Why a Charge Moving Parallel to a Magnetic Field Feels No Force

An electron can travel through a magnetic field at \(4.0\times10^6\ \mathrm{m\,s^{-1}}\) and experience absolutely no magnetic force. It isn’t neutral, and the magnetic field hasn’t disappeared. Its velocity is simply parallel to the field.

Now tilt its direction of travel by \(30^\circ\), keeping its speed and the magnetic field strength unchanged. The magnetic force suddenly becomes half its maximum value.

The reason is that only the component of a charged particle’s velocity perpendicular to the magnetic field produces magnetic force. A component parallel or anti-parallel to the field contributes nothing.

This is more than a mathematical detail. It explains why some charged particles travel in straight lines through magnetic fields, why others follow helical paths, and how physicists can identify a particle’s charge from its curved trajectory.

01Why does parallel motion produce zero magnetic force?

Start with a simple situation.

Imagine a positively charged particle travelling directly to the right. A uniform magnetic field also points to the right.

The particle is moving, it has an electric charge, and it is inside a magnetic field. You might reasonably predict that a magnetic force must act on it.

But it doesn’t.

The magnetic force on a moving charge is described by:

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

where:

• \(F_B\) is the magnitude of the magnetic force, measured in newtons (N).
• \(q\) is the particle’s electric charge, measured in coulombs (C).
• \(v\) is the particle’s speed, measured in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(B\) is the magnetic field strength, measured in tesla (T).
• \(\theta\) is the angle between the particle’s velocity and the magnetic field.

If the particle travels parallel to the magnetic field, the angle is \(0^\circ\).

Substituting:

\[ F_B=|q|vB\sin0^\circ \]

Since \(\sin0^\circ=0\),

\[ \boxed{F_B=0\ \mathrm N} \]

The magnetic force is exactly zero.

There is an important detail here: moving in the opposite direction to the magnetic field also produces zero force.

For anti-parallel motion, \(\theta=180^\circ\), and:

\[ F_B=|q|vB\sin180^\circ=0 \]

So both situations produce zero magnetic force:

The key is that the magnetic field does not produce a force simply because a charged particle is moving. The direction of its motion matters.

This is also why doubling a particle’s speed does not help if it continues travelling exactly parallel to the field. Its magnetic force remains zero.

02Only the perpendicular velocity component matters

The equation \(F_B=|q|vB\sin\theta\) becomes much easier to understand when you separate the particle’s velocity into two components.

Imagine a particle travelling diagonally through a magnetic field.

Part of its velocity points along the field. The rest points across it.

We can resolve its velocity into:

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

These are not two separate velocities belonging to different particles. They are two components of the same velocity vector.

Using basic trigonometry:

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

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

The parallel component contributes no magnetic force because it points along the magnetic field.

The perpendicular component is responsible for the entire magnetic force.

Therefore, we can rewrite the magnetic force equation as:

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

This is often the most useful version of the equation when a particle enters a magnetic field at an angle.

What happens to the two components?

Consider a particle entering a uniform magnetic field with both components present.

The parallel component remains constant. The magnetic force has no component along the magnetic field, so it cannot change the particle’s velocity in that direction.

The perpendicular component behaves differently. The magnetic force continually changes its direction, producing circular motion around the magnetic field direction.

Combining these two motions produces a helical trajectory.

This gives us three distinct possibilities:

Notice something interesting: a particle can have a very large total speed but experience a relatively small magnetic force if most of its velocity is parallel to the field.

The magnetic field responds to \(v_\perp\), not simply to the total speed.

03Testing the prediction: an electron-beam experiment

How could we test whether the magnetic force really depends on the perpendicular component of velocity?

We need an experiment that changes the angle between the particle’s velocity and the magnetic field without accidentally changing other important variables.

Consider an idealised electron-beam experiment.

The apparatus

The apparatus consists of:

1. An electron gun that produces a narrow beam of electrons.
2. Two narrow slits that keep the beam moving in a well-defined direction.
3. A pair of Helmholtz coils that generate an approximately uniform magnetic field in the central region.
4. A fluorescent screen that records where the electron beam arrives.

The electron gun can be rotated to change the angle between the beam’s velocity and the magnetic field.

The accelerating voltage is kept constant so that the electron speed remains approximately constant.

The magnetic field strength is also held constant.

This means the angle is the main variable being investigated.

What should happen?

Start with the electron beam travelling perpendicular to the magnetic field.

Here, \(\theta=90^\circ\), so the perpendicular velocity component equals the total velocity:

\[ v_\perp=v \]

The magnetic force is therefore at its maximum.

