Why Electric Fields Do Work but Static Magnetic Fields Don’t
Discover why electric fields can change a charged particle's kinetic energy while static magnetic fields only change its direction. Learn how perpendicular velocity components, particle curvature, and charge sign explain their different effects, with worked examples and practice questions.
A charged particle enters a magnetic field and starts following a curved path. Its direction changes, but its speed stays exactly the same. Now place the same particle in an electric field. Its speed can increase or decrease. Both fields exert forces on charged particles, so why does only one transfer energy to the particle?
The answer comes down to the direction of the force. An electric field can exert a force with a component parallel to the particle's motion, transferring energy through work. A magnetic field exerts a force perpendicular to the particle's instantaneous velocity, changing its direction but not its kinetic energy.
This gives us two important ideas to keep separate: **a force can change an object's motion without changing its speed, and a curved particle path can reveal the sign of its charge.**
## Why a magnetic force cannot do work
Imagine pushing a shopping trolley forwards. Your force points in the direction the trolley moves, so you transfer energy to it and increase its kinetic energy.
Now imagine pushing sideways while the trolley rolls forwards. Your push changes its direction, but an ideal sideways force does not increase its speed.
A magnetic force behaves like the second push. It acts sideways relative to the particle's instantaneous motion.
[IMAGE: A charged particle moving from left to right through a uniform magnetic field directed into the page. Show the particle's curved path and a magnetic force arrow perpendicular to its instantaneous velocity, with a small velocity tangent to the path. Include a second inset showing an electric force parallel to velocity and the resulting increase in speed.]
The magnetic force on a moving charged particle is described by:
\[
F_B=|q|vB\sin\theta
\]
Here:
- \(F_B\) is the magnitude of the magnetic force, measured in newtons (N).
- \(q\) is the particle's charge, measured in coulombs (C).
- \(v\) is its speed, measured in metres per second (m s\(^{-1}\)).
- \(B\) is the magnetic field strength, measured in tesla (T).
- \(\theta\) is the angle between the velocity and magnetic field directions.
The crucial feature is that the magnetic force is perpendicular to the velocity whenever it is non-zero.
Work depends on the component of force in the direction of displacement:
\[
W=Fs\cos\phi
\]
where \(s\) is the displacement and \(\phi\) is the angle between the force and displacement.
For a magnetic force, the angle between the force and the instantaneous velocity is \(90^\circ\). Therefore:
\[
W_B=Fs\cos90^\circ=0
\]
The magnetic field does no work on the particle.
This remains true even when the particle follows a curved path. The velocity direction changes continuously, but the magnetic force remains perpendicular to that changing velocity at every instant.
**The important distinction:** zero work does not mean zero force. A magnetic field can change momentum and direction without changing kinetic energy.
## Why an electric field can do work
An electric field behaves differently because its force does not have to be perpendicular to the particle's velocity.
The electric force is:
\[
F_E=qE
\]
where \(E\) is electric field strength, measured in newtons per coulomb (N C\(^{-1}\)).
If the electric force acts along the particle's displacement, it does positive work. If it acts against the displacement, it does negative work.
For a uniform electric field and a straight displacement:
\[
W_E=qEs\cos\phi
\]
Here, \(\phi\) is the angle between the electric force and displacement.
For example, a positive charge moving in the direction of an electric field accelerates. The field does positive work, increasing the particle's kinetic energy.
A negative charge experiences a force opposite to the electric field. It can still gain kinetic energy, but its acceleration is in the opposite direction.
[IMAGE: Two side-by-side parallel-plate electric field setups. In the first, a positive particle accelerates towards the negative plate, with electric force and velocity aligned. In the second, a negative particle accelerates towards the positive plate. Show field direction, charge signs, force arrows, and increasing speed.]
This is the central difference:
| Feature | Electric force | Magnetic force |
|---|---|---|
| Direction | Along or against the electric field for a positive or negative charge respectively | Perpendicular to velocity |
| Can change direction? | Yes | Yes |
| Can change speed? | Yes | No, by itself |
| Can change kinetic energy? | Yes, through work | No |
| Can do work on a moving particle? | Yes, depending on displacement | No |
A useful qualification: an electric field does not automatically do work in every situation. If its force is perpendicular to the displacement, its work is zero too. The difference is that an electric force *can* have a component along the displacement, whereas a magnetic force cannot.
## Resolving velocity: why the perpendicular component matters
There is another detail that makes magnetic motion more interesting. A particle does not have to enter a magnetic field at right angles.
