Chapter 03 · Moving Charges & Magnetism · Full book

Moving Charges and Magnetism

Core intuition: Chapter Roadmap

Electrostatics dealt with charge at rest. This chapter is about charge in motion—and the new force that motion unlocks. A moving charge both feels a magnetic force and creates a magnetic field; the whole chapter is the working-out of those two facts. The development runs in five movements:

The Magnetic Force. Why magnetism is perpendicular and motion-dependent; the Lorentz force \(\vec{F}=q\vec{v}\times\vec{B}\); circular and helical trajectories and the cyclotron radius \(r=mv/|q|B\).
Forces on Currents. A current is moving charge, so a wire feels \(\vec{F}=I\vec{L}\times\vec{B}\); two parallel wires force each other—and that force defines the ampere.
The Current Loop as a Magnet. Torque \(\tau = NIAB\sin\theta\) on a loop, the magnetic dipole moment \(\vec{\mu}=NI\vec{A}\), and \(\vec{\tau}=\vec{\mu}\times\vec{B}\)—the seed of every magnetometer and motor.
Fields from Currents. The Biot–Savart law and its harvest: the straight wire, the circular loop, and—through Ampère's circuital law—the solenoid and toroid.
Reading Currents: the Galvanometer. How loop-torque becomes a pointer deflection, and how one shunt or one series resistor turns a galvanometer into an ammeter or a voltmeter.

Throughout, the calculus derivations (Biot–Savart, Ampère) serve JEE-Advanced and AP-C E&M; the boxed standard results serve NEET and AP Physics 2 directly. Bar magnets, magnetic materials, and Earth's field are taken up in the next chapter, Magnetism and Matter.

Perplexing Questions

Perplexing questions

Before we begin, sit with these questions. Do not reach for formulas. Each one is resolved explicitly at the relevant point in this chapter.

The crawling electron and the instant light. Electrons in a wire drift at roughly \(10^{-4}\,\mathrm{m/s}\)—slower than a tortoise. Yet when you close a circuit, the bulb lights almost instantaneously. In Chapter 2 we asked this about current; now ask it again for the field: what actually travels down the wire at close to the speed of light?
The magnet that does nothing to a charge at rest. Bring a powerful permanent magnet close to a stationary electron. Nothing happens. Move the electron, and suddenly a force appears. How can the same magnet be powerless in one situation and potent in another?
The force that bends but does not speed up. A charged particle enters a region of uniform magnetic field. Its path curves dramatically. Yet when you measure the speed before and after, it is identical. The particle was pushed the entire time, and still its kinetic energy did not change. How is this possible?
The helical path. Two identical protons enter the same magnetic field: one head-on, one at a slight angle. One circles endlessly in the same plane; the other drills forward in a helix, advancing even though the field does not exert a forward component. Why do these two trajectories differ so dramatically?
The current-carrying wire that pulls another. Two long parallel wires carry current in the same direction. By Coulomb's logic, the wires are neutral—no net charge—so there should be no force between them. Yet they attract. What pulls them together if charge is the only source of electric force?
The compass that tilts. A compass needle lies flat and points north. Hold a horizontal current-carrying wire directly above it: the needle swings perpendicular to the wire. There is no magnet anywhere near it. What has changed?

Every question is answered at the relevant section.
Look for the [Resolution: PQ N] callout as you read.

Why Magnetism Feels Different

Electrostatics gives us a reassuring picture of forces. Two charges exist, a line connects them, and along that line a force acts. The magnitude obeys Coulomb's inverse-square law; the direction is radial. The framework is clean, symmetric, and geometrically transparent.

Magnetism refuses to behave this way.

Drag a bar magnet across a table: it does not pull iron filings toward itself along the line joining them; it twists them, orients them, and aligns them along curved arcs. A current-carrying wire deflects a compass needle to the side—not toward the wire, and not away from it, but perpendicular to both the wire and the line from wire to needle. A charged particle fired between the poles of a magnet curves sideways, not backward or forward. Every one of these behaviours is at ninety degrees to what electrical intuition predicts.

The reason is not that magnetism is mysterious in principle. The reason is that magnetic force has a fundamentally different geometric character from electric force. Recognising that difference—before writing a single formula—is the whole point of this section.