Now rotate the electron gun until the beam travels at \(30^\circ\) to the field.

The perpendicular component becomes:

\[ v_\perp=v\sin30^\circ \]

\[ v_\perp=0.5v \]

The magnetic force should be half its previous value.

Finally, align the electron beam directly with the magnetic field.

The perpendicular component becomes zero, so the magnetic force disappears.

Why the first experimental design is not enough

Suppose the beam travels parallel to the magnetic field and the fluorescent screen shows no magnetic deflection.

Can we immediately conclude that the magnetic force is zero?

Not from that observation alone.

The beam might be neutral, the magnetic field might not have been established properly, or the apparatus might be misaligned.

A better experimental design includes control measurements.

First, switch off the magnetic field and record the beam’s normal position for each gun angle.

Next, switch the field on and repeat the measurements.

Then reverse the magnetic field direction and repeat again.

A known charged reference beam can also be used to check that the field and detector are working correctly.

The expected observations are:

• At \(90^\circ\), a charged beam experiences a magnetic force.
• Reversing the magnetic field reverses the direction of the magnetic deflection.
• At \(0^\circ\) and \(180^\circ\), the magnetic field produces no deflection.
• At intermediate angles, the magnetic effect depends on the perpendicular velocity component.

The field-off measurements help distinguish magnetic deflection from any background deflection caused by the apparatus.

A small calculation: how much does the force change?

Consider an electron travelling at \(2.0\times10^6\ \mathrm{m\,s^{-1}}\) through a uniform magnetic field of \(0.10\ \mathrm T\).

Use the electron’s charge magnitude:

\[ |q|=1.60\times10^{-19}\ \mathrm C \]

First, calculate the maximum possible magnetic force. This occurs when the velocity is perpendicular to the field.

\[ F_{\max}=|q|vB \]

\[ F_{\max} =(1.60\times10^{-19})(2.0\times10^6)(0.10) \]

\[ F_{\max}=3.2\times10^{-14}\ \mathrm N \]

Now compare the predicted forces at different angles.

The following values are illustrative calculations, not recorded experimental measurements.

The result is clear: the force depends on the perpendicular component, not simply on whether the particle is moving.

The experiment also highlights an important limitation. A real apparatus cannot establish perfect alignment or a perfectly uniform field with infinite precision. A tiny residual angle can produce a small force even when the beam appears parallel.

04How does the perpendicular component help identify charge sign?

The perpendicular velocity component is also important when identifying whether a particle is positively or negatively charged from its curved path.

The direction of the magnetic force is given by:

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

The symbol \(\times\) represents the vector cross product.

For a positive charge, use the right-hand rule to determine the force direction.

For a negative charge, reverse that direction.

When the particle has both parallel and perpendicular velocity components, use \(v_\perp\) when applying the right-hand rule.

Why?

Because the parallel component produces no magnetic force. It cannot affect the direction of the magnetic force or the direction of the particle’s initial curvature.

A useful reference arrangement

Suppose a particle initially travels to the right, while the magnetic field points into the page.

The direction of the magnetic force depends on the charge sign.

This gives us a useful method:

1. Identify the magnetic field direction.
2. Establish the particle’s initial direction of travel.
3. Consider only the velocity component perpendicular to the field.
4. Use the right-hand rule for a positive charge.
5. Reverse the predicted force direction if the particle curves the opposite way.

There is one important limitation.

A curved track alone does not always tell you the charge sign. You must know the direction in which the particle travelled. Otherwise, reversing the direction of travel can create an ambiguity.

A straight track is also not proof that a particle is neutral. A charged particle travelling parallel to a magnetic field can move in a straight line.

For a more detailed treatment of interpreting particle tracks, see How to Find a Particle’s Charge Sign from Its Curved Path.

05Worked examples

06Questions and solutions

Question 1

A proton travels at \(5.0\times10^6\ \mathrm{m\,s^{-1}}\) at an angle of \(30^\circ\) to a uniform magnetic field of strength \(0.20\ \mathrm T\).

Calculate the magnitude of the magnetic force acting on the proton.

Solution 1

The magnetic force has a magnitude of \(8.0\times10^{-14}\ \mathrm N\).

Only the component of velocity perpendicular to the field contributes to the force.

Step 1: Calculate the perpendicular velocity.

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

\[ v_\perp=(5.0\times10^6)\sin30^\circ \]

\[ v_\perp=2.5\times10^6\ \mathrm{m\,s^{-1}} \]

Step 2: Calculate the magnetic force.

