Why Magnetic Force Changes Direction Without Changing Speed

In a 1932 cloud-chamber photograph, a cosmic-ray particle leaves a curved track that becomes more tightly curved after passing through a thin lead plate. Did the magnetic field slow the particle down? No. The lead plate did. A magnetic force changes a particle’s direction without changing its speed because it always acts perpendicular to the particle’s instantaneous velocity. The important detail is that only the component of velocity perpendicular to the magnetic field is affected.

This distinction explains two things that often cause trouble in HSC Physics: why a charged particle can follow a curved path at constant speed, and how the direction of that curve reveals whether its charge is positive or negative.

01Why the magnetic force changes direction, not speed

Imagine a car travelling around a bend at a constant speed. Its direction changes continuously, but its speed does not. The tyres provide a sideways force that turns the car without pushing it forwards or braking it.

A magnetic force does something similar to a charged particle. It acts sideways relative to the particle’s instantaneous motion.

The magnetic force is described by

\[ \vec F_B=q(\vec v\times\vec B) \]

where:

• \(F_B\) is the magnetic force, measured in newtons (N).
• \(q\) is the signed charge of the particle, measured in coulombs (C).
• \(\vec v\) is its velocity, measured in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(\vec B\) is the magnetic field, measured in tesla (T).
• The cross product determines the force’s direction.

The important feature is that the magnetic force is always perpendicular to the instantaneous velocity.

This means the magnetic force cannot do work on the particle.

The rate at which a force does work is its power:

\[ P_B=\vec F_B\cdot\vec v=0 \]

The dot product is zero because the force and velocity are perpendicular.

Since the magnetic force does no work, it cannot change the particle’s kinetic energy. Assuming there are no electric fields, collisions or other forces doing work, the particle’s speed remains constant.

Its momentum still changes because momentum has both magnitude and direction. The magnitude stays constant, but its direction rotates.

The car analogy has one limitation: a car uses friction between its tyres and the road, whereas a charged particle experiences a magnetic force determined by its charge, velocity and the magnetic field. The shared idea is simply that a perpendicular force can change direction without changing speed.

02Only the perpendicular velocity component is affected

There is another detail that becomes important when a particle enters a magnetic field at an angle.

A particle does not have to travel entirely across the magnetic field. It might also have some velocity along the field.

To handle this, split its velocity into two components:

• \(v_\parallel\): the component parallel to the magnetic field.
• \(v_\perp\): the component perpendicular to the magnetic field.

Let \(\theta\) be the angle between the velocity vector and the magnetic field.

The components are

\[ v_\parallel=v\cos\theta \]

\[ v_\perp=v\sin\theta \]

The magnetic force depends only on \(v_\perp\).

\[ \boxed{F_B=|q|v_\perp B} \]

Substituting the velocity component gives

\[ F_B=|q|vB\sin\theta \]

The parallel component contributes nothing to the magnetic force because it points in the same direction as the magnetic field.

Consider three possible entry angles.

The third case is particularly important.

The perpendicular component rotates in a circle, while the parallel component continues along the magnetic field. Together, they produce a helix.

Neither component changes its magnitude in a uniform magnetic field. Therefore, the total speed remains

\[ v=\sqrt{v_\perp^2+v_\parallel^2} \]

A particle can therefore travel along a helical path without speeding up or slowing down.

This is why you should never automatically use the particle’s total velocity when calculating magnetic force or orbital radius. You need the perpendicular component.

03The experiment: how a curved track revealed the charge sign

The relationship between magnetic curvature and charge sign played an important role in one of the most significant discoveries in particle physics.

In 1932, physicist Carl Anderson used a cloud chamber to investigate cosmic-ray particles. His work led to the identification of the positron, the positively charged counterpart of the electron.

The experiment also provides a useful example of why knowing the direction of particle motion is essential when interpreting a curved track.

The apparatus

A cloud chamber contains a gas with supersaturated vapour. When a charged particle travels through it, the particle ionises the gas along its path. Tiny droplets form around these ions, creating a visible trail that can be photographed.

Anderson placed his cloud chamber inside a large electromagnet. The magnetic field caused charged particles travelling through the chamber to follow curved paths.

A thin lead plate was then placed horizontally across the middle of the chamber.

The original problem: which way was the particle travelling?

Suppose a photograph shows a particle following a curved path in a magnetic field.

You can observe the curvature, but that does not automatically tell you the direction in which the particle travelled.

This creates an ambiguity.

A negatively charged particle travelling in one direction can produce the same geometrical curvature as a positively charged particle travelling in the opposite direction.

