How to Find the Speed Selected by Crossed Electric and Magnetic Fields

Imagine a beam of charged particles travelling to the right through a region containing an upward electric field and a magnetic field pointing out of the page. The electric force pushes a positive particle upwards. The magnetic force pushes it downwards.

Some particles pass straight through without changing direction. Others curve upwards or downwards.

What determines which particles pass straight through?

For the standard arrangement, the answer is:

\[ \boxed{v_\perp=\frac{E}{B}} \]

Here, \(v_\perp\) is the component of velocity perpendicular to the magnetic field. If the particle travels at \(90^\circ\) to the magnetic field, this is its total speed. If it enters at an angle, the distinction becomes important.

This is the principle behind a velocity selector. It allows charged particles with a particular perpendicular velocity component to pass through undeflected.

01How crossed electric and magnetic fields select a speed

Start by picturing the arrangement.

A beam of charged particles enters a region between two parallel plates. The plates produce a uniform electric field. A magnetic field is applied perpendicular to both the electric field and the beam’s initial direction of travel.

For the arrangement below, assume:

• The particles initially travel to the right.
• The electric field points upwards.
• The magnetic field points out of the page.
• The particle’s initial velocity is perpendicular to both fields.

Why do some particles pass straight through?

Consider a positively charged particle first.

The electric field exerts an upward force. The magnetic field exerts a downward force because of the particle’s motion through the field.

If these forces have equal magnitudes, they cancel.

The particle continues in a straight line, even though both fields are acting on it.

A negatively charged particle behaves differently in terms of force directions. Its electric force points downwards, and its magnetic force points upwards.

However, the forces still oppose each other.

This gives us an interesting prediction: positive and negative particles can pass through the same velocity selector at the same speed.

The charge sign does not change the selected speed.

Deriving the velocity selector equation

The magnitude of the electric force on a charged particle is:

\[ F_E=|q|E \]

The magnitude of the magnetic force is:

\[ F_B=|q|vB \]

These expressions assume the particle’s velocity is perpendicular to the magnetic field.

For the particle to travel straight through, the two forces must balance:

\[ F_E=F_B \]

Therefore:

\[ |q|E=|q|vB \]

Cancel the common charge magnitude:

\[ E=vB \]

Rearranging gives:

\[ v=\frac{E}{B} \]

This is the standard velocity selector equation.

The symbols represent:

Notice that neither particle mass nor charge magnitude appears in the final equation.

The selector therefore does not distinguish between particles based on their mass or charge magnitude. It selects the appropriate velocity component.

There is one important qualification: the particle must enter in the correct direction so that the electric and magnetic forces oppose each other. Having the correct speed magnitude alone is not sufficient.

02Why you must use the perpendicular velocity component

Now consider a particle entering the magnetic field at an angle.

Its velocity is no longer entirely perpendicular to the magnetic field. Does the entire velocity contribute to the magnetic force?

No.

The magnetic force depends only on the component of velocity perpendicular to the magnetic field.

Resolve the velocity into two components:

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

If \(\theta\) is the angle between the particle’s total velocity and the magnetic field:

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

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

The perpendicular component produces the magnetic force:

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

Equivalently:

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

The parallel component contributes nothing to the magnetic force.

It is useful to picture a particle travelling diagonally through the magnetic field. Its perpendicular component produces circular motion, while its parallel component carries it along the field.

The combination produces a helical trajectory.

The important point for a velocity selector is that the electric force must balance the magnetic force produced by \(v_\perp\), not the total velocity.

Therefore:

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

Cancelling the charge magnitude:

\[ \boxed{v_\perp=\frac{E}{B}} \]

Since \(v_\perp=v\sin\theta\), we can also write:

\[ \boxed{v=\frac{E}{B\sin\theta}} \]

Use this second equation when the question gives the angle between the total velocity and the magnetic field and asks for the total speed.

Three useful cases

The final case is particularly important.

If a particle travels exactly parallel to the magnetic field, the magnetic force is zero. A non-zero electric force cannot be balanced, so the particle cannot pass straight through the selector under these conditions.

04How to identify charge sign from curvature

Charge identification requires careful attention to the direction of the magnetic field and the particle’s motion.

The magnetic force is described by:

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

The cross product gives the force direction for a positive charge. For a negative charge, reverse the direction.

Case 1: Curvature in a magnetic-only region

Consider a particle travelling to the right through a magnetic field pointing into the page.