Suppose a proton enters a magnetic field at an angle of \(30^\circ\). Its velocity has two components:
- \(v_\perp\): the component perpendicular to the magnetic field.
- \(v_\parallel\): the component parallel to the magnetic field.
Using trigonometry:
\[
v_\perp=v\sin\theta
\]
\[
v_\parallel=v\cos\theta
\]
Only \(v_\perp\) contributes to the magnetic force:
\[
F_B=|q|v_\perp B
\]
The parallel component experiences no magnetic force. It continues along the magnetic field direction at constant velocity, assuming a uniform field and no other forces.
[IMAGE: Three-dimensional scientific illustration of a positively charged particle entering a uniform magnetic field at 30 degrees. Show parallel magnetic field lines, the total velocity vector, perpendicular and parallel velocity components, and the resulting helical trajectory. Include a separate end-on view showing the circular projection. Do not include a completed force-direction answer.]
This produces three possible types of motion:
| Velocity relative to magnetic field | Particle's path |
|---|---|
| Entirely perpendicular | Circle |
| Both perpendicular and parallel components | Helix |
| Entirely parallel | Straight line |
The key is that the perpendicular component changes direction but not magnitude. The parallel component stays constant.
Therefore, the total speed remains constant:
\[
v=\sqrt{v_\perp^2+v_\parallel^2}
\]
The magnetic field changes the direction of the velocity vector without changing its magnitude.
### How to identify the charge from curvature
A curved particle track can also tell you whether a particle is positively or negatively charged.
The vector equation for magnetic force is:
\[
\vec F_B=q(\vec v\times\vec B)
\]
For a positive charge, use the right-hand rule to determine the direction of \(\vec v\times\vec B\).
For a negative charge, reverse the predicted direction.
Consider a particle whose perpendicular velocity component points east, while the magnetic field points into the page.
- A positive particle experiences a force northwards.
- A negative particle experiences a force southwards.
[IMAGE: Side-by-side comparison of two particle tracks through the same magnetic field directed into the page. Both particles initially travel east. Show the positive particle curving north and the negative particle curving south. Include the into-page cross symbols and velocity arrows, but no force arrows.]
Use this decision process when interpreting a particle track:
1. Identify the direction of travel. A curved line without an arrow is not enough.
2. Identify the magnetic field direction.
3. Resolve the velocity into its perpendicular component if necessary.
4. Apply the right-hand rule as though the particle were positive.
5. Compare the predicted force direction with the observed initial curvature.
6. If the observed curvature is opposite to the prediction, the particle is negative.
Be careful with the first step. If you do not know which direction the particle travelled, you generally cannot determine its charge sign from the curve alone.
### Worked example: a proton entering a magnetic field at an angle
A proton travels at \(5.0\times10^6\text{ m s}^{-1}\) into a uniform magnetic field of \(0.20\text{ T}\). Its velocity makes an angle of \(30^\circ\) with the magnetic field, which points into the page. The perpendicular velocity component initially points east.
The proton has a charge of \(+1.60\times10^{-19}\text{ C}\) and a mass of \(1.67\times10^{-27}\text{ kg}\).
Calculate:
- The parallel and perpendicular velocity components.
- The magnetic force.
- The radius of the circular component of its motion.
- Whether the proton's kinetic energy changes.
[IMAGE: A positively charged proton entering a magnetic field directed into the page. Show its total velocity at 30 degrees to the field, with the perpendicular component pointing east and parallel component pointing into the page. Include the resulting helical trajectory, but leave force arrows out of the question illustration.]
**Step 1**: Resolve the velocity into its components.
The perpendicular component is:
\[
v_\perp=v\sin\theta
\]
\[
v_\perp=5.0\times10^6\sin30^\circ
\]
\[
\boxed{v_\perp=2.5\times10^6\text{ m s}^{-1}}
\]
The parallel component is:
\[
v_\parallel=v\cos\theta
\]
\[
v_\parallel=5.0\times10^6\cos30^\circ
\]
\[
\boxed{v_\parallel=4.33\times10^6\text{ m s}^{-1}}
\]
**Step 2**: Calculate the magnetic force.
Only the perpendicular velocity contributes:
\[
F_B=|q|v_\perp B
\]
\[
F_B=(1.60\times10^{-19})(2.5\times10^6)(0.20)
\]
\[
\boxed{F_B=8.0\times10^{-14}\text{ N}}
\]
**Step 3**: Calculate the radius of the circular component.
The magnetic force supplies the centripetal force:
\[
|q|v_\perp B=\frac{mv_\perp^2}{r}
\]
Rearranging:
\[
r=\frac{mv_\perp}{|q|B}
\]
Substitute:
\[
r=\frac{(1.67\times10^{-27})(2.5\times10^6)}
{(1.60\times10^{-19})(0.20)}
\]
\[
\boxed{r=0.13\text{ m}}
\]
**Step 4**: Determine the effect on kinetic energy.
The proton moves east in a magnetic field directed into the page.