Anderson initially had to consider this uncertainty when interpreting his photographs.

He needed an independent way to establish the particle’s direction of travel.

The revision: adding a lead plate

The lead plate provided the missing information.

When a charged particle passes through lead, it loses energy through interactions with the material. Its momentum decreases.

For a given magnetic field and charge magnitude, a particle with less perpendicular momentum follows a tighter curve.

In other words:

• Before the plate, the particle has a larger momentum and a larger radius of curvature.
• After the plate, it has a smaller momentum and a smaller radius of curvature.

The magnetic field itself has not slowed the particle. Its interactions with the lead have.

This difference is crucial.

The observation and conclusion

In Anderson’s famous photograph, the track was more tightly curved above the lead plate than below it.

This showed that the particle had travelled upwards: it passed through the plate and lost energy before producing the more tightly curved upper section.

Once the direction of travel was established, Anderson could combine it with the known magnetic-field direction to determine the sign of the charge.

The track indicated a positively charged particle. Further evidence showed that its mass was comparable to that of an electron.

It was the positron.

The American Physical Society’s historical account describes how the lead plate resolved the ambiguity in the particle’s direction. The Caltech Archives records the famous photograph, dated 2 August 1932.

Sources: American Physical Society: Discovery of the Positron and Caltech Archives: Cloud Chamber Photo of a Positron.

The key experimental lesson is that curvature alone is not enough to identify charge sign. You also need to know the direction of travel and the magnetic-field direction.

04How to identify charge sign from curvature

You can use the right-hand rule to determine whether a particle is positive or negative.

The safest approach is to establish a consistent coordinate system.

For a diagram viewed directly from the front:

• Right is the positive \(x\)-direction.
• Up is the positive \(y\)-direction.
• Out of the page is the positive \(z\)-direction.
• A cross (\(\times\)) represents a magnetic field directed into the page.
• A dot (\(\bullet\)) represents a magnetic field directed out of the page.

Consider a particle entering a magnetic field from the left, moving horizontally to the right.

The magnetic field points into the page.

For a positive charge, apply the right-hand rule to the cross product \(\vec v\times\vec B\).

The result is an upward magnetic force.

A negative charge experiences a force in the opposite direction, so it bends downwards.

For this particular arrangement:

If the magnetic field points out of the page instead, these results reverse.

A reliable decision process

When an exam question asks you to identify a particle’s charge from its trajectory, use this sequence:

Step 1: Identify the particle’s direction of travel. Use the initial tangent to the path, not the direction in which the curve eventually points.

Step 2: Identify the magnetic-field direction. Look carefully for crosses and dots.

Step 3: Apply the right-hand rule to a positive charge using \(\vec v\times\vec B\).

Step 4: Compare your predicted force direction with the observed curvature.

The magnetic force points towards the centre of curvature. If the observed curvature is opposite to the direction predicted for a positive charge, the particle must be negative.

One common trap is to identify a particle as positive simply because its track curves clockwise. Clockwise and anticlockwise are not enough by themselves. You must also know the direction of the magnetic field, the direction of travel and the viewing orientation.

05Calculating magnetic force and orbital radius

Once you have identified the perpendicular velocity component, two equations become particularly useful.

The magnetic force magnitude is

\[ F_B=|q|v_\perp B \]

For a particle moving in a circular path perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force:

\[ F_B=\frac{mv_\perp^2}{r} \]

Equating these expressions gives

\[ \boxed{r=\frac{mv_\perp}{|q|B}} \]

where:

• \(r\) is the radius of the circular motion, in metres.
• \(m\) is the particle’s mass, in kilograms.
• \(v_\perp\) is its perpendicular velocity component, in metres per second.
• \(|q|\) is the magnitude of its charge, in coulombs.
• \(B\) is the magnetic-field strength, in tesla.

For helical motion, this equation gives the radius of the circular component of the trajectory.

Notice that the equation uses \(v_\perp\), not necessarily the total speed.

All numerical examples below assume non-relativistic motion in a uniform, static magnetic field, with no electric fields or collisions.

06Questions and solutions

Question 1

An ion with charge \(+2.0\times10^{-19}\,\mathrm C\) enters a uniform magnetic field of \(0.25\,\mathrm T\) at \(4.0\times10^5\,\mathrm{m\,s^{-1}}\).

Its velocity is perpendicular to the magnetic field. At the entry point, it moves to the right and the magnetic field points into the page.

Calculate:

a) The magnitude of the initial magnetic force.

b) The direction of the magnetic force.

c) The particle’s speed \(2.0\,\mu\mathrm s\) later, assuming it remains in the field and no other force acts.