Core intuition: The Two Great Differences from Electrostatics

Difference 1: Magnetic force requires motion. A stationary charge in a magnetic field experiences no magnetic force, regardless of the field strength. The electric field acts on charge; the magnetic field acts on moving charge. Motion is not a detail: it is a prerequisite. Difference 2: Magnetic force is not along the line of action. The electric force on a charge at position \(\vec{r}\) from a source points along \(\vec{r}\). The magnetic force on a moving charge points perpendicular to both the velocity and the field. It is a sideways force—it deflects without pushing the particle either toward or away from the field source. These two differences together mean that magnetic interactions are inherently more three-dimensional and direction-dependent than electric ones. Understanding them physically is more important than memorising the formula.

Historically, the study of magnetism began with permanent magnets, such as lodestones and compass needles, and for centuries it seemed like a completely separate branch of physics from electricity. The connection came in 1820 when Hans Christian Ørsted observed, apparently by accident, that a current-carrying wire deflected a nearby compass needle. A wire with moving charges could mimic the effect of a magnet. Within weeks Ampère had established quantitative laws for the force between parallel current-carrying wires. Within years Faraday had found the reverse: changing magnetic fields produce electric effects.

The deeper lesson took another four decades to emerge, when Maxwell unified electricity and magnetism into a single framework. We now understand that there is one electromagnetic field, and the split into “electric” and “magnetic” components is partly a matter of the observer's reference frame. A charge at rest sees only an electric field. The same configuration, viewed from a moving frame, presents a magnetic component. Magnetism is, in a precise sense, a relativistic consequence of electricity. We will not develop this perspective here—it belongs to a course on special relativity—but keeping it in the background prevents a common conceptual error.

◈ Resolution · PQ 1 The drifting electrons crawl, but the field they set up races down the wire at nearly light speed—it is the field, not the charge carriers, that carries the signal. How that field propagates is taken up fully with electromagnetic waves.

Misconception alert

Trap: “Magnetism is just electricity with magnets instead of charges.” Students who carry over their Coulomb-force intuition expect a magnetic force to act along the line joining source and target, to be attractive or repulsive depending on sign, and to act on any charge regardless of its state of motion. None of these is true. Magnetic force is perpendicular to the connecting line, it neither attracts nor repels in the simple Coulomb sense, and it does not act on stationary charges at all. JEE/NEET trap: A positive charge placed at rest in a region of uniform magnetic field \(\vec{B}\) experiences zero magnetic force. Students who think “charge + field = force” will incorrectly choose a nonzero answer. The correct answer is zero: no velocity, no magnetic force.

The productive way to enter magnetism is not to ask “where is the source charge?” but to ask “what is the velocity of the charge I am interested in?” That velocity, combined with the local magnetic field, determines both the magnitude and the direction of the force. This is the logic that organises the rest of the chapter.

A second organisational point: throughout this chapter we treat the magnetic field \(\vec{B}\) as a given—a vector field in space that we can measure with a test particle—and ask what it does to moving charges and current-carrying wires. How \(\vec{B}\) is created by currents (Biot–Savart, Ampère's Law) comes in Sections 5 and 6. This separation is deliberate: the force law and the field law are conceptually independent, and keeping them separate avoids the confusion of trying to learn both at once.

✓ Checkpoint · Why Magnetism Feels Different

A proton and an electron are both stationary in a region where a uniform magnetic field \(\vec{B}\) points vertically upward. Describe the magnetic force on each.
Zero for both. Magnetic force requires the charge to be in motion (\(\vec{F} = q\vec{v}\times\vec{B}\), and \(\vec{v} = \vec{0}\)). No velocity means no magnetic force, regardless of charge sign or field strength.
A compass needle brought near a stationary charged rubber rod shows no deflection. The same needle brought near a current-carrying wire deflects visibly. State the physical reason for this difference in terms of the nature of magnetic sources.
A stationary charged rod produces only an electric field. A compass needle (magnetic dipole) does not respond to a uniform static electric field. A current consists of moving charges, which create a magnetic field; it is this field that exerts a torque on the compass needle's magnetic dipole.
Give a physical reason why the magnetic force on a charged particle is always perpendicular to the particle's displacement, and state what this implies about the particle's kinetic energy as it moves through a static magnetic field.
The magnetic force \(\vec{F} = q\vec{v}\times\vec{B}\) is perpendicular to \(\vec{v}\) by the definition of the cross product, so \(\vec{F}\cdot\vec{v} = 0\) at every instant. Power delivered equals \(\vec{F}\cdot\vec{v}\), which is zero. Therefore the magnetic force does no work and the particle's kinetic energy (hence speed) remains constant.
State two observable differences between the electric force on a charge and the magnetic force on the same charge, as it moves through a region where both fields are present.
(1) The electric force is independent of the charge's velocity; the magnetic force exists only if the charge is moving. (2) The electric force is generally along (or opposite to) the electric field direction; the magnetic force is always perpendicular to both the velocity and the magnetic field—it can have no component along the velocity.