Using the right-hand rule, its magnetic force initially points north. Its perpendicular velocity therefore changes direction, producing circular motion.
However, the parallel component continues unchanged, and the magnitude of the perpendicular component also stays constant.
The proton follows a helix, but its total speed remains \(5.0\times10^6\text{ m s}^{-1}\).
Therefore:
\[
\boxed{W_B=0,\qquad \Delta K=0}
\]
The magnetic field changes the proton's trajectory without transferring energy to it.
## Testing the idea: measuring speed before and after a magnetic field
So far, we have used the equations to predict what should happen. But how could an experiment distinguish a particle that has changed direction from one that has changed speed?
A photograph of a curved track is not enough. We need to measure the particle's speed independently.
### The experimental problem
Imagine a narrow beam of protons travelling through an evacuated tube.
The beam enters a region containing a uniform magnetic field. A detector records its curved trajectory.
The unanswered question is whether the proton leaves the magnetic region with the same speed it had on entering.
### Building the apparatus
A suitable experimental arrangement has four main stages.
[IMAGE: A labelled scientific apparatus diagram showing a proton source at the left, two entrance timing detectors separated by 0.10 m, a uniform magnetic-field region that bends the proton beam through 90 degrees, and two exit timing detectors separated by 0.10 m along the outgoing tangent. Show the beam direction, magnetic field into the page, and detector positions. Do not show measured results.]
1. An ion source produces a narrow, pulsed proton beam.
2. Two timing detectors sit in the field-free region before the magnet. Their known separation allows the initial speed to be calculated.
3. The beam enters a uniform magnetic field directed perpendicular to its velocity. It follows a circular arc.
4. Two more timing detectors sit beyond the magnetic region, aligned with the outgoing beam. They measure the final speed.
The detectors measure the time taken for the proton pulse to travel between two known positions.
### What should the experiment show?
If the magnetic field does no work, the proton's speed should be identical before and after the magnetic region.
Its direction will have changed, but its kinetic energy will not.
For example, suppose the separation between each pair of detectors is \(0.10\text{ m}\).
The illustrative measurements are:
| Measurement | Before magnet | After magnet |
|---|---:|---:|
| Detector separation | 0.10 m | 0.10 m |
| Transit time | 25 ns | 25 ns |
| Calculated speed | \(4.0\times10^6\text{ m s}^{-1}\) | \(4.0\times10^6\text{ m s}^{-1}\) |
These are illustrative data, not measurements from a historical experiment.
### Worked measurement: calculate the initial and final speeds
Use:
\[
v=\frac{d}{t}
\]
where \(d\) is detector separation and \(t\) is transit time.
Before the magnetic field:
\[
v_i=\frac{0.10}{25\times10^{-9}}
\]
\[
\boxed{v_i=4.0\times10^6\text{ m s}^{-1}}
\]
After the magnetic field:
\[
v_f=\frac{0.10}{25\times10^{-9}}
\]
\[
\boxed{v_f=4.0\times10^6\text{ m s}^{-1}}
\]
Within the measurement uncertainty, the speeds are identical.
The particle has changed direction without changing speed.
### Why the first experimental design is incomplete
Suppose we had used only a fluorescent screen to record the curved beam.
We would know that the magnetic field changed the particle's direction. We would not independently know whether its speed changed.
A curved track is evidence of acceleration because the velocity vector changes direction. It is not evidence of an increase or decrease in kinetic energy.
There is another possible problem. If the proton enters at an angle to the magnetic field, its path becomes helical rather than circular.
The revised experiment controls this by arranging the entry velocity perpendicular to the field. The timing detectors also provide an independent measurement of speed on either side of the magnetic region.
The conclusion follows from two separate observations:
- The particle's trajectory changes.
- Its measured speed remains constant.
This is precisely what the magnetic-force model predicts.
The idealised experiment assumes a stable magnetic field during the measurement, negligible collisions, and no significant energy loss through radiation. These are appropriate simplifying assumptions for the HSC treatment.
## What changes when electric and magnetic fields act together?
Now consider a particle moving through a region containing both fields.