Solution 1

The magnetic force is \(2.0\times10^{-14}\,\mathrm N\), directed upwards, and the particle’s speed remains \(4.0\times10^5\,\mathrm{m\,s^{-1}}\).

Part a: Force magnitude

Since the velocity is perpendicular to the field,

\[ F_B=|q|vB \]

Substitute:

\[ \begin{aligned} F_B&=(2.0\times10^{-19}) (4.0\times10^5)(0.25)\\ &=2.0\times10^{-14}\,\mathrm N \end{aligned} \]

Part b: Force direction

The particle is positive, so use the right-hand rule.

Point your fingers to the right, then curl them into the page. Your thumb points upwards.

Therefore, the magnetic force is upwards.

Part c: Speed after \(2.0\,\mu\mathrm s\)

The magnetic force is perpendicular to the particle’s velocity, so it does no work.

Its speed cannot change.

\[ \boxed{v=4.0\times10^5\,\mathrm{m\,s^{-1}}} \]

The particle’s direction will have changed, but its speed will remain constant.

Question 2

An ion has a mass of \(6.64\times10^{-27}\,\mathrm{kg}\) and a charge magnitude of \(3.20\times10^{-19}\,\mathrm C\).

It enters a uniform magnetic field of \(0.15\,\mathrm T\) with a speed of \(5.0\times10^6\,\mathrm{m\,s^{-1}}\). Its velocity makes an angle of \(60^\circ\) with the magnetic field.

Calculate:

a) The parallel velocity component.

b) The perpendicular velocity component.

c) The magnetic force magnitude.

d) The radius of the circular component of its trajectory.

e) The type of path followed by the particle.

Solution 2

The particle follows a helix, with a perpendicular velocity of \(4.33\times10^6\,\mathrm{m\,s^{-1}}\), a parallel velocity of \(2.50\times10^6\,\mathrm{m\,s^{-1}}\), a magnetic force of \(2.08\times10^{-13}\,\mathrm N\), and a circular-component radius of approximately \(0.60\,\mathrm m\).

Part a: Parallel velocity

\[ \begin{aligned} v_\parallel&=v\cos60^\circ\\ &=(5.0\times10^6)(0.5)\\ &=2.50\times10^6\,\mathrm{m\,s^{-1}} \end{aligned} \]

Part b: Perpendicular velocity

\[ \begin{aligned} v_\perp&=v\sin60^\circ\\ &=(5.0\times10^6)(0.866)\\ &=4.33\times10^6\,\mathrm{m\,s^{-1}} \end{aligned} \]

Part c: Magnetic force

Use the perpendicular component:

\[ \begin{aligned} F_B&=|q|v_\perp B\\ &=(3.20\times10^{-19}) (4.33\times10^6)(0.15)\\ &=2.08\times10^{-13}\,\mathrm N \end{aligned} \]

Part d: Radius

\[ r=\frac{mv_\perp}{|q|B} \]

Substitute:

\[ \begin{aligned} r&=\frac{(6.64\times10^{-27})(4.33\times10^6)} {(3.20\times10^{-19})(0.15)}\\ &=0.599\,\mathrm m \end{aligned} \]

Therefore,

\[ \boxed{r\approx0.60\,\mathrm m} \]

Part e: Path shape

The particle has a non-zero parallel velocity component, so it continues moving along the magnetic-field direction while its perpendicular component rotates.

Its path is helical, not circular.

Its total speed remains \(5.0\times10^6\,\mathrm{m\,s^{-1}}\).

Question 3

An unknown charged particle enters a uniform magnetic field from the left. Its initial velocity is directed to the right, the magnetic field points into the page, and the particle’s track curves downwards.

a) Determine whether the particle is positively or negatively charged.

b) If the magnetic field is reversed but the particle’s velocity remains unchanged, which direction will the magnetic force point?

c) If both the magnetic field and the particle’s velocity are reversed from their original directions, which direction will the magnetic force point?

Solution 3

The particle is negatively charged. Reversing the magnetic field alone makes the force point upwards, while reversing both the field and velocity makes the force point downwards again.

Part a: Charge sign

For a positive charge, a particle moving right through a magnetic field directed into the page experiences an upward force.

However, the observed particle curves downwards.

Its force is therefore opposite to the positive-charge prediction.

The particle must be negatively charged.

Part b: Reverse the magnetic field only

The magnetic force is

\[ \vec F_B=q(\vec v\times\vec B) \]

Reversing the magnetic field reverses the cross product.

The magnetic force therefore reverses direction.