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

Using the proton’s charge magnitude:

\[ |q|=1.60\times10^{-19}\ \mathrm C \]

\[ \begin{aligned} F_B&=(1.60\times10^{-19})\\ &\quad\times(2.5\times10^6)(0.20)\\ &=8.0\times10^{-14}\ \mathrm N \end{aligned} \]

Therefore:

\[ \boxed{F_B=8.0\times10^{-14}\ \mathrm N} \]

The force is half the maximum possible force because the perpendicular velocity component is half the total speed.

Question 2

Two singly charged positive ions travel through the same uniform magnetic field of strength \(0.25\ \mathrm T\).

Ion A travels at \(6.0\times10^6\ \mathrm{m\,s^{-1}}\) at \(30^\circ\) to the field.

Ion B travels at \(5.0\times10^6\ \mathrm{m\,s^{-1}}\) at \(36.9^\circ\) to the field.

Determine:

a) The magnetic force acting on each ion.

b) Whether the forces are equal.

c) Whether either particle follows a straight, circular, or helical trajectory.

A student claims that ion A must experience a greater magnetic force because it has a greater total speed. Evaluate this claim.

Solution 2

Both ions experience the same magnetic force of \(1.2\times10^{-13}\ \mathrm N\), and both follow helical trajectories.

The student’s prediction is incorrect because the magnetic force depends on perpendicular velocity, not total velocity.

Step 1: Calculate the perpendicular velocity of ion A.

\[ v_{\perp A}=v_A\sin30^\circ \]

\[ v_{\perp A}=(6.0\times10^6)(0.5) \]

\[ v_{\perp A}=3.0\times10^6\ \mathrm{m\,s^{-1}} \]

Step 2: Calculate the perpendicular velocity of ion B.

\[ v_{\perp B}=v_B\sin36.9^\circ \]

Using \(\sin36.9^\circ\approx0.600\):

\[ v_{\perp B}=(5.0\times10^6)(0.600) \]

\[ v_{\perp B}=3.0\times10^6\ \mathrm{m\,s^{-1}} \]

The perpendicular components are identical.

Step 3: Calculate the magnetic force on each ion.

Both ions have a charge magnitude of:

\[ |q|=1.60\times10^{-19}\ \mathrm C \]

For ion A:

\[ F_A=|q|v_{\perp A}B \]

\[ F_A=(1.60\times10^{-19})(3.0\times10^6)(0.25) \]

\[ F_A=1.20\times10^{-13}\ \mathrm N \]

For ion B:

\[ F_B=(1.60\times10^{-19})(3.0\times10^6)(0.25) \]

\[ F_B=1.20\times10^{-13}\ \mathrm N \]

Therefore:

\[ \boxed{F_A=F_B=1.20\times10^{-13}\ \mathrm N} \]

Step 4: Determine the trajectories.

Both particles have non-zero parallel and perpendicular velocity components.

Therefore, both follow helical trajectories.

Their total speeds differ, but this does not change the equality of their magnetic force magnitudes.

The important distinction is that ion A has a greater parallel velocity component, while both ions have the same perpendicular component.

Question 3

An unknown particle travels at \(4.0\times10^6\ \mathrm{m\,s^{-1}}\) at an angle of \(30^\circ\) to a uniform magnetic field of \(0.50\ \mathrm T\), directed into the page.

The perpendicular velocity component points east.

The particle has a charge magnitude of \(3.2\times10^{-19}\ \mathrm C\), and its trajectory initially curves southwards.

Determine:

a) The magnitude of the magnetic force.

b) Whether the particle is positively or negatively charged.

c) Why the curvature alone would not be sufficient to identify the charge sign if the direction of travel were unknown.

Solution 3

The particle experiences a magnetic force of \(3.2\times10^{-13}\ \mathrm N\) and is negatively charged.

Step 1: Resolve the velocity into its perpendicular component.

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

\[ v_\perp=(4.0\times10^6)\sin30^\circ \]

\[ v_\perp=2.0\times10^6\ \mathrm{m\,s^{-1}} \]

Step 2: Calculate the magnetic force magnitude.

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

\[ \begin{aligned} F_B&=(3.2\times10^{-19})\\ &\quad\times(2.0\times10^6)(0.50)\\ &=3.2\times10^{-13}\ \mathrm N \end{aligned} \]

Therefore:

\[ \boxed{F_B=3.2\times10^{-13}\ \mathrm N} \]

Step 3: Apply the right-hand rule.

The perpendicular velocity points east, and the magnetic field points into the page.

For a positive charge, the right-hand rule predicts a force upwards (northwards).

Step 4: Compare with the observed curvature.

The actual trajectory initially curves southwards.