Apply the right-hand rule to a positive charge:

• Point your fingers in the direction of the particle’s velocity.
• Orient them towards the magnetic field.
• Your thumb gives the magnetic force direction.

The force points upwards.

Therefore:

This table applies specifically to a particle travelling rightwards through a field directed into the page.

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

There is another important condition: you must know the direction of travel.

A positively charged particle travelling rightwards can produce the same curvature as a negatively charged particle travelling leftwards. A curved track without a known direction of travel is therefore insufficient to determine the charge sign.

Case 2: Curvature inside the velocity selector

There is an additional complication when both electric and magnetic fields are present.

The net force is:

\[ \vec F=q(\vec E+\vec v\times\vec B) \]

The electric and magnetic forces oppose each other in our standard arrangement.

Consequently, the direction of curvature depends on whether the particle is travelling faster or slower than the selected perpendicular speed.

Let the selected perpendicular speed be:

\[ v_s=\frac{E}{B} \]

For a particle initially travelling rightwards, with the electric field upwards and the magnetic field out of the page, the following applies.

Why does this happen?

When \(v_\perp\) is below the selected speed, the magnetic force is too weak to balance the electric force. The electric force determines the direction of the net force.

When \(v_\perp\) is above the selected speed, the magnetic force is stronger, so the net force points in the opposite direction.

This is why you should not automatically use the magnetic right-hand rule alone when interpreting a trajectory inside a velocity selector.

You must first establish which force dominates.

For a more detailed explanation of the underlying force directions, see Electric Fields vs Magnetic Fields: HSC Physics Explained.

For additional practice identifying charge signs from magnetic trajectories, see How to Find a Particle’s Charge Sign from Its Curved Path.

05A reliable method for velocity selector questions

When you encounter a velocity selector question in an HSC Physics assessment, work through these steps.

1. Identify the field directions.

Draw the electric field, magnetic field, and initial particle velocity.

Check that the two forces oppose each other for the chosen direction of travel.

2. Resolve the velocity if necessary.

Find the angle between the velocity and the magnetic field.

If the particle enters at an angle, calculate:

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

3. Apply the velocity selector equation.

For undeflected motion:

\[ v_\perp=\frac{E}{B} \]

If the velocity is perpendicular to the magnetic field, you can use the total speed directly.

4. Compare the forces if the particle curves.

If the particle is travelling below the selected perpendicular speed, the electric force dominates.

If it is travelling above the selected perpendicular speed, the magnetic force dominates.

5. Determine the charge sign.

Use the right-hand rule for a positive charge, then reverse the direction for a negative charge.

Always account for the direction of travel and whether an electric field is still present.

06Questions and solutions

The following questions build from straightforward velocity selection to more challenging combinations of velocity components, force directions, and helical motion.

Question 1

A positively charged particle enters a velocity selector travelling horizontally to the right.

The electric field strength is \(1.8\times10^4\ \mathrm{N\,C^{-1}}\), directed upwards.

The magnetic field strength is \(3.0\times10^{-2}\ \mathrm T\), directed out of the page.

The particle travels perpendicular to the magnetic field.

Calculate the speed required for the particle to pass through undeflected.

Would a negatively charged particle require a different speed?

Solution 1

The required speed is \(6.0\times10^5\ \mathrm{m\,s^{-1}}\), and a negatively charged particle requires exactly the same speed.

Step 1: Identify the velocity selector equation.

The particle travels perpendicular to the magnetic field, so:

\[ v=\frac{E}{B} \]

Step 2: Substitute the values.

\[ v=\frac{1.8\times10^4}{3.0\times10^{-2}} \]

\[ v=6.0\times10^5\ \mathrm{m\,s^{-1}} \]

Step 3: Explain the charge independence.

The charge magnitude cancels when the electric and magnetic forces are equated.

A negative particle experiences both forces in the opposite directions to those experienced by a positive particle.

The forces still balance at the same speed.

Therefore, both particles pass straight through at:

\[ v=6.0\times10^5\ \mathrm{m\,s^{-1}} \]

The selector distinguishes velocity, not charge sign.

Question 2

A particle enters a velocity selector with a total speed of \(6.0\times10^5\ \mathrm{m\,s^{-1}}\).

The angle between its velocity and the magnetic field is \(45^\circ\).

The fields are:

• \(E=1.2\times10^4\ \mathrm{N\,C^{-1}}\), upwards.
• \(B=3.0\times10^{-2}\ \mathrm T\), out of the page.