Since the original force was downwards, the new force points upwards.

Part c: Reverse both velocity and magnetic field

Reversing both vectors produces

\[ (-\vec v)\times(-\vec B) =\vec v\times\vec B \]

The cross product is unchanged.

The particle’s charge is also unchanged, so its magnetic force points in the same direction as originally.

Therefore, the force points downwards.

This is a useful test of your understanding: reversing one vector reverses the force, but reversing both leaves the force unchanged.

Question 4

The same charged particle enters the same uniform magnetic field in two separate experiments.

Its mass, charge magnitude and magnetic-field strength are unchanged.

• Run A: speed \(6.0\times10^5\,\mathrm{m\,s^{-1}}\), with velocity at \(30^\circ\) to the magnetic field.
• Run B: speed \(4.0\times10^5\,\mathrm{m\,s^{-1}}\), with velocity perpendicular to the magnetic field.

Determine:

a) The perpendicular velocity in each run.

b) The ratio of the magnetic force in Run B to that in Run A.

c) The ratio of the circular radii in Run B and Run A.

d) Which run produces the smaller radius?

Explain why the run with the larger magnetic force does not necessarily have the smaller radius.

Solution 4

Run B has a magnetic force and circular radius both \(4/3\) times those of Run A. However, Run A has the smaller radius and therefore the tighter circular path.

Part a: Perpendicular velocities

For Run A:

\[ \begin{aligned} v_{\perp A}&=v_A\sin30^\circ\\ &=(6.0\times10^5)(0.5)\\ &=3.0\times10^5\,\mathrm{m\,s^{-1}} \end{aligned} \]

For Run B, the velocity is already perpendicular to the field:

\[ v_{\perp B}=4.0\times10^5\,\mathrm{m\,s^{-1}} \]

Part b: Magnetic force ratio

For the same particle in the same magnetic field,

\[ F_B=|q|v_\perp B \]

Therefore,

\[ \begin{aligned} \frac{F_B^{(B)}}{F_B^{(A)}} &=\frac{v_{\perp B}}{v_{\perp A}}\\ &=\frac{4.0\times10^5}{3.0\times10^5}\\ &=\frac43 \end{aligned} \]

Run B experiences a magnetic force \(4/3\) times as large.

Part c: Radius ratio

The circular radius is

\[ r=\frac{mv_\perp}{|q|B} \]

Since mass, charge magnitude and field strength are identical,

\[ \begin{aligned} \frac{r_B}{r_A} &=\frac{v_{\perp B}}{v_{\perp A}}\\ &=\frac43 \end{aligned} \]

Run B has the larger radius.

Part d: Explain the apparent contradiction

Run A has the smaller radius, so it follows the tighter circular path.

At first, this might seem strange. Run B experiences a greater magnetic force, so you might expect it to curve more tightly.

The missing consideration is that Run B also has a greater perpendicular speed.

The radius depends directly on perpendicular speed:

\[ r\propto v_\perp \]

A faster-moving particle needs a larger circular radius to maintain its motion under the magnetic force.

In fact, the angular speed of circular motion is

\[ \omega=\frac{v_\perp}{r} =\frac{|q|B}{m} \]

For the same particle in the same magnetic field, this angular speed is unchanged.

The particle with the greater perpendicular speed travels around a larger circle at the same angular rate.

This is why a larger magnetic force does not automatically mean a smaller orbital radius.

Question 5

An unknown charged particle travels through a uniform magnetic field of \(0.20\,\mathrm T\).

The field points out of the page towards a camera. The camera is positioned on the positive \(z\)-side, looking back towards the origin. In the camera’s view:

• The positive \(x\)-direction is to the right.
• The positive \(y\)-direction is upwards.
• The positive \(z\)-direction points towards the camera.

The particle has a mass of \(6.64\times10^{-27}\,\mathrm{kg}\) and a charge magnitude of \(3.20\times10^{-19}\,\mathrm C\).

The camera records a projected circular track with radius \(4.0\,\mathrm{cm}\). A timing marker establishes that, at a particular point, the particle’s projected velocity points to the right. Its projected path curves downwards at that point.

A second detector plane, located \(0.12\,\mathrm m\) further along the positive \(z\)-direction, records the particle after exactly one complete revolution.

Assume the field is uniform, the particle experiences no collisions or electric forces, and its motion is non-relativistic.

Determine:

a) The sign of the particle’s charge.

b) Its perpendicular velocity component.

c) Its parallel velocity component.

d) Its total speed.

e) The angle between its velocity and the magnetic field.