The total electromagnetic force is:
\[
\vec F=q(\vec E+\vec v\times\vec B)
\]
The two forces can act simultaneously, but they have different effects on kinetic energy.
The magnetic force changes the trajectory without doing work. The electric force can transfer energy.
To calculate the work correctly, keep the sign convention consistent:
\[
W_E=q\vec E\cdot\Delta\vec r
\]
For a uniform field, if \(\theta_E\) is the angle between the electric field and displacement:
\[
W_E=qEs\cos\theta_E
\]
Alternatively, if the angle is measured from the electric force, use:
\[
W_E=|q|Es\cos\phi_F
\]
Here, \(\phi_F\) is the angle between the electric force and displacement.
The work-energy theorem gives:
\[
W_{\text{net}}=\Delta K
\]
Since the magnetic field contributes no work:
\[
\boxed{\Delta K=W_E}
\]
provided no other forces transfer energy.
### Worked example: a charged particle curves while gaining kinetic energy
An alpha particle enters a region where an electric field of \(1.20\times10^4\text{ N C}^{-1}\) points east. A uniform magnetic field of \(0.30\text{ T}\) points into the page.
The particle initially travels east at \(1.50\times10^6\text{ m s}^{-1}\).
Its charge is \(+3.20\times10^{-19}\text{ C}\), and its mass is \(6.64\times10^{-27}\text{ kg}\).
While inside the fields, it moves through a net horizontal displacement of \(0.20\text{ m}\) eastwards.
Calculate its final speed.
[IMAGE: A positive alpha particle entering an overlapping electric and magnetic field region. Show the electric field directed east, magnetic field directed into the page, initial velocity east, and the curved trajectory bending northwards. Mark the horizontal displacement of 0.20 m. Do not include the calculated force magnitudes or final speed.]
**Step 1**: Identify which field does work.
The alpha particle is positively charged.
Its electric force points east, in the same direction as its net horizontal displacement. The electric field therefore does positive work.
The magnetic force initially points north, perpendicular to the velocity. It remains perpendicular to the instantaneous velocity throughout the motion.
Therefore:
\[
W_B=0
\]
**Step 2**: Calculate the work done by the electric field.
Because the electric field is uniform and the particle moves \(0.20\text{ m}\) eastwards:
\[
W_E=qE\Delta x
\]
\[
W_E=(3.20\times10^{-19})
(1.20\times10^4)(0.20)
\]
\[
\boxed{W_E=7.68\times10^{-16}\text{ J}}
\]
**Step 3**: Calculate the initial kinetic energy.
\[
K_i=\frac12mv_i^2
\]
\[
K_i=\frac12(6.64\times10^{-27})
(1.50\times10^6)^2
\]
\[
\boxed{K_i=7.47\times10^{-15}\text{ J}}
\]
**Step 4**: Calculate the final kinetic energy.
Using the work-energy theorem:
\[
\Delta K=W_E
\]
Therefore:
\[
K_f=K_i+W_E
\]
\[
K_f=7.47\times10^{-15}
+7.68\times10^{-16}
\]
\[
\boxed{K_f=8.238\times10^{-15}\text{ J}}
\]
**Step 5**: Calculate the final speed.
Use:
\[
K_f=\frac12mv_f^2
\]
Rearrange:
\[
v_f=\sqrt{\frac{2K_f}{m}}
\]
Substitute:
\[
v_f=\sqrt{
\frac{2(8.238\times10^{-15})}
{6.64\times10^{-27}}
}
\]
\[
\boxed{v_f=1.58\times10^6\text{ m s}^{-1}}
\]
The alpha particle has gained kinetic energy and increased its speed.
Notice what happened: the magnetic field curved its trajectory, while the electric field increased its kinetic energy. Both fields acted, but only one transferred energy.
## An important distinction: electric fields do not always do work
It would be easy to take the previous examples and conclude that electric fields always change a particle's kinetic energy.
That is not correct.
An electric field can do work, but the work depends on the displacement relative to the electric force.
For example, consider a negative particle moving through a uniform electric field directed east. Its electric force points west.
If its net displacement is northwards, the electric force is perpendicular to the displacement.
Therefore:
\[
W_E=Fs\cos90^\circ=0
\]
The electric field does no net work over that displacement either.
The difference between the fields is not that one always does work and the other never does.
It is that a magnetic force is *always* perpendicular to the instantaneous velocity, whereas an electric force can point along, against, or perpendicular to the motion.
### A subtle limitation: changing magnetic fields
When we say a magnetic field cannot do work, we are talking about the magnetic component of the Lorentz force.