Magnetic Force on a Moving Charge

◈ Lorentz Force

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Set v, B and the sign of the charge — F = qv×B always comes out perpendicular to both, and reversing the charge flips it.

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The previous section established that magnetic force needs motion and acts sideways. But “sideways” is not a formula, and the question of which side still needs an answer. If you fire a proton northward through a field pointing east, which way does the force point? Upward? Downward? Does it depend on whether the charge is positive or negative?

These questions require a precise, quantitative statement of the force law.

Begin with observation. When experimenters measure the force on a charge \(q\) moving with velocity \(\vec{v}\) through a magnetic field \(\vec{B}\), three things emerge consistently:

The magnitude of the force is proportional to \(|q|\), to the speed \(v\), and to the field strength \(B\).
The force is zero when \(\vec{v}\) is parallel to \(\vec{B}\), and maximum when they are perpendicular.
The force is always perpendicular to both \(\vec{v}\) and \(\vec{B}\); reversing the sign of \(q\) reverses the direction of the force.

These three observations are precisely encoded by the vector cross product. The magnetic force on the charge is \[ \vec{F} = q\,\vec{v} \times \vec{B}. \] In magnitude: \[ F = |q|\,v\,B\sin\theta, \] where \(\theta\) is the angle between \(\vec{v}\) and \(\vec{B}\). This is called the Lorentz magnetic force. (The full Lorentz force, combining electric and magnetic contributions, is \(\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})\); here we focus on the magnetic part alone.)

◈ Resolution · PQ 2 This is why a magnet does nothing to a charge at rest: with \(\vec{v}=\vec{0}\) the force \(q\vec{v}\times\vec{B}\) is zero. Set the charge moving and the force appears—motion is the switch.

Birth of the formula: Reading the Cross Product Physically

The expression \(\vec{v}\times\vec{B}\) defines a vector perpendicular to both \(\vec{v}\) and \(\vec{B}\), with magnitude \(vB\sin\theta\) and direction given by the right-hand rule. Right-hand rule (curl form): Point the fingers of your right hand along \(\vec{v}\). Curl them toward \(\vec{B}\) through the smaller angle. Your extended thumb points in the direction of \(\vec{v}\times\vec{B}\). For a positive charge \(q \gt 0\), this is also the direction of \(\vec{F}\). For a negative charge \(q \lt 0\), the force is opposite to this direction. Three cases worth committing to physical intuition: \(\theta = 0\) (velocity parallel to field, \(\vec{v}\parallel\vec{B}\)): \(\sin 0 = 0\), so \(F = 0\). The particle moves along the field line without deflection. \(\theta = \pi\) (velocity anti-parallel to field): \(\sin\pi = 0\), so \(F = 0\) again. A head-on charge heading backward along a field line also feels nothing. \(\theta = \pi/2\) (velocity perpendicular to field): \(\sin(\pi/2) = 1\), so \(F = |q|vB\)—this is the maximum possible force for given \(q\), \(v\), \(B\). The force varies as \(\sin\theta\) between these extremes. A particle travelling at \(30^\circ\) to the field experiences half the maximum force (\(\sin 30^\circ = 0.5\)).

The SI unit of magnetic field is the tesla (T), defined precisely by this force law: \[ 1\,\mathrm{T} = \frac{1\,\mathrm{N}}{1\,\mathrm{C}\cdot 1\,\mathrm{m/s}} = 1\,\frac{\mathrm{kg}}{\mathrm{A\cdot s^2}}. \] For reference, the Earth's magnetic field near the surface is roughly \(30\)–\(60\,\mu\mathrm{T}\). A strong laboratory electromagnet might reach \(1\)–\(2\,\mathrm{T}\). MRI machines operate at \(1.5\)–\(3\,\mathrm{T}\).