Finally, explain why the circular appearance of the track does not prove that the particle’s actual three-dimensional path is circular.

Solution 5

The particle is positively charged and follows a helical path. Its perpendicular velocity is approximately \(3.86\times10^5\,\mathrm{m\,s^{-1}}\), its parallel velocity is approximately \(1.84\times10^5\,\mathrm{m\,s^{-1}}\), and its total speed is approximately \(4.27\times10^5\,\mathrm{m\,s^{-1}}\). Its velocity makes an angle of approximately \(64.5^\circ\) with the magnetic field.

This question contains an important trap: a circular projection does not necessarily represent a circular three-dimensional trajectory.

Part a: Determine the charge sign

The magnetic field points out of the page, in the positive \(z\)-direction.

At the specified point, the perpendicular component of velocity points to the right, in the positive \(x\)-direction.

For a positive charge,

\[ \vec v_\perp\times\vec B =+\hat x\times+\hat z =-\hat y \]

Therefore, a positive particle experiences a downward magnetic force.

The observed track curves downwards, so

\[ \boxed{q>0} \]

The parallel velocity component does not affect this direction because it is parallel to the magnetic field.

Part b: Calculate the perpendicular velocity

We first need the time taken for one complete revolution.

The perpendicular component travels around a circle, so

\[ T=\frac{2\pi r}{v_\perp} \]

Using

\[ r=\frac{mv_\perp}{|q|B} \]

gives

\[ T=\frac{2\pi m}{|q|B} \]

Substitute the known values:

\[ \begin{aligned} T&=\frac{2\pi(6.64\times10^{-27})} {(3.20\times10^{-19})(0.20)}\\ &=6.52\times10^{-7}\,\mathrm s \end{aligned} \]

Now calculate the perpendicular velocity using the measured projected radius:

\[ v_\perp=\frac{2\pi r}{T} \]

\[ \begin{aligned} v_\perp&=\frac{2\pi(0.040)} {6.52\times10^{-7}}\\ &=3.86\times10^5\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore,

\[ \boxed{v_\perp=3.86\times10^5\,\mathrm{m\,s^{-1}}} \]

Part c: Calculate the parallel velocity

The particle travels \(0.12\,\mathrm m\) along the magnetic-field direction during one complete revolution.

This distance is the pitch of the helical path.

The parallel component of velocity is therefore

\[ v_\parallel=\frac{\text{pitch}}{T} \]

Substitute:

\[ \begin{aligned} v_\parallel&=\frac{0.12}{6.52\times10^{-7}}\\ &=1.84\times10^5\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore,

\[ \boxed{v_\parallel=1.84\times10^5\,\mathrm{m\,s^{-1}}} \]

Part d: Calculate the total speed

The parallel and perpendicular components are at right angles.

Use Pythagoras:

\[ v=\sqrt{v_\perp^2+v_\parallel^2} \]

\[ \begin{aligned} v&=\sqrt{(3.86\times10^5)^2+ (1.84\times10^5)^2}\\ &=4.27\times10^5\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore,

\[ \boxed{v=4.27\times10^5\,\mathrm{m\,s^{-1}}} \]

Part e: Calculate the angle to the magnetic field

The angle is measured between the total velocity and the magnetic field.

Since the parallel component is adjacent to the angle and the perpendicular component is opposite,

\[ \tan\theta=\frac{v_\perp}{v_\parallel} \]

\[ \begin{aligned} \theta&=\tan^{-1} \left(\frac{3.86\times10^5}{1.84\times10^5}\right)\\ &=64.5^\circ \end{aligned} \]

Therefore,

\[ \boxed{\theta=64.5^\circ} \]

Why the circular image is misleading

The camera looks directly along the magnetic-field direction.

Any movement parallel to the field occurs towards or away from the camera. It is therefore hidden in the end-on projection.

The particle is actually following a helix, not a circle. Its perpendicular component produces the circular projection, while its parallel component produces the \(0.12\,\mathrm m\) advance along the field during each revolution.

The important lesson is that a two-dimensional track can hide part of a particle’s velocity. You must consider the viewing direction and any motion along the magnetic field before deciding what trajectory the particle actually follows.

Once you can separate \(v_\perp\) from \(v_\parallel\), you can tackle the next decision: whether a particle follows a circular or helical trajectory. That distinction is essential when calculating orbital radius, period and magnetic-field strength.

For the broader explanation of magnetic force and energy, see Why Magnetic Fields Change a Charged Particle’s Direction, Not Speed.

For the next step, see How to Decide Whether a Magnetic Trajectory Is Circular or Helical.