Even if a magnetic field changes with time, its magnetic force on an individual moving charge remains perpendicular to the velocity.
However, a changing magnetic field can induce an electric field. That induced electric field can do work on charged particles.
The energy transfer is still associated with the electric force, not the magnetic component.
For the HSC model, the important case is a charged particle moving through a steady magnetic field, with no other energy-transfer mechanisms.
## A quick decision method
When a question involves charged particles in electric and magnetic fields, separate the trajectory problem from the energy problem.
| What are you trying to determine? | What you should use |
|---|---|
| Magnetic force magnitude | \(F_B=|q|v_\perp B\) |
| Magnetic force direction | Right-hand rule for positive charge; reverse for negative |
| Circular orbit radius | \(r=\frac{mv_\perp}{|q|B}\) |
| Work done by magnetic field | \(W_B=0\) |
| Work done by a uniform electric field | \(W_E=q\vec E\cdot\Delta\vec r\) |
| Change in kinetic energy | \(W_{\text{net}}=\Delta K\) |
The most common mistake is to see a particle accelerating and assume it must be speeding up.
Remember that acceleration means a change in velocity. Velocity includes both speed and direction.
A magnetic field can continuously change direction without changing speed.
---
## Questions and solutions
### Question 1
A charged particle travels eastwards at \(2.0\times10^6\text{ m s}^{-1}\) through a uniform magnetic field of \(0.15\text{ T}\), directed into the page.
Its charge magnitude is \(1.60\times10^{-19}\text{ C}\), and its trajectory initially curves northwards.
Determine:
a) The sign of the particle's charge.
b) The magnitude of the magnetic force.
c) Whether its kinetic energy changes.
[IMAGE: Top-down view of a particle travelling east into a uniform magnetic field directed into the page. Show its initial trajectory curving northwards, the velocity arrow, and the magnetic field crosses. Do not show a force arrow.]
### Solution 1
The particle is positively charged, experiences a magnetic force of \(4.8\times10^{-14}\text{ N}\), and undergoes no change in kinetic energy.
**a) Determine the charge sign.**
For a positive charge, the right-hand rule applied to an eastward velocity and a magnetic field into the page predicts a northward force.
The observed trajectory initially curves northwards, matching this prediction.
Therefore, the particle is positive.
**b) Calculate the magnetic force.**
The velocity is perpendicular to the magnetic field, so:
\[
F_B=|q|vB
\]
Substitute:
\[
F_B=(1.60\times10^{-19})
(2.0\times10^6)(0.15)
\]
\[
\boxed{F_B=4.8\times10^{-14}\text{ N}}
\]
**c) Determine the change in kinetic energy.**
The magnetic force is perpendicular to the instantaneous velocity.
Therefore, it does no work:
\[
W_B=0
\]
Using the work-energy theorem:
\[
\Delta K=W_B=0
\]
The particle changes direction but maintains its original speed.
### Question 2
An electron enters a uniform magnetic field of \(0.25\text{ T}\), directed into the page.
Its total speed is \(4.0\times10^6\text{ m s}^{-1}\), and its velocity makes an angle of \(30^\circ\) with the magnetic field.
The perpendicular velocity component initially points east, while the parallel component points into the page.
The observed circular projection of its helical trajectory curves southwards.
Given:
- Electron charge magnitude: \(1.60\times10^{-19}\text{ C}\)
- Electron mass: \(9.11\times10^{-31}\text{ kg}\)
Calculate:
a) The parallel and perpendicular velocity components.
b) The magnetic force magnitude.
c) The radius of the circular component.
d) The particle's charge sign and whether its kinetic energy changes.
[IMAGE: Three-dimensional electron trajectory entering a uniform magnetic field directed into the page. Show the total velocity at 30 degrees to the field, its eastward perpendicular component, its into-page parallel component, and the helical path. Include an end-on circular projection curving south. Do not show the magnetic force.]
### Solution 2
The electron has a negative charge, a perpendicular speed of \(2.0\times10^6\text{ m s}^{-1}\), a parallel speed of \(3.46\times10^6\text{ m s}^{-1}\), and a circular component radius of approximately \(45.6\text{ μm}\). Its kinetic energy remains constant.