Misconception alert

Trap 1: “The force is along the magnetic field.” The magnetic force \(q\vec{v}\times\vec{B}\) is perpendicular to \(\vec{B}\) by the nature of the cross product. It can never have a component along \(\vec{B}\). A charge moving parallel to a field line experiences exactly zero force. JEE trap: Questions sometimes ask whether \(\vec{B}\) can “speed up” a particle. Since \(\vec{F} \perp \vec{v}\) always, the field cannot change the particle's speed—only its direction. Trap 2: “The force is along the velocity.” Equally wrong and equally common. \(\vec{F} = q\vec{v}\times\vec{B}\) is perpendicular to \(\vec{v}\). A force along \(\vec{v}\) would accelerate or decelerate the particle (change its speed); a magnetic force does neither. Trap 3: “Reversing the field direction reverses the force, and that is the only way to reverse it.” Reversing the sign of the charge also reverses the force. A proton and an electron moving in the same direction through the same field experience forces in opposite directions.

✓ Checkpoint · Magnetic Force on a Moving Charge

An alpha particle (\(q = +2e\)) moves with velocity \(2.0\times10^6\,\mathrm{m/s}\) at an angle of \(30^\circ\) to a uniform magnetic field of magnitude \(0.40\,\mathrm{T}\). Find the magnitude of the magnetic force on the alpha particle.
\(F = |q|vB\sin\theta = (2\times1.6\times10^{-19})(2.0\times10^6)(0.40)\sin30^\circ = 3.2\times10^{-19}\times2.0\times10^6\times0.40\times0.5 = 1.28\times10^{-13}\,\mathrm{N}\).
A particle with charge \(q = -3.2\times10^{-19}\,\mathrm{C}\) moves along the \(+x\) axis in a field \(\vec{B} = B_0\,\hat{z}\) (pointing out of the page). Using the right-hand rule on \(\vec{v}\times\vec{B}\) and then applying the sign of \(q\), determine the direction of the magnetic force.
\(\vec{v}\times\vec{B} = v\hat{x}\times B_0\hat{z} = vB_0(\hat{x}\times\hat{z}) = -vB_0\hat{y}\) (since \(\hat{x}\times\hat{z} = -\hat{y}\)). Multiplying by the negative charge: \(\vec{F} = (-|q|)(-vB_0\hat{y}) = +|q|vB_0\hat{y}\), i.e., in the \(+y\) direction.
A proton moves through a region of uniform magnetic field and follows a curved path. At a certain instant the proton's speed is \(v_0\). Without any other force acting, what is the proton's speed one second later?
Still \(v_0\). The magnetic force is always perpendicular to \(\vec{v}\), so \(\vec{F}\cdot\vec{v} = 0\) and no work is done. The kinetic energy is constant, so the speed is constant.
Two particles—a proton and an antiproton (same mass as a proton, but charge \(-e\))—are fired with the same velocity into the same uniform magnetic field. Compare the magnitude and direction of the force on each.
The magnitudes are equal: \(F = |q|vB\sin\theta\) is the same since \(|{+e}| = |{-e}|\), \(v\) and \(B\) are identical. The directions are opposite: the antiproton is deflected to the opposite side of its trajectory.

Worked Examples

Example 1 · Velocity and Field Given in Components

A proton moves with velocity \(\vec v = (3.0\,\hat x + 4.0\,\hat y)\times10^6\,\mathrm{m/s}\) in a uniform field \(\vec B = 0.20\,\hat z\,\mathrm{T}\). Find the magnetic force, as a vector and as a magnitude.

Solution. Given: \(\vec v = (3.0,\,4.0,\,0)\times10^6\,\mathrm{m/s}\), \(\vec B=(0,0,0.20)\,\mathrm{T}\), \(q=+1.6\times10^{-19}\,\mathrm{C}\).
Find: \(\vec F\) and \(|\vec F|\).
Setup: \(\vec F=q\,\vec v\times\vec B\); with \(\vec B=B\hat z\), \(\vec v\times\vec B=(v_yB,\,-v_xB,\,0)\).
Solve: \[ \vec v\times\vec B=\big[(4.0\times10^6)(0.20)\,\hat x-(3.0\times10^6)(0.20)\,\hat y\big] =(8.0\,\hat x-6.0\,\hat y)\times10^5, \] \[ \vec F=q(\vec v\times\vec B)=(1.28\,\hat x-0.96\,\hat y)\times10^{-13}\,\mathrm{N}. \] Answer: \(\boxed{\vec F=(1.28\,\hat x-0.96\,\hat y)\times10^{-13}\,\mathrm{N},\quad |\vec F|=1.6\times10^{-13}\,\mathrm{N}}\).
Check: \(\vec v\) lies in the \(xy\)-plane and \(\vec B\) points along \(\hat z\), so \(\vec v\perp\vec B\) and the magnitude must equal \(qvB\) with \(v=5.0\times10^6\,\mathrm{m/s}\): \(qvB=(1.6\times10^{-19})(5.0\times10^6)(0.20)=1.6\times10^{-13}\,\mathrm{N}\), matching \(\sqrt{1.28^2+0.96^2}\times10^{-13}\). The force has no \(\hat z\) component—it cannot, being perpendicular to \(\vec B\). ✓