The perpendicular component of the particle’s velocity points to the right.

Determine:

(a) The perpendicular velocity component.

(b) The parallel velocity component.

(c) Whether the particle passes straight through.

(d) The initial direction of deflection for a positive and a negative particle.

Solution 2

The particle does not pass straight through. A positive particle initially curves downwards, while a negative particle curves upwards.

Step 1: Calculate the perpendicular velocity component.

Use:

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

Substitute:

\[ v_\perp=(6.0\times10^5)\sin45^\circ \]

\[ v_\perp=4.24\times10^5\ \mathrm{m\,s^{-1}} \]

Step 2: Calculate the parallel velocity component.

Use:

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

\[ v_\parallel=(6.0\times10^5)\cos45^\circ \]

\[ v_\parallel=4.24\times10^5\ \mathrm{m\,s^{-1}} \]

Step 3: Calculate the selected perpendicular speed.

\[ v_s=\frac{E}{B} \]

\[ v_s=\frac{1.2\times10^4}{3.0\times10^{-2}} \]

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

Compare the two values:

\[ v_\perp=4.24\times10^5 \]

\[ v_s=4.0\times10^5 \]

The particle’s perpendicular velocity is greater than the selected value.

Therefore, the magnetic force is stronger than the electric force.

Step 4: Determine the deflection directions.

For a positive particle:

• Electric force acts upwards.
• Magnetic force acts downwards.
• Magnetic force is stronger.

The net force points downwards.

For a negative particle, both forces reverse, so the net force points upwards.

The particle therefore cannot pass straight through.

Notice that comparing the total speed with \(E/B\) would have been incorrect. Only the perpendicular component determines whether the forces balance.

Question 3

An unknown charged particle enters a velocity selector travelling to the right.

The fields are:

• Electric field strength \(9.0\times10^3\ \mathrm{N\,C^{-1}}\), directed upwards.
• Magnetic field strength \(3.0\times10^{-2}\ \mathrm T\), directed out of the page.

The particle’s total speed is \(4.0\times10^5\ \mathrm{m\,s^{-1}}\), and its velocity makes an angle of \(30^\circ\) with the magnetic field.

The particle initially curves downwards.

Determine:

(a) Its perpendicular velocity component.

(b) Whether the electric or magnetic force is stronger.

(c) The sign of its charge.

(d) Its parallel velocity component.

Solution 3

The particle has a negative charge. Its perpendicular velocity is \(2.0\times10^5\ \mathrm{m\,s^{-1}}\), and the electric force is stronger than the magnetic force.

Step 1: Calculate the perpendicular velocity.

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

\[ v_\perp=(4.0\times10^5)\sin30^\circ \]

\[ v_\perp=2.0\times10^5\ \mathrm{m\,s^{-1}} \]

Step 2: Compare the selected speed with the actual perpendicular speed.

The selected speed is:

\[ v_s=\frac{E}{B} \]

\[ v_s=\frac{9.0\times10^3}{3.0\times10^{-2}} \]

\[ v_s=3.0\times10^5\ \mathrm{m\,s^{-1}} \]

Since:

\[ v_\perp<v_s \]

the electric force is stronger.

We can also confirm this by comparing their magnitudes:

\[ F_E=|q|E \]

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

Therefore:

\[ \frac{F_B}{F_E} =\frac{v_\perp B}{E} \]

\[ =\frac{(2.0\times10^5)(3.0\times10^{-2})} {9.0\times10^3} \]

\[ =0.667 \]

The magnetic force is approximately two-thirds of the electric force.

Step 3: Determine the charge sign.

For a positive particle:

• The electric force points upwards.
• The magnetic force points downwards.
• The electric force dominates.

The net force would point upwards.

However, the observed trajectory curves downwards.

Therefore, the particle must be negatively charged, reversing both force directions.

Step 4: Calculate the parallel velocity component.

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

\[ v_\parallel=(4.0\times10^5)\cos30^\circ \]

\[ v_\parallel=3.46\times10^5\ \mathrm{m\,s^{-1}} \]

This component does not contribute to the magnetic force.

The particle therefore has a substantial velocity component along the magnetic field, even though its transverse motion is being deflected.

Question 4

Three identical charged particles enter the same velocity selector.

The electric field strength is \(1.0\times10^4\ \mathrm{N\,C^{-1}}\), and the magnetic field strength is \(2.5\times10^{-2}\ \mathrm T\).