**a) Resolve the velocity.**
The perpendicular component is:
\[
v_\perp=v\sin30^\circ
\]
\[
\boxed{v_\perp=2.0\times10^6\text{ m s}^{-1}}
\]
The parallel component is:
\[
v_\parallel=v\cos30^\circ
\]
\[
\boxed{v_\parallel=3.46\times10^6\text{ m s}^{-1}}
\]
**b) Calculate the magnetic force.**
Only the perpendicular component contributes:
\[
F_B=|q|v_\perp B
\]
\[
F_B=(1.60\times10^{-19})
(2.0\times10^6)(0.25)
\]
\[
\boxed{F_B=8.0\times10^{-14}\text{ N}}
\]
**c) Calculate the circular radius.**
Use:
\[
r=\frac{mv_\perp}{|q|B}
\]
Substitute:
\[
r=\frac{(9.11\times10^{-31})(2.0\times10^6)}
{(1.60\times10^{-19})(0.25)}
\]
\[
\boxed{r=4.56\times10^{-5}\text{ m}}
\]
This is approximately \(45.6\text{ μm}\).
**d) Determine the charge sign and energy change.**
For a positive particle travelling east through a magnetic field directed into the page, the magnetic force points north.
However, the observed circular projection curves southwards.
The actual force must therefore be opposite to the positive-charge prediction.
The particle is negative.
Its non-zero parallel velocity component means its overall trajectory is helical rather than circular.
However, neither velocity component changes magnitude in this uniform magnetic field. The magnetic force does no work, so the electron's kinetic energy remains constant.
### Question 3
An electron moves through a region containing:
- A uniform electric field of \(2.0\times10^3\text{ N C}^{-1}\), directed east.
- A static magnetic field directed into the page.
The electron follows a curved path of length \(0.12\text{ m}\) from point A to point B.
The two endpoints have the same eastward coordinate and are separated by \(0.080\text{ m}\) in the north-south direction.
The electron's charge magnitude is \(1.60\times10^{-19}\text{ C}\).
Assume only the electric and magnetic forces act.
a) Calculate the work done by the electric field.
b) Calculate the work done by the magnetic field.
c) Determine the change in kinetic energy.
d) Explain why using the curved path length in \(W=Fs\) gives an incorrect answer.
[IMAGE: Top-down illustration of a curved electron trajectory from A to B through a region with an eastward electric field and magnetic field into the page. Show A and B on the same north-south line, the 0.080 m north-south displacement, and the 0.12 m curved path length. Do not show force arrows.]
### Solution 3
Both fields do zero net work, so the electron's kinetic energy is unchanged.
**a) Calculate the electric work.**
The electron's displacement has no eastward component.
Since the electric field points east, the electric force on the electron points west.
The force is perpendicular to the net displacement.
Using the dot product:
\[
W_E=qE\Delta x
\]
Since:
\[
\Delta x=0
\]
we obtain:
\[
\boxed{W_E=0\text{ J}}
\]
**b) Calculate the magnetic work.**
The magnetic force is always perpendicular to the instantaneous velocity.
Therefore:
\[
\boxed{W_B=0\text{ J}}
\]
**c) Calculate the change in kinetic energy.**
The work-energy theorem gives:
\[
\Delta K=W_E+W_B
\]
\[
\Delta K=0+0
\]
\[
\boxed{\Delta K=0\text{ J}}
\]
The electron may change direction along its path, but its final speed equals its initial speed.
**d) Explain the incorrect use of path length.**
The student might calculate:
\[
W=Fs
\]
using the full curved path length of \(0.12\text{ m}\).
This ignores the angle between force and displacement.
Work depends on the component of displacement along the force, not simply the total distance travelled.
The electron's curved trajectory does not mean the electric field has transferred energy overall. Any positive and negative work contributions along the path cancel.
This question exposes an important distinction: even an electric field does not necessarily change kinetic energy.
### Question 4
Two identical protons enter the same uniform magnetic field of \(0.20\text{ T}\).
Particle A enters perpendicular to the magnetic field at \(2.0\times10^6\text{ m s}^{-1}\).
Particle B enters at an angle of \(30^\circ\) to the magnetic field at \(4.0\times10^6\text{ m s}^{-1}\).
Both particles have:
\[
q=+1.60\times10^{-19}\text{ C}
\]
\[
m=1.67\times10^{-27}\text{ kg}
\]
Determine:
a) The perpendicular velocity component of each particle.
b) Whether their magnetic forces have the same magnitude.
c) Whether their circular orbit radii are equal.
d) The ratio of their kinetic energies.
e) The shape of each trajectory.
[IMAGE: Comparison of two proton trajectories in the same uniform magnetic field. Particle A travels perpendicular to the field and follows a circular path. Particle B enters at 30 degrees and follows a helix. Show the same perpendicular velocity component magnitude for both, but a larger total velocity for B. Include an end-on projection illustrating equal circular radii. Do not include force arrows.]