Example 2 · Reading the Angle from the Force

A proton travelling at \(5.0\times10^5\,\mathrm{m/s}\) through a \(0.80\,\mathrm{T}\) field is measured to feel a magnetic force of \(3.2\times10^{-14}\,\mathrm{N}\). At what angle to the field is it moving?

Solution. Given: \(v=5.0\times10^5\,\mathrm{m/s}\), \(B=0.80\,\mathrm{T}\), \(F=3.2\times10^{-14}\,\mathrm{N}\), \(q=e\).
Find: the angle \(\theta\) between \(\vec v\) and \(\vec B\).
Setup: \(F=qvB\sin\theta\Rightarrow\sin\theta=F/(qvB)\).
Solve: \[ \sin\theta=\frac{3.2\times10^{-14}}{(1.6\times10^{-19})(5.0\times10^5)(0.80)} =\frac{3.2\times10^{-14}}{6.4\times10^{-14}}=0.50,\qquad\theta=30^\circ. \] Answer: \(\boxed{\theta=30^\circ\ (\text{or }150^\circ)}\).
Check: The maximum possible force here is \(qvB=6.4\times10^{-14}\,\mathrm{N}\); the measured value is exactly half, so \(\sin\theta=0.5\). The magnitude alone cannot separate \(30^\circ\) from \(150^\circ\)—both give the same \(\sin\theta\). ✓

Motion of a Charged Particle in a Uniform Magnetic Field

◈ Circular Orbit in a Field

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The gyration: change v, B, q and m and watch r = mv/qB resize while the period T = 2πm/qB holds fixed.

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We now have the force law. The next question is geometrical: if a force always acts perpendicular to the velocity, what trajectory results?

The answer is not obvious from ordinary experience. Gravity acts downward and produces parabolas. A constant sideways push produces a curve that spirals outward. But a force that is always perpendicular to the velocity is different from both of these. It can never accelerate or decelerate the particle. It can only redirect it. And because the redirected particle immediately acquires a new velocity, the perpendicular force changes direction too—always chasing the velocity vector, always ninety degrees behind it.

This is precisely the condition that produces uniform circular motion.

◈ Resolution · PQ 3 Here is the force that bends without speeding up: being perpendicular to \(\vec{v}\), it does no work, so the speed leaving the field equals the speed entering—only the direction has changed.

Perpendicular Entry: Circular Motion

Consider a particle with charge \(q\) and mass \(m\) entering a region of uniform magnetic field \(\vec{B}\) with its velocity \(\vec{v}\) entirely perpendicular to \(\vec{B}\). The magnetic force is then always of magnitude \(F = |q|vB\) and always directed toward the centre of the resulting circular orbit. It plays the role of the centripetal force.

Birth of the formula: The Cyclotron Radius

Setup. A charge \(q\), mass \(m\), speed \(v\), with \(\vec{v}\perp\vec{B}\). The magnetic force \(F=|q|vB\) is constant in magnitude and always points to the centre of the path—exactly a centripetal force. Solve. Equate magnetic and centripetal force: \[ |q|vB = \frac{mv^2}{r}. \] Result: \[ \boxed{r = \frac{mv}{|q|B}.} \] Read it. \(r\) grows with momentum \(mv\) (faster or heavier \(\Rightarrow\) gentler curve) and shrinks with \(|q|B\) (stronger field or larger charge \(\Rightarrow\) tighter curve).

This quantity is the cyclotron radius (or Larmor radius / radius of gyration, depending on context).

Core intuition: What the Radius Formula Tells You

\[ r = \frac{mv}{|q|B} \] \(r\) increases with momentum \(mv\): a faster or heavier particle curves more gently. \(r\) decreases with \(|q|B\): a stronger field or larger charge curves the particle more tightly. \(r\) is independent of the direction of \(\vec{v}\) (so long as \(\vec{v}\perp\vec{B}\)): every particle with the same momentum-to-charge ratio orbits at the same radius. This ratio \(p/q = mv/q\) is called the magnetic rigidity of the particle. Mass spectrometers exploit equation () directly: particles of the same charge but different masses land at different radii and are thereby separated.

