Magnetic field of a moving charge
Use the magnetic-field expression for a moving point charge and apply angle and direction logic with cross products.
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Guided steps
- Subtract the source position from the observation position in one fixed laboratory frame. The displacement points from charge to observation point. This expression uses prescribed uniform motion at low speed in vacuum. It does not say that changes of motion propagate instantly, and it excludes acceleration and radiation fields.
- For positive-x velocity, a positive-y displacement gives positive-z geometric cross product; a positive-z displacement gives negative-y geometric cross product. Apply the signed source charge after finding those directions. The field is perpendicular to both velocity and displacement. A magnitude is non-negative even when a component is negative.
- Calculate each source field using its own charge, velocity and displacement, then add components at the same point. When all source velocities are the same, their common velocity can be pulled outside the electric-field cross-product sum. With different source velocities, that factorisation is not generally valid. Different velocities can leave a magnetic field even where electric fields cancel. A separate test charge uses a separate force law.
- The original transverse-point example uses the source-to-point distance in metres and the charge magnitude in coulombs. Its field is two hundred forty microtesla. At the same distance the velocity-axis field would be zero, while the electric field need not vanish there.
- Use the offset source position before calculating any distance or angle. Form the velocity crossed with displacement, then apply the negative source charge to every component. Check the resultant magnitude after the signed components are known.
- Find the signed y and z field components and the positive magnitude. Keep the source displacement distinct from the observation coordinates. Explain the vector steps separately.
- Explain the source subtraction, cross product and effect of the negative charge. Compare your reasoning yourself; three checked values do not mark that argument. Paper work is fine.
- Compare the two source-to-origin displacements and the two source velocities separately. Decide the electric-field cancellation from the charge geometry, but decide the magnetic field from the two cross products. Do not factor out a velocity that is not common to the sources.
- Find the combined signed z field component and choose whether the electric field vanishes in the stated approximation. Keep both source contributions and explain their comparison separately.
- Show the two separate source cross products, explain the electric-field cancellation and why no common source velocity can be taken out. Compare that reasoning yourself; the numeric and choice checks do not mark it.
- Predict the field magnitude when source-to-point distance doubles with charge, speed and angle fixed. Keep this point-source model distinct from an integrated infinite wire.
- Find the magnetic field directly ahead of the uniformly moving source. Check the angle between source velocity and source-to-point displacement; a magnetic-field zero does not by itself imply an electric-field zero.
- Predict magnitude and direction after reversing the source charge sign while keeping its velocity and position fixed. Separate the absolute charge magnitude from the signed vector expression.
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A moving charge’s field depends on its motion and where you observe it. Start with the displacement from the source to the observation point; then form a cross product and apply the charge’s sign. A field calculation does not require a second test charge, and finding B is a different job from finding a force on a test charge.
Use vector components and the cross-product direction in the current-element field law. The mathematics toolkit is available for vector notation; integral theorems are not a prerequisite for this point-source calculation. We use prescribed uniform motion in vacuum, with speed much smaller than c, and exclude acceleration and radiation fields. Treat the examples as source-field models, not a prediction that interacting charges maintain their velocities without constraints.
Choose the source displacement and low-speed model
In one fixed laboratory frame, let the charge’s instantaneous position be rq and the observation point be rP. Define s = rP-rq, s = |s| > 0, and s hat = s/s. To leading order at low speed, its magnetic field is
Here q is signed charge in coulombs and v is that source’s velocity in metres per second. Field components are in tesla. For calculations use μ₀/(4π) ≈ 10⁻⁷ T m/A. The equivalent unit-vector form is B ≈ (μ₀/4π)q(v × s hat)/s²: the extra factor of s in the first numerator accounts for the different denominator. Do not use the full displacement vector with a denominator of s².
The present-position displacement is valid in this uniform-motion expression. It does not mean a change of source motion propagates instantly. At relativistic speed the angular dependence of the field changes; uniformly moving sources can be treated by transforming their fields, while general accelerated motion needs a more complete treatment. The uniformly moving-charge field derivation also gives B = v × E/c² for that source. Its low-speed electric field is approximately qs hat/(4πε₀s²); using 1/(ε₀c²) = μ₀ recovers the expression above.
Take the cross product before the charge sign
For right-handed axes and a source moving in + x, write v = vx hat with v > 0. If s = sₓx hat + syy hat + szz hat, then
For a positive charge, a point on + y has field + z, while a point on + z has field -y. A negative charge reverses these directions. Reversing both q and v leaves qv unchanged, so it does not reverse the field. Use a single fixed coordinate frame throughout; changing the observer’s viewpoint must not silently reverse the axes.
If θ is the angle between v and the source-to-point displacement, the magnitude is
This is non-negative. It is largest at 90° for fixed s, |q| and v, and zero on either side of the velocity axis. A zero magnetic field there does not imply a zero electric field. Away from the axis, B is perpendicular to both v and s; these dot-product checks can catch a mistaken component or cross-product order.
The units are (T m/A)(A s)(m/s)/m² = T. At fixed angle the single-source magnitude falls as 1/s², unlike the integrated 1/ρ field of an ideal infinite straight wire. The latter already sums many source elements at different distances and angles.
Superpose source fields before interpreting the result
For several moving charges, compute each qᵢvᵢ × sᵢ/sᵢ³ from its own position, velocity and charge sign, then add field components at the common observation point. If every source has the same velocity, that common velocity can be taken outside the cross-product sum in B total = v × E total/c² for these uniform-motion fields. If velocities differ, cancellation of electric fields alone does not decide whether magnetic fields cancel.