### Solution 4
Both protons have the same magnetic force magnitude and circular radius, but particle B has four times the kinetic energy of particle A. Particle A follows a circle, while particle B follows a helix.
**a) Calculate the perpendicular velocity components.**
For particle A, all its velocity is perpendicular to the field:
\[
v_{\perp A}=2.0\times10^6\text{ m s}^{-1}
\]
For particle B:
\[
v_{\perp B}=v_B\sin30^\circ
\]
\[
v_{\perp B}=4.0\times10^6(0.5)
\]
\[
v_{\perp B}=2.0\times10^6\text{ m s}^{-1}
\]
Therefore:
\[
\boxed{v_{\perp A}=v_{\perp B}}
\]
**b) Compare the magnetic forces.**
The magnetic force depends on perpendicular velocity:
\[
F_B=|q|v_\perp B
\]
For both particles:
\[
F_B=(1.60\times10^{-19})
(2.0\times10^6)(0.20)
\]
\[
\boxed{F_{BA}=F_{BB}=6.4\times10^{-14}\text{ N}}
\]
Their magnetic forces have equal magnitudes.
**c) Compare the circular radii.**
Use:
\[
r=\frac{mv_\perp}{|q|B}
\]
Since both particles have the same mass, charge magnitude, magnetic field strength, and perpendicular velocity:
\[
r_A=r_B
\]
Numerically:
\[
r=\frac{(1.67\times10^{-27})(2.0\times10^6)}
{(1.60\times10^{-19})(0.20)}
\]
\[
\boxed{r_A=r_B=0.104\text{ m}}
\]
**d) Compare kinetic energies.**
Kinetic energy depends on total speed, not just perpendicular speed.
For particle A:
\[
K_A=\frac12mv_A^2
\]
\[
K_A=\frac12(1.67\times10^{-27})(2.0\times10^6)^2
\]
\[
\boxed{K_A=3.34\times10^{-15}\text{ J}}
\]
For particle B:
\[
K_B=\frac12mv_B^2
\]
\[
K_B=\frac12(1.67\times10^{-27})(4.0\times10^6)^2
\]
\[
\boxed{K_B=1.336\times10^{-14}\text{ J}}
\]
Therefore:
\[
\frac{K_B}{K_A}
=
\frac{1.336\times10^{-14}}
{3.34\times10^{-15}}
\]
\[
\boxed{\frac{K_B}{K_A}=4}
\]
Particle B has four times the kinetic energy.
**e) Determine the trajectory shapes.**
Particle A has no parallel velocity component, so it follows a circle.
Particle B has a non-zero parallel component:
\[
v_{\parallel B}=4.0\times10^6\cos30^\circ
\]
\[
v_{\parallel B}=3.46\times10^6\text{ m s}^{-1}
\]
It therefore follows a helix.
This is the important result: equal circular radii do not imply equal total speeds or equal kinetic energies.
The radius depends on \(v_\perp\), not the total velocity.
Neither particle gains or loses kinetic energy in the static magnetic field.
### Question 5
A proton enters a region containing two uniform fields.
The magnetic field is:
\[
B=0.20\text{ T}
\]
The electric field is:
\[
E=4.0\times10^3\text{ N C}^{-1}
\]
Both fields point in the positive \(z\)-direction.
At the entrance, the proton has:
\[
v_\perp=2.0\times10^6\text{ m s}^{-1}
\]
\[
v_\parallel=1.0\times10^6\text{ m s}^{-1}
\]
The proton moves \(0.50\text{ m}\) in the positive \(z\)-direction before leaving the field region.
Its charge and mass are:
\[
q=+1.60\times10^{-19}\text{ C}
\]
\[
m=1.67\times10^{-27}\text{ kg}
\]
Assume uniform fields, negligible edge effects, and no collisions.
Determine:
a) The work done by the electric field.
b) The work done by the magnetic field.
c) The initial and final kinetic energies.
d) The initial and final total speeds.
e) Whether the radius of the circular component changes.
f) Whether the particle continues along a constant-pitch helix.
[IMAGE: Three-dimensional apparatus view of a proton entering a cylindrical field region. Show parallel electric and magnetic field arrows along the positive z-axis, the initial velocity vector with perpendicular and parallel components, and the region's 0.50 m axial length. Do not show the subsequent trajectory shape or any calculated changes.]
### Solution 5
The proton gains kinetic energy from the electric field, even though its circular orbit radius remains unchanged. Its parallel velocity increases, so its helical trajectory develops an increasing pitch.
