question illustrates why the direction of travel matters. If you reversed the assumed direction of motion, the same geometric curvature would imply the opposite charge sign.
Question 4
A particle travels through a uniform magnetic field.
A camera positioned directly along the magnetic field axis records a circular track with a constant radius of \(5.0\text{ cm}\).
A second camera, positioned perpendicular to the field axis, records that the particle advances \(0.18\text{ m}\) along the magnetic field during each complete revolution.
Determine whether the particle follows a circular, helical, or spiral trajectory.
Explain why the first camera’s photograph is insufficient to establish that the particle follows a true circular path.
Solution 4
The particle follows a helical trajectory, not a purely circular trajectory.
The second camera provides the crucial evidence: the particle advances \(0.18\text{ m}\) along the magnetic field during every revolution.
Therefore,
\[ v_\parallel\neq0 \]
Since the particle also follows a circular path around the field axis,
\[ v_\perp\neq0 \]
Both components are present, so the actual trajectory must be a helix.
The first camera records a circular projection because it looks directly along the magnetic field axis.
The particle’s movement along that axis is towards or away from the camera, so the camera cannot directly show the axial advance.
This is a projection effect, not evidence that the particle has no parallel velocity.
If the second camera recorded no axial advance, the trajectory would be a true circle, assuming the perpendicular motion continued.
It is also not a spiral: the measured radius remains constant rather than progressively decreasing.
Question 5
A particle of mass \(m\) and charge magnitude \(|q|\) moves through a uniform magnetic field.
Before entering a thin foil, its velocity components are:
\[ v_{\perp,1}=4.0\times10^6\text{ m s}^{-1} \]
\[ v_{\parallel,1}=3.0\times10^6\text{ m s}^{-1} \]
The particle passes through the foil, which reduces its kinetic energy and may change its direction through scattering.
Immediately after leaving the foil, measurements show:
- The radius of its helical trajectory is half its original value.
- The pitch of its trajectory is \(4/3\) of its original value.
- The particle continues advancing along the magnetic field.
Assume the particle’s mass and charge remain unchanged, and the magnetic field is uniform and unchanged. Treat the motion before and after the foil as non-relativistic helical motion.
An end-on camera is positioned on the positive magnetic-field side. The magnetic field points towards the camera, the initial perpendicular velocity points right, and the observed curvature is clockwise.
Determine:
(a) Whether the particle follows a helix before and after the foil.
(b) The sign of its charge.
(c) Its perpendicular and parallel velocity components after passing through the foil.
(d) Its initial and final total speeds.
(e) Whether the measured increase in pitch is consistent with the particle losing kinetic energy.
Solution 5
The particle remains helical, has a positive charge, and loses 20% of its kinetic energy despite its pitch increasing.
The key is that radius and pitch measure different velocity components. A reduction in radius does not require a reduction in the parallel velocity.
(a) Determine the trajectory
Before the foil, both velocity components are non-zero:
\[ v_{\perp,1}=4.0\times10^6 \]
\[ v_{\parallel,1}=3.0\times10^6 \]
Therefore, the particle initially follows a helix.
After the foil, the radius and pitch are both still non-zero.
This means both perpendicular and parallel motion remain present.
The particle therefore continues following a helix.
(b) Determine the charge sign
The magnetic field points towards the camera, and the perpendicular velocity points right.
For a positive charge, the right-hand rule gives a magnetic force downwards.
This produces clockwise curvature when viewed from the positive magnetic-field side.
The observed curvature is clockwise.
Therefore, the particle has a positive charge.
(c) Calculate the final velocity components
Start with the radius relationship:
\[ r=\frac{mv_\perp}{|q|B} \]
The mass, charge magnitude, and magnetic field are unchanged.
Therefore,
\[ r\propto v_\perp \]
The final radius is half the original radius:
\[ \frac{r_2}{r_1}=\frac12 \]
Hence,
\[ v_{\perp,2}=\frac12v_{\perp,1} \]
Substituting:
\[ v_{\perp,2} =\frac12(4.0\times10^6) \]
\[ \boxed{v_{\perp,2}=2.0\times10^6\text{ m s}^{-1}} \]
Now consider the pitch.
For the same particle in the same magnetic field,
\[ T=\frac{2\pi m}{|q|B} \]
The period is unchanged.
Since
\[ h=v_\parallel T \]
the pitch is directly proportional to the parallel velocity.
We are told that
\[ \frac{h_2}{h_1}=\frac43 \]
Therefore,
\[ v_{\parallel,2}=\frac43v_{\parallel,1} \]
Substituting:
\[ v_{\parallel,2} =\frac43(3.0\times10^6) \]
\[ \boxed{v_{\parallel,2}=4.0\times10^6\text{ m s}^{-1}} \]
Notice the unusual result: the perpendicular velocity has halved, but the parallel velocity has increased.
(d) Compare the initial and final total speeds
The initial speed is
\[ v_1=\sqrt{v_{\perp,1}^2+v_{\parallel,1}^2} \]
\[ \begin{aligned} v_1&=\sqrt{(4.0\times10^6)^2+(3.0\times10^6)^2}\\ &=5.0\times10^6\text{ m s}^{-1} \end{aligned} \]
The final speed is
\[ v_2=\sqrt{v_{\perp,2}^2+v_{\parallel,2}^2} \]
\[ \begin{aligned} v_2&=\sqrt{(2.0\times10^6)^2+(4.0\times10^6)^2}\\ &=4.47\times10^6\text{ m s}^{-1} \end{aligned} \]
Therefore,
\[ v_2<v_1 \]
The particle is moving more slowly overall after passing through the foil.
(e) Check the kinetic energy
For an unchanged particle mass,
\[ K=\frac12mv^2 \]
Therefore,
\[ \frac{K_2}{K_1}=\frac{v_2^2}{v_1^2} \]
Substituting:
\[ \begin{aligned} \frac{K_2}{K_1} &=\frac{(2.0)^2+(4.0)^2}{(5.0)^2}\\ &=\frac{20}{25}\\ &=0.80 \end{aligned} \]
The particle retains 80% of its original kinetic energy.
It has therefore lost 20% of its kinetic energy.
This is the important conceptual result.
A larger pitch does not necessarily mean a greater total speed. It means a greater parallel velocity.
The foil has scattered the particle, changing the direction of its velocity while reducing its overall speed. A greater proportion of its remaining velocity is now directed along the magnetic field.
The tempting assumption is that a particle losing energy must have both velocity components decrease. That would only follow if its direction remained unchanged.
Here, the direction changes, so the components do not have to decrease together.
The magnetic field itself does no work on the particle. The energy loss occurs through its interaction with the foil.
08What to remember when interpreting a magnetic trajectory
When a question gives you a charged particle moving through a magnetic field, work through these decisions in order:
- Identify the direction of the magnetic field.
- Identify the particle’s initial velocity and resolve it into parallel and perpendicular components.
- If only the perpendicular component is present, the trajectory is circular.
- If both components are present, the trajectory is helical.
- If only the parallel component is present, the particle travels in a straight line.
- Use the right-hand rule for a positive charge, reversing the force direction for a negative charge.
- Make sure the direction of particle travel is known before identifying its charge from curvature.
The next useful step is learning to extract particle properties from measured trajectories. The radius tells you about perpendicular momentum, while the pitch reveals parallel motion. Together, they turn a curved detector track into information about the particle’s full velocity.