IB Physics HLCharges Moving in FieldsPaper 1 & 2F = Bqv sin θ~16 min read
Force on a Moving Charge
A current is only ever charges in motion. So when a magnetic field pushes a wire, it is not really pushing the wire at all — it is pushing each individual electron drifting inside, and the wire simply comes along because the electrons are trapped in it. Take one charge out of the wire, fire it through the field on its own, and it is still pushed. Same physics, one particle at a time.
📘 What you need to know
A moving charge produces its own magnetic field, so in an applied field it feels a force
F = Bqv sin θ, where θ is the angle between the velocity and the field
Maximum when θ = 90°: F = Bqv. Zero when the charge moves parallel to the field
F, B and v are mutually perpendicular
Directions come from Fleming’s left-hand rule, using the conventional current
For a positive charge, the current points the same way as the velocity
For a negative charge, the current points the opposite way to the velocity
F = BIL is for a current-carrying conductor; F = Bqv is for an isolated moving charge
Because F is always perpendicular to v, the particle’s speed never changes — it turns in a circle
The equation
Force on a moving chargeF = Bqv sin θB = flux density (T) • q = charge (C) • v = speed (m s−1)θ = angle between the velocity and the field
Move across the field and you are pushed hardest. Move along it and you are not pushed at all. Everything in between is the sine of the angle.
Notice the force is perpendicular to v, which means it can never speed the particle up or slow it down — it only ever steers. A magnetic field does no work on a moving charge. The kinetic energy is untouchable. All the field can do is bend the path, and if it bends it steadily, the path becomes a circle. That is the next page, and it follows from this one sentence.
The trap: which way for a negative charge?
Fleming’s second finger points along the conventional current — the flow of positive charge. So for an electron, moving one way, the current points the other way, and the force flips.
Read the diagonals. Top-left and bottom-right agree; top-right and bottom-left agree. Reversing two things reverses the force twice, which is to say not at all.
The safest habit in the whole topic: before you lift a finger, write down the direction of the conventional current. If the question says “a beam of electrons travels east”, write “current: west” on the paper. Then point your fingers. Students who skip that one line get exactly half of these questions wrong, and always the same half.
Bqv and BIL are the same equation
Last page’s F = BIL was never a separate law. It is F = Bqv, added up over every charge carrier in the wire.
Each electron feels a tiny Bqv, of order 10−24 N. Multiply by 1022 of them and you get a force you can feel in your hand.
F = BIL sin θ
F = Bqv sin θ
Use it for
A current-carrying conductor
An isolated moving charge
θ is between
The wire and the field
The velocity and the field
Maximum when
θ = 90°, giving BIL or Bqv
Zero when
θ = 0° — motion parallel to the field
Direction
Fleming’s left hand, using conventional current
Charge q at speed v
across a field B
F = Bqv sin θ
and F ⊥ v always
so it turns, never speeds up
⚡ Working a moving-charge question
Write the conventional current first. Positive charge → along v. Negative → opposite to v.
Find θ. Between the velocity and the field. Perpendicular → F = Bqv. Parallel → F = 0.
Use the magnitude of q. For an electron, q = 1.60 × 10−19 C. The sign lives in the direction.
Direction? Fleming’s left hand: first finger B, second finger I, thumb F.
Ratio questions?Fθ / F⊥ = sin θ. Everything else cancels.
Conductor or particle?BIL for a wire, Bqv for a lone charge.
WE 1
An electron travels at 4.0 × 106 m s−1 perpendicular to a uniform magnetic field of flux density 0.25 T. (a) Calculate the magnetic force on it. (b) Determine the angle to the field at which the force would fall to 40% of this value. (c) State the force if the electron moved parallel to the field. (qe = 1.60 × 10−19 C)
(a) Step 1 — perpendicular, so sin 90° = 1F = Bqv = (0.25)(1.60 × 10⁻¹⁹)(4.0 × 10⁶)F = 1.6 × 10⁻¹³ N(b) Step 2 — take the ratio, and everything cancelsFθ / F⊥ = Bqv sin θ / Bqv = sin θsin θ = 0.40 → θ = sin⁻¹(0.40)θ = 24°(c) Step 3 — parallel means θ = 0°F = 0 NPart (b) never needed a single number from part (a). Write the ratio, watch B, q and v all cancel, and you are left with sin θ = the fraction. It works for any fraction, any particle, any field.