This is the important distinction: the circular radius depends on perpendicular velocity, while the total kinetic energy depends on both velocity components.
**a) Calculate the work done by the electric field.**
The proton is positively charged, and its electric force points along the positive \(z\)-direction.
Its displacement is also in this direction.
Therefore:
\[
W_E=qE\Delta z
\]
Substitute:
\[
W_E=(1.60\times10^{-19})
(4.0\times10^3)(0.50)
\]
\[
\boxed{W_E=3.2\times10^{-16}\text{ J}}
\]
The electric field does positive work.
**b) Calculate the work done by the magnetic field.**
The magnetic force is perpendicular to the instantaneous velocity at every point.
Therefore:
\[
\boxed{W_B=0\text{ J}}
\]
**c) Calculate the initial and final kinetic energies.**
The initial total speed satisfies:
\[
v_i^2=v_\perp^2+v_\parallel^2
\]
Therefore:
\[
v_i^2=(2.0\times10^6)^2
+(1.0\times10^6)^2
\]
\[
v_i^2=5.0\times10^{12}
\]
Initial kinetic energy:
\[
K_i=\frac12mv_i^2
\]
\[
K_i=\frac12(1.67\times10^{-27})
(5.0\times10^{12})
\]
\[
\boxed{K_i=4.175\times10^{-15}\text{ J}}
\]
The work-energy theorem gives:
\[
K_f=K_i+W_E
\]
\[
K_f=4.175\times10^{-15}
+3.2\times10^{-16}
\]
\[
\boxed{K_f=4.495\times10^{-15}\text{ J}}
\]
**d) Calculate the initial and final total speeds.**
Initially:
\[
v_i=\sqrt{v_\perp^2+v_\parallel^2}
\]
\[
\boxed{v_i=2.24\times10^6\text{ m s}^{-1}}
\]
Finally, using the final kinetic energy:
\[
v_f=\sqrt{\frac{2K_f}{m}}
\]
\[
v_f=\sqrt{
\frac{2(4.495\times10^{-15})}
{1.67\times10^{-27}}
}
\]
\[
\boxed{v_f=2.32\times10^6\text{ m s}^{-1}}
\]
The proton has increased its total speed.
But which velocity component changed?
The electric field points along the magnetic field. Its force therefore changes the parallel component, not the perpendicular component.
Since \(v_\perp\) remains \(2.0\times10^6\text{ m s}^{-1}\), the final parallel component is:
\[
v_{\parallel f}
=\sqrt{v_f^2-v_\perp^2}
\]
\[
v_{\parallel f}
=\sqrt{5.383\times10^{12}
-4.0\times10^{12}}
\]
\[
\boxed{v_{\parallel f}=1.18\times10^6\text{ m s}^{-1}}
\]
The parallel velocity has increased from \(1.0\times10^6\) to approximately \(1.18\times10^6\text{ m s}^{-1}\).
**e) Determine whether the circular radius changes.**
The circular radius is:
\[
r=\frac{mv_\perp}{|q|B}
\]
The electric field is parallel to the magnetic field, so it does not change \(v_\perp\).
The magnetic field also changes only the direction of \(v_\perp\), not its magnitude.
Therefore, the radius remains constant:
\[
r=\frac{(1.67\times10^{-27})(2.0\times10^6)}
{(1.60\times10^{-19})(0.20)}
\]
\[
\boxed{r=0.104\text{ m}}
\]
**f) Determine whether the helix has constant pitch.**
It does not.
The parallel velocity increases as the proton travels through the electric field.
The magnetic field remains uniform, so the time for each circular revolution stays constant. However, the proton travels further along the magnetic field direction during each revolution as its parallel speed increases.
Consequently, the helix has an increasing pitch.
This is the trap in the question: a constant circular radius does not prove that the particle's total speed or kinetic energy is constant.
The magnetic field still does no work. The electric field supplies the additional kinetic energy.
---
The next step is using the perpendicular velocity component to calculate orbit radii, compare particle mass-to-charge ratios, and interpret more complicated trajectories.
For the broader comparison of electric and magnetic forces, see [Electric Fields vs Magnetic Fields: HSC Physics Explained](https://tutorgum.com/learn/resources/hsc-physics-electric-fields-versus-magnetic-fields/).
For the energy bookkeeping behind electric-field acceleration, continue with [Electric Potential Energy and Work in Electric Fields](https://tutorgum.com/learn/resources/hsc-physics-electric-potential-energy-and-work/).