From the radius we derive the remaining orbital quantities. The circumference is \(2\pi r\); the speed is \(v\). Therefore the time for one full revolution—the period—is: \[ T = \frac{2\pi r}{v} = \frac{2\pi m}{|q|B}. \] The corresponding cyclotron frequency (number of revolutions per second) is: \[ f = \frac{1}{T} = \frac{|q|B}{2\pi m}, \] and the angular frequency: \[ \omega_c = 2\pi f = \frac{|q|B}{m}. \] Notice that \(T\), \(f\), and \(\omega_c\) all depend only on the field and the charge-to-mass ratio, not on the speed. A slow proton and a fast proton in the same field complete orbits of different sizes in exactly the same time. This speed-independence of the period is the physical principle underlying the cyclotron accelerator.

Misconception alert

Trap 1: “A stronger magnetic force means the particle moves faster.” The speed is fixed by the initial conditions and never changes. A stronger field \(B\) means a smaller radius \(r\) (tighter circle) and a larger cyclotron frequency, but the speed is identical. The particle is being turned, not accelerated. Trap 2: “The period depends on speed, so a faster particle completes more orbits per second.” From equation (): \(T = 2\pi m/(|q|B)\)—no \(v\) appears. Period and frequency are entirely independent of speed. This is not obvious and runs counter to the intuition developed from pendulums or springs, where period depends on energy. Circular motion in a magnetic field is different: the radius adjusts to accommodate any speed, and the time to complete one orbit stays constant. Trap 3: “The magnetic force is centripetal, so it does centripetal work.” Work requires \(\vec{F}\cdot d\vec{s}\ne 0\). For circular motion, \(\vec{F}\) points radially inward and \(d\vec{s}\) is tangential—they are perpendicular, so \(\vec{F}\cdot d\vec{s} = 0\) everywhere. The magnetic force sustains the circular motion without doing work.

Oblique Entry: Helical Motion

What happens when the velocity is not perpendicular to \(\vec{B}\)?

Suppose the particle enters with velocity \(\vec{v}\) at an angle \(\alpha\) to the field direction \(\hat{B}\). Decompose \(\vec{v}\) into two components: \[\begin{aligned} v_\parallel &= v\cos\alpha \quad \text{(parallel to } \vec{B}),\\ v_\perp &= v\sin\alpha \quad \text{(perpendicular to } \vec{B}). \end{aligned}\]

The magnetic force is \(\vec{F} = q\vec{v}\times\vec{B}\). The parallel component \(v_\parallel\,\hat{B}\) is parallel to \(\vec{B}\), so \(v_\parallel\,\hat{B}\times\vec{B} = \vec{0}\)—it contributes nothing to the force. The perpendicular component \(v_\perp\) contributes the full cross product, giving a force perpendicular to both \(v_\perp\) and \(\vec{B}\).

The result is a superposition of two independent motions:

Uniform circular motion in the plane perpendicular to \(\vec{B}\), driven by \(v_\perp\), with radius \[ r = \frac{mv_\perp}{|q|B} = \frac{mv\sin\alpha}{|q|B}; \]
Uniform linear motion along \(\vec{B}\) at constant speed \(v_\parallel = v\cos\alpha\), since no force acts along the field.

The combination of a circle and a straight translation perpendicular to the circle's plane is a helix. The particle drills forward along the field direction while rotating in the transverse plane.

◈ Resolution · PQ 4 This settles the two protons: the head-on one has no parallel velocity and orbits in a plane, while the tilted one keeps its \(v_\parallel\) and drills forward in a helix, even though no force pushes it along the field.

The pitch of the helix, the distance the particle advances along \(\vec{B}\) in one complete transverse revolution, is: \[ p = v_\parallel \cdot T = v\cos\alpha \cdot \frac{2\pi m}{|q|B}. \] Note that the pitch does depend on speed (via \(v_\parallel\)), unlike the period. A faster particle with the same \(\alpha\) and \(B\) advances further per revolution, producing a helix with larger pitch but the same radius.