If a separate test charge Q with velocity u is introduced, its magnetic force is Qu × B(P). Do not insert that test velocity into the source-field formula; the Lorentz-force lesson uses the field after it has been found.
Field beside a moving charge
Worked example 1
Field at a transverse point
Problem
A charge q = +2.0 μ C moves uniformly in vacuum at v = 3.0 × 10⁶ m s⁻¹. Find the magnetic-field magnitude at a point r = 5.0 cm from the charge where the angle between velocity and the source-to-point vector is 90°. Use μ₀/(4π) ≈ 10⁻⁷ T m A⁻¹.
Show full solution
Convert units and use the transverse factor
Method
Use q = 2.0 × 10⁻⁶ C, r = 0.050 m and sin 90° = 1.Reason
The non-relativistic magnitude formula uses the inverse square of source-to-point distance and the component of velocity transverse to that vector.Working
Answer
B = (10⁻⁷(2.0 × 10⁻⁶)(3.0 × 10⁶))/((5.0 × 10⁻²)²) = (6.0 × 10⁻⁷)/(2.5 × 10⁻³) = 2.4 × 10⁻⁴ T.State the field magnitude
Working
Answer
B = 2.4 × 10⁻⁴ T. This is the largest magnitude at that distance for the stated charge and speed.
Quick check
Along the line of motion, at 0° or 180°, the field is zero.
Resolve a negative charge’s field in components
Guided practice 1
A negative source away from the origin
Problem
In fixed right-handed laboratory axes, a charge q = -4.0 nC moves uniformly with v = (2.0 × 10⁵,0,0) m/s. At the instant it is at C = (0.010,-0.020,0) m, find By, Bz and |B| at P = (0.010,0.010,0.040) m in vacuum. Explain the source-to-P displacement, cross product and charge sign. Use the leading low-speed uniform-motion field with μ₀/(4π) ≈ 10⁻⁷ T m/A. Calculate this source’s field only.
Use a displacement from C, then keep the field components signed
Hints
Hint 1
Locate the source
The observation coordinates are not the source-to-P displacement unless the source is at the origin.Hint 2
Separate geometry from charge sign
For +x velocity, use x cross y = z and x cross z = -y. Apply the negative charge after this geometric step.Show solution step by step
Form the source-to-P displacement
Method
Subtract C from P.Reason
The distance and angle in the source law refer to the charge, not an arbitrary coordinate origin.Working
Answer
s = (0,0.030,0.040) m, so s = 0.050 m. The displacement is perpendicular to the velocity, but the field has two nonzero components in the selected axes.Cross the vectors and apply the negative charge
Method
Use v × s = (0,-8.0 × 10³, + 6.0 × 10³) m²/s.Reason
Multiplication by q = -4.0 × 10⁻⁹ C reverses both geometric components.Working
Answer
By = +2.6 × 10⁻⁸ T and Bz = -1.9 × 10⁻⁸ T to two significant figures. Keeping guard digits for the magnitude, B = square root of ((2.56 × 10⁻⁸)² + (1.92 × 10⁻⁸)²) = 3.2 × 10⁻⁸ T. Compare the source subtraction and sign reasoning yourself; the checker verifies the three values only.
Quick check
Cancel electric fields without cancelling magnetic fields
Check your understanding 1
Two prescribed uniformly moving charges, each + 3.0 nC, are instantaneously at (-0.040,0,0) m and (+0.040,0,0) m. Their velocities are respectively (0, + 2.0 × 10⁵,0) and (0,-2.0 × 10⁵,0) m/s. At the origin in vacuum, find their combined Bz in the low-speed source-field model and decide whether their combined electric field vanishes to this order. Explain whether a single source velocity can be taken outside the field sum. Use μ₀/(4π) ≈ 10⁻⁷ T m/A; neglect changes of the prescribed source velocities and radiation.
Show hint
Show answer
Each source gives Bz = -3.75 × 10⁻⁸ T, so Bz = -7.5 × 10⁻⁸ T overall. The electric fields have equal magnitudes and opposite x directions, so they cancel to this order. The source velocities differ: ∑ᵢvᵢ × Eᵢ cannot be replaced by one source velocity crossed with ∑ᵢEᵢ. Compare the separate-source reasoning yourself; the checker verifies the component and electric-field choice only.
Check your understanding 2
For a uniformly moving charge, if source-to-point distance doubles while q, v and the angle stay fixed, what happens to the field magnitude?
Show hint
Compare 1/(2r)² with 1/r².
Show answer
It becomes one quarter of its original value because B ∝ r⁻². Doubling distance divides the magnitude by 2² = 4.
Check your understanding 3
In the non-relativistic uniform-motion model, what magnetic field does a moving charge produce at a point directly ahead of it on its velocity axis?
Show hint
The velocity and source-to-point vector are parallel.
Show answer
B = 0 because θ = 0° and vector v × r hat = 0. This does not mean the electric field is zero there.
Check your understanding 4
At a point with nonzero magnetic field, what changes if a moving charge is replaced by an equal-magnitude charge of opposite sign, with velocity and position unchanged?
Show hint
Compare the signed vector formula with its magnitude form.
Show answer
The magnitude is unchanged because it depends on |q|. The field direction reverses because vector B is proportional to the signed charge q.
Keep source motion separate from the observer’s field
Subtract source position from observation position, cross the source velocity with that displacement, then apply the source charge’s sign. Add field components source by source. Test perpendicularity, units and the fixed-angle inverse-square limit before using the field in a separate force calculation. The uniform, low-speed assumptions are part of the model, not optional labels.
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