WE 2
A beam of electrons travels horizontally to the right through a region where the magnetic field is directed into the page. (a) State the direction of the conventional current in the beam. (b) Determine the direction of the magnetic force on the electrons. (c) State the direction of the force if protons, travelling at the same velocity, replaced the electrons.
(a) the current comes first
Electrons are negative, so the conventional current is opposite to their motion.
the current is to the left(b) now Fleming’s left hand
First finger (field): into the page.
Second finger (current): to the left.
the thumb points downwards — the force is downwards(c) swap to protons
Protons are positive, so the current now points to the right.
Reversing the current reverses the force.
the force on the protons is upwardsSame field, same velocity, opposite deflection. This is precisely how a mass spectrometer separates positive from negative ions — you simply let the field sort them.
WE 3
A wire of length 0.30 m carries a current of 2.4 A at right angles to a field of 0.40 T. (a) Calculate the force on the wire. (b) The length of wire contains 4.5 × 1022 free electrons. Calculate the average magnetic force on a single electron. (c) Use your answer to estimate the drift speed of the electrons, and comment.
(a) Step 1 — the whole wireF = BIL = (0.40)(2.4)(0.30)F = 0.29 N(b) Step 2 — share it outf = F / N = 0.288 / (4.5 × 10²²)f = 6.4 × 10⁻²⁴ N per electron(c) Step 3 — each electron obeys f = Bqvv = f / Bq = (6.4 × 10⁻²⁴) / [(0.40)(1.60 × 10⁻¹⁹)]v = 1.0 × 10⁻⁴ m s⁻¹, about 0.1 mm per secondThat drift speed is real. Electrons in a copper wire crawl along at a fraction of a millimetre per second, and yet the lamp comes on instantly — because the field that pushes them travels at nearly the speed of light. The force on the wire is nothing more than 4.5 × 10²² tiny Bqv nudges, all pointing the same way.
💡 Top tips
Write the conventional current down before you use your hand. Negative charge → reverse it.
θ is between the velocity and the field, not between anything else.
Ratio questions: Fθ/F⊥ = sin θ. No numbers required.
Parallel to the field → no force at all. Say it explicitly; it is a mark.
A magnetic field does no work on a charge. Speed constant, direction changing.
BIL is for a wire; Bqv is for a charge. They are the same law.
Use the left hand. Fleming’s right hand belongs to generators.
⚠ Common mistakes
Pointing the second finger along an electron’s velocity. It must point along the conventional current
Using cos θ. The force is maximum when perpendicular, so it is a sine
Putting a negative value of q into the formula. Use the magnitude; the sign shows up in the direction
Mixing up F = BIL and F = Bqv. Wire, or lone charge?
Saying the magnetic field speeds the particle up. It does no work — only the direction changes
Reaching for Fleming’s right-hand rule, which is for induced currents
Forgetting that a charge moving parallel to the field feels zero force
Quick recap: A charge q moving at speed v through a field B feels F = Bqv sin θ, where θ is the angle between the velocity and the field. It is maximum (Bqv) when they are perpendicular and zero when they are parallel. F, B and v are mutually perpendicular, and directions come from Fleming’s left-hand rule using the conventional current — which runs opposite to the velocity of a negative charge. F = BIL is simply Bqv summed over every carrier in the wire. And because F is always perpendicular to v, the field does no work: the speed never changes.
Hold that last sentence up to the light. A force of constant size, always at right angles to the velocity, never changing the speed — where have you met that before? Circular motion. The magnetic force is behaving exactly like a centripetal force, and if you set Bqv equal to mv2/r and cancel one v, out drops the radius of the circle a particle will travel in. Next page: Charged Particles in Magnetic Fields.
Electrons deflecting the wrong way?
Book a free meeting and we’ll drill the conventional-current reversal, the sin θ ratio trick and the BIL-versus-Bqv choice.