Core intuition: The Two Limiting Cases of Helical Motion

When \(\alpha = 90^\circ\): \(v_\parallel = 0\), \(v_\perp = v\). The helix degenerates to a circle. The particle goes nowhere along the field; it simply orbits in a plane. When \(\alpha = 0^\circ\) (or \(180^\circ\)): \(v_\perp = 0\), \(v_\parallel = v\). The helix degenerates to a straight line. The particle sails along the field direction with no deflection. All other angles produce genuine helical motion. The closer \(\alpha\) is to \(90^\circ\), the tighter the helix; the closer to \(0^\circ\), the more stretched it becomes.

Helical motion is not a laboratory curiosity. Charged particles from solar storms are captured by Earth's magnetic field (which runs roughly from geographic south to geographic north) and spiral along field lines toward the poles. The resulting collisions with atmospheric molecules produce the aurora borealis and aurora australis. In fusion reactors (tokamaks), helical trajectories confine plasma particles within a toroidal magnetic field, preventing them from striking the reactor wall.

✓ Checkpoint · Motion of a Charged Particle in a Magnetic Field

A deuteron (mass \(= 2\times m_p\), charge \(= +e\)) and a proton enter the same uniform magnetic field perpendicularly with the same speed. Compare their orbital radii and periods.
Radius: \(r = mv/(eB)\). For the deuteron, \(m_d = 2m_p\), so \(r_d = 2m_p v/(eB) = 2r_p\). The deuteron orbits at twice the radius. Period: \(T = 2\pi m/(eB)\). \(T_d = 2\pi(2m_p)/(eB) = 2T_p\). The deuteron takes twice as long to complete one revolution. (The period depends on mass, not on speed.)
A proton moves in a circle of radius \(4.0\,\mathrm{cm}\) in a magnetic field of \(0.50\,\mathrm{T}\). Find its speed.
From \(r = mv/(eB)\): \(v = erB/m = (1.6\times10^{-19})(0.04)(0.50)/(1.67\times10^{-27}) = 3.2\times10^{-21}/1.67\times10^{-27} \approx 1.92\times10^6\,\mathrm{m/s}\).
A charged particle enters a uniform magnetic field at \(45^\circ\) to the field direction and follows a helical path. (a) How does the radius of the helix compare to that of a particle of the same mass, charge, and speed entering perpendicularly? (b) How does the pitch of the helix change if the speed is doubled while the angle and field are kept the same?
(a) \(r = mv_\perp/(|q|B) = mv\sin45^\circ/(|q|B)\). Compared to perpendicular entry (\(r_\perp = mv/(|q|B)\)): \(r_{45} = r_\perp/\sqrt{2} \approx 0.707\,r_\perp\). The oblique particle has a smaller radius. (b) Pitch \(p = v_\parallel T = v\cos45^\circ \cdot 2\pi m/(|q|B) \propto v\). Doubling \(v\) doubles both \(v_\parallel\) and \(v_\perp\), so the pitch doubles. (The radius also doubles, but \(T\) is unchanged.)
An electron spirals along the Earth's magnetic field lines toward the north magnetic pole, producing part of the aurora. The field at a particular point is \(5.0\times10^{-5}\,\mathrm{T}\) and the electron's speed is \(2.0\times10^7\,\mathrm{m/s}\) at \(80^\circ\) to the field. Find the radius of the circular component of its helical motion.
\(v_\perp = v\sin80^\circ \approx 1.97\times10^7\,\mathrm{m/s}\). \(r = m_e v_\perp/(eB) = (9.11\times10^{-31})(1.97\times10^7)/((1.6\times10^{-19})(5.0\times10^{-5})) = 1.79\times10^{-23}/8.0\times10^{-24} \approx 2.2\,\mathrm{m}\). The spiralling aurora electron sweeps circles of about \(2\,\mathrm{m}\) radius as it advances along a field line that spans thousands of kilometres.

Worked Examples

Example 3 · How Many Times Around?

An electron circulates in a uniform field \(B=1.2\,\mathrm{mT}\). Find (a) the period of its motion and (b) the number of complete revolutions it makes in \(1.0\,\mathrm{ms}\).