All three particles have:

• The same total speed of \(8.0\times10^5\ \mathrm{m\,s^{-1}}\).
• A perpendicular velocity component directed to the right.

Their angles to the magnetic field are:

• Particle A: \(30^\circ\).
• Particle B: \(90^\circ\).
• Particle C: \(150^\circ\).

Determine:

(a) Which particles pass straight through the selector.

(b) The parallel velocity component of particles A and C, including its direction relative to the magnetic field.

(c) Whether the selector can distinguish particles A and C.

(d) For a positively charged particle B, determine its initial direction of deflection.

Solution 4

Particles A and C pass straight through, while particle B is deflected downwards if it is positively charged.

The key is that the selector selects the perpendicular velocity component, not the total speed.

Step 1: Calculate the selected perpendicular speed.

\[ v_s=\frac{E}{B} \]

\[ v_s=\frac{1.0\times10^4}{2.5\times10^{-2}} \]

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

Each particle must have this perpendicular component to pass straight through.

Step 2: Calculate the perpendicular components.

For particle A:

\[ v_{\perp A}=(8.0\times10^5)\sin30^\circ \]

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

For particle B:

\[ v_{\perp B}=(8.0\times10^5)\sin90^\circ \]

\[ v_{\perp B}=8.0\times10^5\ \mathrm{m\,s^{-1}} \]

For particle C:

\[ v_{\perp C}=(8.0\times10^5)\sin150^\circ \]

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

Compare these with the selected speed:

Step 3: Calculate the parallel components of A and C.

For particle A:

\[ v_{\parallel A} =(8.0\times10^5)\cos30^\circ \]

\[ v_{\parallel A}=+6.93\times10^5\ \mathrm{m\,s^{-1}} \]

The positive sign indicates motion along the magnetic field.

For particle C:

\[ v_{\parallel C} =(8.0\times10^5)\cos150^\circ \]

\[ v_{\parallel C}=-6.93\times10^5\ \mathrm{m\,s^{-1}} \]

The negative sign indicates motion against the magnetic field.

Step 4: Explain why A and C both pass straight through.

Particles A and C have identical perpendicular velocity components.

Their parallel velocity components are equal in magnitude but opposite in direction.

The magnetic force depends only on the perpendicular component, so both particles experience the same force balance.

The electric force also has no component along the magnetic field.

Therefore, the selector cannot distinguish between A and C based on their parallel velocity components.

This is a useful limitation of velocity selectors: they select a particular perpendicular velocity, not one unique three-dimensional velocity vector.

Step 5: Determine the deflection of particle B.

Particle B has twice the selected perpendicular speed.

Its magnetic force is therefore twice the electric force in magnitude.

For a positive particle:

• Electric force acts upwards.
• Magnetic force acts downwards.
• Magnetic force dominates.

The net force points downwards.

Particle B therefore initially curves downwards.

Question 5

A hypothetical singly charged ion has a mass of \(1.6\times10^{-27}\ \mathrm{kg}\) and a charge magnitude of \(1.6\times10^{-19}\ \mathrm C\).

It enters a velocity selector containing:

• An electric field of \(8.0\times10^3\ \mathrm{N\,C^{-1}}\), directed upwards.
• A magnetic field of \(2.0\times10^{-2}\ \mathrm T\), directed out of the page.

Its perpendicular velocity component points to the right, and it also has a non-zero velocity component along the magnetic field.

The ion passes straight through the selector.

It then enters a second region where the electric field is absent, but the same magnetic field continues.

In this second region, the ion follows a helical trajectory with:

• A circular radius of \(0.20\ \mathrm m\).
• A pitch of \(0.94\ \mathrm m\) per revolution along the direction of the magnetic field.
• An initial downward curvature in the plane perpendicular to the magnetic field.

Determine:

(a) The perpendicular velocity component selected by the first region.

(b) The time taken for one complete revolution in the second region.

(c) The parallel velocity component.

(d) The total speed of the ion.

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

(f) Whether the ion is positively or negatively charged.

A student argues that the radius of the helical trajectory gives the total particle speed, so the parallel velocity must be zero.

Explain precisely why this conclusion is incorrect.

Solution 5

The ion is positively charged, and its total speed is approximately \(5.0\times10^5\ \mathrm{m\,s^{-1}}\). The student’s conclusion is incorrect because the radius determines the perpendicular velocity component, not the total speed.

This question combines velocity selection with the geometry of helical motion.

Step 1: Calculate the perpendicular velocity component.

The particle passes straight through the first region.