This is opposite to the predicted direction for a positive charge.

Therefore:

\[ \boxed{q<0} \]

Step 5: Consider the importance of the direction of travel.

If the photograph showed a curved track but did not establish which direction the particle travelled, the charge sign would be ambiguous.

A particle travelling in the opposite direction would produce the opposite magnetic force for the same charge sign.

Therefore, identifying charge sign requires knowledge of both the magnetic field direction and the particle’s direction of travel.

Question 4

An unknown beam of particles travels through a uniform magnetic field of \(0.20\ \mathrm T\).

The particles travel at \(3.0\times10^6\ \mathrm{m\,s^{-1}}\).

Each particle is either neutral or has a charge magnitude of \(1.60\times10^{-19}\ \mathrm C\). Assume the beam contains only one particle type, so all charged particles have the same charge sign.

Initially, the beam travels parallel to the magnetic field. No magnetic deflection is observed.

A student concludes that the particles must be neutral.

a) Explain why this conclusion is not justified.

b) Describe an experimental test that would distinguish a charged beam from a neutral beam.

c) Calculate the magnetic force expected if the particles are charged and the beam is redirected perpendicular to the field.

d) Explain what should happen if the magnetic field is reversed.

Solution 4

The student’s conclusion is not justified. A charged particle travelling parallel to a magnetic field also experiences zero magnetic force.

Part a: Why the conclusion is incorrect

The initial angle is:

\[ \theta=0^\circ \]

Therefore:

\[ F_B=|q|vB\sin0^\circ=0 \]

This result applies to both positive and negative charged particles.

Consequently, observing no magnetic deflection cannot distinguish a neutral particle from a charged particle travelling parallel to the field.

Part b: Design an experimental test

Redirect the beam so that it travels perpendicular to the magnetic field.

Keep the particle speed and magnetic field strength constant.

Compare the beam’s position with the magnetic field switched on and switched off.

If the particles are charged, they should experience a magnetic force and their trajectories should curve.

If the particles are neutral, there should be no magnetic deflection.

Reverse the magnetic field while keeping the beam direction unchanged.

A charged beam should curve in the opposite direction, while a neutral beam should remain unaffected by the magnetic field.

A field-off measurement also helps identify any deflection caused by stray electric fields or apparatus misalignment.

Part c: Calculate the expected magnetic force

If the particles are charged and travel perpendicular to the field:

\[ \theta=90^\circ \]

Therefore:

\[ F_B=|q|vB \]

Substituting:

\[ \begin{aligned} F_B&=(1.60\times10^{-19})\\ &\quad\times(3.0\times10^6)(0.20)\\ &=9.6\times10^{-14}\ \mathrm N \end{aligned} \]

Therefore:

\[ \boxed{F_B=9.6\times10^{-14}\ \mathrm N} \]

A neutral particle would experience zero magnetic force.

Part d: Reverse the magnetic field

Reversing the magnetic field reverses the direction of the magnetic force on a charged particle.

This follows from:

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

Changing \(\vec B\) to \(-\vec B\) changes the force to \(-\vec F_B\).

A neutral particle experiences no magnetic force before or after the reversal.

The combination of changing the beam angle and reversing the field provides much stronger evidence than the original observation alone.

Question 5

A positively charged ion enters region A with two velocity components:

\[ v_\parallel=3.0\times10^6\ \mathrm{m\,s^{-1}} \]

directed east, and:

\[ v_\perp=4.0\times10^6\ \mathrm{m\,s^{-1}} \]

directed north.

The magnetic field in region A is uniform, has a strength of \(0.20\ \mathrm T\), and points east.

The ion’s charge is:

\[ q=+1.60\times10^{-19}\ \mathrm C \]

The ion leaves region A after exactly one-quarter of the circular period of its perpendicular motion.

It then travels through a field-free gap and enters region B. The magnetic field in region B has the same strength, \(0.20\ \mathrm T\), but points west.

Assume:

• The magnetic field is uniform within each region.
• There are no electric fields or other forces.
• The field-free gap does not change the particle’s velocity.
• The particle enters region B with the velocity it had when leaving region A.

Use east as \(+x\), north as \(+y\), and vertically upwards as \(+z\).

Determine:

a) The particle’s total speed and its initial magnetic force magnitude and direction in region A.

b) The direction of its perpendicular velocity when it leaves region A.

c) The magnetic force magnitude and direction immediately after it enters region B.

d) Whether the magnetic force can become zero later in region B.

