Magnetic force on charge — F = qv × B: Question bank

Two ideas do all the heavy lifting below, so pin them up first:

Recall The two facts every answer leans on

Fact A — sideways only. $F = q v \times B$ is always perpendicular to $v$ . A Cross product of two vectors is perpendicular to both of them. Fact B — needs perpendicular motion. Magnitude is $∣ F ∣ = ∣ q ∣ v B sin θ$ , where $θ$ is the angle between $v$ and $B$ . Only the part of $v$ across $B$ matters; the part along $B$ contributes nothing.

True or false — justify

A magnetic field can hold a stationary electron in place against gravity.

False. With

v = 0

F = q v \times B = 0

: a magnetic field ignores a charge that isn't moving, so it can't support anything at rest.

A magnetic force can change the kinetic energy of a charged particle.

False. Power is

F \cdot v

, and since

F ⊥ v

(Fact A), this dot product is always

0

; kinetic energy

\frac{1}{2} m v^{2}

is constant.

If you double the speed of a proton in a fixed field, its circular period doubles.

False.

T = \frac{2 π m}{q B}

has no

v

in it; doubling

v

doubles the radius (

r = m v / q B

) but keeps the period identical.

A magnetic force can change a particle's velocity.

True. Velocity is a vector: the force turns its direction even though it never changes its magnitude (speed). "Can't change speed" is not "can't change velocity."

A proton and an electron fired with the same

v

into the same

B

curve the same way.

False. The sign of

q

flips

F = q v \times B

, so they curve in opposite directions (though the shape, a circle, is the same).

v

is exactly parallel to

B

, the particle travels in a straight line.

True.

sin θ = sin 0 = 0

, so

F = 0

and there is nothing to bend the path.

Reversing the direction of

B

reverses the direction of the force.

True.

v \times (- B) = - (v \times B)

; the cross product is linear, so flipping

B

flips

F

A faster particle in the same field always feels a stronger force.

False in general.

∣ F ∣ = ∣ q ∣ v B sin θ

grows with

v

only if

sin θ \neq = 0

. If

v ∥ B

, the force stays zero no matter how fast it goes.

Spot the error

F = q v \cdot B

, so the force is a number pointing along

v

The law uses a cross product, not a dot. A dot gives a scalar; force is a vector, and it points perpendicular to

v

, not along it.

"The force is largest when

v

is parallel to

B

Backwards:

sin θ

is maximum at

θ = 9 0^{\circ}

(perpendicular) and zero at

θ = 0^{\circ}

(parallel). Maximum force needs

v ⊥ B

"For an electron, point fingers along

v

, curl to

B

, thumb gives the force."

The thumb gives

v \times B

. For a negative charge you must flip it: the force points opposite the thumb.

"Since force is

q v B sin θ

, an electron feels a negative force."

Magnitude

∣ q ∣ v B sin θ

is never negative — the

∣ q ∣

is an absolute value. The sign of

q

decides direction, handled by the vector form, not the magnitude.

"The period is

T = 2 π m / q B

, so a heavier particle laps faster."

Opposite.

T \propto m

: a heavier particle takes longer per lap in the same field.

"A magnetic field speeds a particle up as it enters, then it circles."

There's no speed-up phase. The instant the charge moves, the force is purely sideways (Fact A); speed is constant from the first moment.

Why questions

Why must a magnetic force be perpendicular to the velocity?

Because it is

q v \times B

, and a cross product is by definition perpendicular to both input vectors — including

v

(Fact A).

Why does the cyclotron period not depend on speed?

Faster speed enlarges the radius

r = m v / q B

in exact proportion, so the longer path is covered at the higher speed in the same time:

T = 2 π r / v = 2 π m / q B

, and

v

cancels.

Why does a charge moving along

B

feel no force?

Only the component of

v

across

B

matters, and that component is zero when they're parallel:

sin 0 = 0

, giving

∣ F ∣ = 0

(Fact B).

Why can only an electric field, never a magnetic one, do work on a charge?

The electric force

q E

can have a component along

v

, so

F \cdot v \neq = 0

; the magnetic force is always

⊥ v

, so its power is always

0

Why does a velocity with both parallel and perpendicular parts trace a helix?

The perpendicular part circles (constant sideways force), the parallel part drifts freely (no force); circle + steady drift = a helix along

B

Why does flipping the sign of the charge flip the curving direction but not the radius?

The sign reverses

F

's direction (curve the other way), but

r = m v / (∣ q ∣ B)

uses only the magnitude of

q

, so the circle is the same size.

Edge cases

What happens to the radius

r = m v / q B

as the field

B \to 0

r \to \infty

: the circle becomes infinitely large, i.e. the path straightens into a line — no field, no bending.

What is the force if the charge is neutral (

q = 0

Zero.

F = 0 \cdot (v \times B) = 0

; a magnetic field acts only on charge in motion, and there's no charge here.

A charge moves so that

v

and

B

are antiparallel (

θ = 18 0^{\circ}

). What's the force?

Zero.

sin 18 0^{\circ} = 0

, so

∣ F ∣ = 0

— antiparallel is just as "force-free" as parallel.

As the angle

θ

between

v

and

B

goes from

0^{\circ}

9 0^{\circ}

, how does the force change?

It grows smoothly with

sin θ

: from

0

(parallel) up to its maximum

∣ q ∣ v B

(perpendicular), tracing the shape of a sine curve.

In the limit of very strong

B

(with

v

q

m

fixed), what happens to the circle?

r = m v / q B \to

tiny and

T = 2 π m / q B \to

tiny: the particle spirals in a tight, fast little loop — tightly "trapped" by the field.

A particle enters a field region moving exactly perpendicular to

B

and exits it. Has its speed changed?

No. Throughout,

F ⊥ v

so no work is done; it leaves with the same speed it entered, only aimed in a new direction.

Connections

Magnetic force on charge — F = qv × B (index 1.8.20) — the parent law these traps probe.
Cross product — why the force is perpendicular to both inputs (Fact A).
Lorentz force law — add the electric part $q E$ to see who actually does the work.
Centripetal force and circular motion — the "sideways force ⇒ circle" machinery.
Cyclotron — exploits the speed-independent period $T = 2 π m / q B$ .
Mass spectrometer — uses $r = m v / q B$ to sort masses.
Velocity selector — balances electric and magnetic forces (only $E$ does work).
Magnetic force on a current-carrying wire — the same law summed over many moving charges.