Therefore:

\[ v_\perp=\frac{E}{B} \]

Substitute:

\[ v_\perp=\frac{8.0\times10^3}{2.0\times10^{-2}} \]

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

This component remains unchanged in the second region because a magnetic field does no work on the particle.

Step 2: Calculate the time for one complete revolution.

In the second region, the particle follows a helix.

The perpendicular component produces circular motion around the magnetic field direction.

For one complete revolution, the particle travels a circular circumference of:

\[ d=2\pi r \]

The time for one revolution is therefore:

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

Substitute:

\[ T=\frac{2\pi(0.20)}{4.0\times10^5} \]

\[ T=3.14\times10^{-6}\ \mathrm s \]

Step 3: Calculate the parallel velocity component.

The pitch is the distance travelled along the magnetic field during one complete revolution.

Therefore:

\[ p=v_\parallel T \]

Rearrange:

\[ v_\parallel=\frac{p}{T} \]

Substitute:

\[ v_\parallel= \frac{0.94}{3.14\times10^{-6}} \]

\[ v_\parallel=2.99\times10^5\ \mathrm{m\,s^{-1}} \]

The particle moves along the magnetic field at approximately \(3.0\times10^5\ \mathrm{m\,s^{-1}}\).

Step 4: Calculate the total speed.

The perpendicular and parallel velocity components are at right angles.

Use Pythagoras:

\[ v^2=v_\perp^2+v_\parallel^2 \]

Therefore:

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

Substitute:

\[ v=\sqrt{(4.0\times10^5)^2+ (2.99\times10^5)^2} \]

\[ v\approx5.0\times10^5\ \mathrm{m\,s^{-1}} \]

This is the total speed of the ion.

Step 5: Calculate the angle to the magnetic field.

Recall:

\[ \sin\theta=\frac{v_\perp}{v} \]

Substitute:

\[ \sin\theta= \frac{4.0\times10^5}{5.0\times10^5} \]

\[ \sin\theta=0.80 \]

Therefore:

\[ \theta=\sin^{-1}(0.80) \]

\[ \theta\approx53^\circ \]

The particle travels at approximately \(53^\circ\) to the magnetic field.

Step 6: Determine the charge sign.

The perpendicular velocity points right, and the magnetic field points out of the page.

For a positive particle, the right-hand rule predicts a magnetic force downwards.

For a negative particle, the force would point upwards.

The observed transverse curvature is downwards.

Therefore, the ion is positively charged.

Its parallel velocity component does not affect this conclusion because a velocity component parallel to the magnetic field produces no magnetic force.

Step 7: Explain the student’s mistake.

The student has assumed that the radius of the helical trajectory depends on the total speed.

That is incorrect.

The magnetic force is:

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

The circular motion is produced by this perpendicular component, so the centripetal force equation is:

\[ |q|v_\perp B= \frac{mv_\perp^2}{r} \]

Rearranging:

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

The radius therefore depends on \(v_\perp\), not the total speed.

We can verify this using the given particle data:

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

\[ r= \frac{(1.6\times10^{-27})(4.0\times10^5)} {(1.6\times10^{-19})(2.0\times10^{-2})} \]

\[ r=0.20\ \mathrm m \]

This agrees with the measured radius.

The non-zero pitch independently demonstrates that the particle has a parallel velocity component.

The student’s method would incorrectly discard that component and underestimate the total speed.

The important distinction is that the radius describes the circular part of the motion, while the pitch describes the motion along the magnetic field.

Together, these measurements reveal the particle’s complete velocity.

07What to remember

When finding the speed selected by crossed electric and magnetic fields, the most important question is whether the particle’s velocity is perpendicular to the magnetic field.

For perpendicular motion:

\[ v=\frac{E}{B} \]

For oblique motion:

\[ v_\perp=v\sin\theta=\frac{E}{B} \]

And when interpreting a curved particle trajectory, remember:

• A straight path through a selector means the electric and magnetic forces balance for a charged particle.
• Both positive and negative particles can pass straight through.
• A magnetic field acts only on the perpendicular velocity component.
• In a magnetic-only region, the right-hand rule identifies the charge sign when the direction of travel is known.
• Inside a velocity selector, you must compare the electric and magnetic forces before interpreting the direction of curvature.
• A helical trajectory has both perpendicular and parallel velocity components.

Once you can identify the selected perpendicular speed, the next useful step is understanding how that component determines the radius and period of a particle’s circular or helical motion.