Free Pearson Edexcel International A Level Physics lesson
Magnetic forces and particle paths
A magnetic field turns moving charge without supplying its kinetic energy.
What you’ll learn
- Define magnetic flux density through force.
- Predict forces on charges and wires.
- Connect curvature to momentum.
The angle to the field controls magnetic force
Magnetic flux density B describes the strength and direction of a magnetic field. A moving charge has force magnitude B|q|v sin θ, where θ is the angle between its velocity and B. The force vanishes for motion parallel to the field and is largest for perpendicular motion. A stationary charge feels no magnetic force from this field.
Fleming’s left-hand rule connects field, conventional current and force: first finger follows B, second finger follows positive charge motion, and thumb gives force. Reverse the result for a negative charge moving in the same direction. Equivalently use the vector cross product qv × B. Make the direction conventions explicit instead of treating an electron velocity arrow as conventional current.
The force is always perpendicular to instantaneous velocity, so a static magnetic field alone transfers no kinetic energy. It may change the direction of momentum without changing its magnitude. This distinguishes magnetic deflection from electric acceleration. If both fields are present, evaluate their vector forces separately; they may balance for a particular velocity.
|F| = B|q|v sin θ; vector F = q(v × B)
Symbols and units
- B is magnetic flux density in T
- q is signed charge in C
- v is speed in m s⁻¹
- θ is angle between velocity and field
- F is force magnitude in N.
How the equation explains the idea
- Only the velocity component perpendicular to B contributes: F = B|q|v sin θ. Parallel motion gives zero force; perpendicular motion gives the maximum.
- The vector direction is q(v × B), reversed for negative charge. Magnetic force is perpendicular to velocity and does no work on a point charge in this model, changing direction rather than speed.
Conditions. Particle charge magnitude 1.60 × 10⁻¹⁹ C, speed 1.0 × 10⁶ m s⁻¹ and uniform B = 0.10 T. Force magnitude B|q|v sin θ is shown for 0°–180°.
Angle between velocity and field / degrees; Force magnitude / 10⁻¹⁴ N.
The maximum is 1.60 × 10⁻¹⁴ N at 90°. Parallel and antiparallel velocities both give zero force. The plot gives magnitude only; negative charge reverses vector force direction.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A proton crossing a field
- A proton travels at 2.0 × 10⁶ m s⁻¹ perpendicular to B = 0.30 T.
- F = 0.30 × 1.60 × 10⁻¹⁹ × 2.0 × 10⁶.
- F = 9.6 × 10⁻¹⁴ N, perpendicular to both v and B.
The speed does not change solely because of this magnetic force.
Keep this idea. Use the angle and charge sign before predicting magnetic force direction or magnitude.
A current-carrying wire feels the total charge force
A current-carrying conductor contains many drifting charges. Adding their magnetic forces over a straight wire segment gives F = BIl sin θ, where l is the length actually inside the field. Conventional current, not electron drift, is used with Fleming’s left-hand rule. The field or current direction reversed alone reverses the force; reversing both preserves it.
The relation defines the tesla operationally: B = F/(Il) for perpendicular wire and field. A balance can detect the equal-and-opposite force on a magnet assembly when current passes through a wire between its poles. Subtract the zero-current reading, convert a balance mass difference to force using g, and identify whether you are describing the force on wire or magnet.
A useful investigation varies current while keeping active wire length, orientation and pole gap fixed. A force-against-current graph should have gradient Bl for a uniform perpendicular field. Account for heating at large currents, field nonuniformity and flexible connections that exert unwanted forces. Use a current-limited low-voltage supply and a suitable wire rating.
F = BIl sin θ; perpendicular graph gradient ΔF/ΔI = Bl; 1 T = 1 N A⁻¹ m⁻¹
Symbols and units
- B is magnetic flux density in T
- I is conventional current in A
- l is active wire length in m
- θ is its angle to the field
- F is force in N.
How the equation explains the idea
- Adding forces on moving carriers gives F = BIl sin θ. For a perpendicular wire, fixed B and l give a force–current gradient Bl.
- Thus B = gradient/l. If only 0.040 m of a longer wire lies within the uniform field, use that active length; using the total wire length underestimates B.
Conditions. Straight active wire length l = 0.10 m perpendicular to uniform B = 0.20 T. Current direction is fixed and positive; the graph gives force magnitude.
Current / A; Force magnitude / N.
The slope is Bl = 0.020 N A⁻¹. Dividing the gradient by the 0.10 m active length recovers B = 0.20 T. At 3 A the force is 0.060 N.
Open the interactive lesson to view the plotted curves and their coordinate tables.
Worked example
A balance measures a motor force
- A 0.050 m active wire length carries 2.0 A perpendicular to the field.
- Measured force magnitude is 0.030 N.
- B = F/(Il) = 0.030/(2.0 × 0.050) = 0.30 T.
Use only the wire length inside the measured field.
Keep this idea. Wire force depends on active length, conventional current and orientation.
A curved track is a momentum measurement
For a particle entering perpendicular to a uniform magnetic field, the magnetic force acts as the centripetal resultant. Equating B|q|v = mv²/r and using p = mv gives r = p/(B|q|). The curvature radius measures momentum divided by charge magnitude, not directly mass or kinetic energy.
A higher-momentum particle bends less in the same field. A larger charge magnitude bends more for the same momentum. Opposite signs bend to opposite sides when their incident velocities agree. If a particle loses momentum by ionising a detector medium, its radius can shrink along the track; the direction of travel must be inferred before using curvature to infer charge sign.
The simple circular projection assumes the relevant momentum component is perpendicular to B. A component parallel to B persists and produces a helical path. Neutral particles do not leave magnetic curvature, but may be inferred from missing momentum or secondary charged tracks. Absence of a visible track alone does not establish that no particle was present.
B|q|v = mv²/r ⇒ r = p/(B|q|); p = B|q|r
Symbols and units
- r is circular-track radius in m
- p is momentum magnitude in kg m s⁻¹
- B is field strength in T
- q is charge in C
- m is mass in kg
- v is speed in m s⁻¹.
How the equation explains the idea
- For perpendicular nonrelativistic motion, magnetic force supplies inward force: B|q|v = mv²/r. Cancel v to obtain r = mv/(B|q|) = p/(B|q|).
- At fixed B and charge magnitude, twice the momentum gives twice the radius. Curvature direction gives the sign of charge only when the motion direction and magnetic-field direction are known.
Worked example
Momentum from a detector arc
- A singly charged track has r = 0.25 m in B = 0.80 T.
- p = 0.80 × 1.60 × 10⁻¹⁹ × 0.25.
- p = 3.2 × 10⁻²⁰ kg m s⁻¹.
An energy estimate additionally needs the particle mass and an appropriate relativistic or classical model.
Keep this idea. Use radius for momentum per charge, and use the travel direction when assigning charge sign.
Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.
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