Moving charges in magnetic field — force, circular motion, helical path

Question

What force does a magnetic field exert on a moving charge? When does the charge move in a circle, and when does it follow a helical (spiral) path?

(CBSE 12, JEE Main, NEET — Lorentz force and circular motion of charges in magnetic fields appear in every paper)

Solution — Step by Step

\vec{F} = q(\vec{v} \times \vec{B})

Magnitude: $F = qvB\sin\theta$

where $\theta$ = angle between $\vec{v}$ and $\vec{B}$ .

Key properties:

Force is always perpendicular to both $\vec{v}$ and $\vec{B}$ (use the right-hand rule / cross product)
Since $\vec{F} \perp \vec{v}$ , the magnetic force does no work on the charge — it changes direction but not speed
If charge is negative, the force direction reverses

When $\vec{v} \perp \vec{B}$ ( $\theta = 90°$ ), the force has maximum magnitude ( $F = qvB$ ) and is always perpendicular to velocity. This is exactly the condition for uniform circular motion.

Equating magnetic force to centripetal force:

qvB = \frac{mv^2}{r}

r = \frac{mv}{qB}

Time period: $T = \frac{2\pi m}{qB}$ (independent of velocity and radius!)

Cyclotron frequency: $f = \frac{qB}{2\pi m}$ (depends only on charge-to-mass ratio and field strength)

This is the principle behind the cyclotron particle accelerator — the frequency of the alternating electric field can be kept constant because $T$ does not depend on the particle’s speed.

When $\vec{v}$ makes angle $\theta$ with $\vec{B}$ (neither 0 nor 90 degrees), decompose the velocity:

$v_\perp = v\sin\theta$ — this component creates circular motion (radius $r = mv\sin\theta / qB$ )
$v_\parallel = v\cos\theta$ — this component is along $\vec{B}$ and is unaffected (no force on a charge moving parallel to field)

The result: the charge spirals along the field direction in a helical (spiral) path.

Pitch (distance moved along $\vec{B}$ in one revolution):

p = v_\parallel \cdot T = v\cos\theta \cdot \frac{2\pi m}{qB}

When $\theta = 0$ or $180°$ ( $\vec{v} \parallel \vec{B}$ ):

$F = qvB\sin 0° = 0$

The charge continues in a straight line. The magnetic field has no effect on charges moving along the field lines.

flowchart TD
    A["Charge moving in magnetic field"] --> B{"Angle between v and B?"}
    B -->|"θ = 0° or 180° (parallel)"| C["No force, straight line"]
    B -->|"θ = 90° (perpendicular)"| D["Circular motion<br/>r = mv/qB<br/>T = 2πm/qB"]
    B -->|"0° < θ < 90° (oblique)"| E["Helical path<br/>r = mv sinθ/qB<br/>pitch = v cosθ × T"]
    D --> F["Cyclotron principle:<br/>T is independent of v"]

Why This Works

The magnetic force is always perpendicular to velocity, so it acts as a centripetal force without changing the speed. This is fundamentally different from electric force (which can accelerate or decelerate charges). The perpendicularity condition means magnetic fields can steer charges without adding or removing kinetic energy.

The helical path is simply the superposition of two independent motions: circular motion perpendicular to $\vec{B}$ and uniform linear motion parallel to $\vec{B}$ . Since these are in perpendicular directions, they combine into a helix.

Common Mistake

The most common error: students claim the magnetic force does work and changes the kinetic energy of the charge. Since $\vec{F} \perp \vec{v}$ at all times, the work done $W = \vec{F} \cdot d\vec{s} = 0$ . The speed remains constant; only the direction changes. If a problem says “a charged particle enters a magnetic field and its speed increases,” there MUST be an electric field present too — the magnetic field alone cannot change speed.

For the right-hand rule with negative charges: first apply the rule for a positive charge, then reverse the direction. In JEE, many problems involve electrons — do not forget to flip the force direction. Alternatively, use the formula directly with $q = -e$ and let the negative sign handle the direction.

Question

Solution — Step by Step

Why This Works

Common Mistake

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