Solution 5

The particle’s speed is \(5.0\times10^6\ \mathrm{m\,s^{-1}}\), its initial magnetic force in region A is \(1.28\times10^{-13}\ \mathrm N\) downwards, and its magnetic force immediately after entering region B is the same magnitude but points northwards. The force never becomes zero in region B.

The important detail is that reversing the magnetic field changes the force direction, but it does not make a non-zero perpendicular velocity component disappear.

Part a: Calculate the speed and initial force in region A

The two velocity components are perpendicular, so the total speed is found using Pythagoras’ theorem.

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

Substituting:

\[ \begin{aligned} v&=\sqrt{(3.0\times10^6)^2}\\ &\quad+\left(4.0\times10^6\right)^2 \end{aligned} \]

Therefore:

\[ \boxed{v=5.0\times10^6\ \mathrm{m\,s^{-1}}} \]

Now calculate the magnetic force.

Only the perpendicular velocity contributes:

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

\[ \begin{aligned} F_B&=(1.60\times10^{-19})\\ &\quad\times(4.0\times10^6)(0.20)\\ &=1.28\times10^{-13}\ \mathrm N \end{aligned} \]

For the direction, use:

\[ \vec F_B=q(\vec v_\perp\times\vec B) \]

The perpendicular velocity points north (\(+y\)), and the magnetic field points east (\(+x\)).

Using the right-hand rule:

\[ \hat y\times\hat x=-\hat z \]

Therefore, the force points vertically downwards.

\[ \boxed{F_A=1.28\times10^{-13}\ \mathrm N\text{ downwards}} \]

Part b: Determine the perpendicular velocity direction after one-quarter of a period

The magnetic force continuously changes the direction of the perpendicular velocity.

Because the ion is positively charged, the initial force acts downwards.

After one-quarter of the circular period, the perpendicular velocity has rotated through \(90^\circ\).

It initially pointed north. It now points vertically downwards.

Therefore:

\[ \boxed{\vec v_\perp=-4.0\times10^6\hat z\ \mathrm{m\,s^{-1}}} \]

The parallel component remains directed east.

The ion’s velocity when leaving region A is therefore:

\[ \vec v=3.0\times10^6\hat x -4.0\times10^6\hat z \]

In the field-free gap, no magnetic force acts on the particle.

Its velocity remains constant, so it travels along a straight line with this direction until it enters region B.

Part c: Calculate the force in region B

This is the most important part of the question.

The magnetic field now points west, which is the opposite direction to the original field.

The particle’s eastward velocity component is therefore anti-parallel to the new magnetic field.

Does this produce a magnetic force?

No.

Parallel and anti-parallel motion both produce zero magnetic force.

The perpendicular velocity component still points vertically downwards, however, so it continues to produce a magnetic force.

For a positive charge:

\[ \vec F_B=q(\vec v_\perp\times\vec B) \]

The perpendicular velocity points downwards (\(-z\)), and the new magnetic field points west (\(-x\)).

Therefore:

\[ (-\hat z)\times(-\hat x)=+\hat y \]

The force now points northwards.

Its magnitude is unchanged because the perpendicular speed and magnetic field strength are unchanged.

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

\[ F_B=(1.60\times10^{-19})(4.0\times10^6)(0.20) \]

\[ \boxed{F_B=1.28\times10^{-13}\ \mathrm N\text{ northwards}} \]

The magnetic field reversal has reversed the direction of the force, but not its magnitude.

Part d: Can the magnetic force become zero later?

No. The magnetic force cannot become zero while the perpendicular velocity remains non-zero.

In a uniform magnetic field, the magnetic force is always perpendicular to the perpendicular velocity component.

It changes the direction of \(v_\perp\), but not its magnitude.

Therefore:

\[ v_\perp=4.0\times10^6\ \mathrm{m\,s^{-1}} \]

remains constant.

Since:

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

and the charge magnitude, perpendicular speed, and field strength are all constant, the magnetic force magnitude also remains constant.

The particle continues following a helical trajectory in region B.

This is the subtle distinction: the particle’s total velocity direction changes continuously, but its perpendicular component never disappears. It cannot gradually become parallel to the magnetic field under the action of a uniform magnetic field alone.

07Where to take this next

The perpendicular velocity component explains both the size of the magnetic force and the shape of a charged particle’s trajectory.

Once you can resolve velocity correctly, the next useful step is connecting that perpendicular motion to circular radius and cyclotron period.

For further practice, see How to Decide Whether a Magnetic Trajectory Is Circular or Helical and How to Calculate a Charged Particle’s Period in a Magnetic Field.

The central idea to carry into those problems is simple: the magnetic field changes the perpendicular motion, not the motion along its own direction.