Solution. Given: \(B=1.2\times10^{-3}\,\mathrm{T}\), \(t=1.0\times10^{-3}\,\mathrm{s}\), \(m_e=9.11\times10^{-31}\,\mathrm{kg}\).
Find: \(T\) and \(N=t/T\).
Setup: \(T=2\pi m_e/(eB)\); revolutions \(=t/T=ft\).
Solve: \[ T=\frac{2\pi(9.11\times10^{-31})}{(1.6\times10^{-19})(1.2\times10^{-3})} \approx2.98\times10^{-8}\,\mathrm{s}, \] \[ N=\frac{t}{T}=\frac{1.0\times10^{-3}}{2.98\times10^{-8}}\approx3.4\times10^{4}\ \text{revolutions}. \] Answer: \(\boxed{T\approx30\,\mathrm{ns},\quad N\approx3.4\times10^{4}}\).
Check: The cyclotron period is independent of speed, so this count holds however fast the electron was injected, as long as \(B\) is fixed. A \(30\,\mathrm{ns}\) period is a \(34\,\mathrm{MHz}\) orbit—in the radio band. ✓

Example 4 · Same Energy, Different Ions

A proton, a deuteron (\(m=2m_p\), \(q=+e\)) and an alpha particle (\(m=4m_p\), \(q=+2e\)) are each given the same kinetic energy \(K=1.0\,\mathrm{keV}\) and injected perpendicular to a \(0.10\,\mathrm{T}\) field. Find the proton's orbit radius and compare the three radii.

Solution. Given: \(K=1.0\,\mathrm{keV}=1.6\times10^{-16}\,\mathrm{J}\), \(B=0.10\,\mathrm{T}\).
Find: \(r_p\) and the ratio \(r_p:r_d:r_\alpha\).
Setup: \(r=\dfrac{mv}{qB}=\dfrac{\sqrt{2mK}}{qB}\) (from \(v=\sqrt{2K/m}\)), so at fixed \(K\), \(r\propto\sqrt m/q\).
Solve: \[ r_p=\frac{\sqrt{2m_pK}}{eB} =\frac{\sqrt{2(1.67\times10^{-27})(1.6\times10^{-16})}}{(1.6\times10^{-19})(0.10)} \approx4.6\times10^{-2}\,\mathrm{m}, \] \[ \frac{r_d}{r_p}=\frac{\sqrt{2m_p}/e}{\sqrt{m_p}/e}=\sqrt2,\qquad \frac{r_\alpha}{r_p}=\frac{\sqrt{4m_p}/2e}{\sqrt{m_p}/e}=\frac{2}{2}=1. \] Answer: \(\boxed{r_p\approx4.6\,\mathrm{cm};\quad r_p:r_d:r_\alpha=1:\sqrt2:1}\).
Check: At equal kinetic energy the alpha and proton share one radius—the alpha's larger mass is exactly offset by its doubled charge. Hold momentum fixed instead (\(r\propto1/q\)), or speed fixed (\(r\propto m/q\)), and the ordering changes: what is “bigger” depends entirely on what is held constant. ✓

Solved examples

Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.

Example 1 · Lorentz Force: Direction and Magnitude

A proton (\(q = +1.6\times10^{-19}\,\mathrm{C}\), \(m = 1.67\times10^{-27}\,\mathrm{kg}\)) moves with velocity \(\vec{v} = 3.0\times10^6\,\hat{x}\,\mathrm{m/s}\) in a uniform field \(\vec{B} = 0.50\,\hat{y}\,\mathrm{T}\).

Find the magnetic force on the proton.
Find its initial acceleration.
If the proton were replaced by an electron with the same velocity, in what direction would the force act?

Solution. (a) Force. With \(\hat{x}\times\hat{y} = \hat{z}\), \[ \vec{F} = q\,\vec{v}\times\vec{B} = (1.6\times10^{-19})(3.0\times10^6)(0.50)\,\hat{z} = 2.4\times10^{-13}\,\hat{z}\;\mathrm{N}. \] (b) Acceleration. \[ a = \frac{F}{m} = \frac{2.4\times10^{-13}}{1.67\times10^{-27}} \approx 1.4\times10^{14}\,\mathrm{m/s^2}\quad(+\hat{z}). \] Enormous, but it acts for a vanishing instant before the velocity—and hence the force—rotates. (c) Electron. The cross product \(\vec{v}\times\vec{B}\) is unchanged (\(+\hat{z}\)), but the negative charge flips the force to \(-\hat{z}\). Proton and electron curve to opposite sides—the principle of the mass spectrometer. Check. The force \(2.4\times10^{-13}\,\mathrm{N}\) is perpendicular to both \(\vec{v}\) (\(\hat{x}\)) and \(\vec{B}\) (\(\hat{y}\)), so it lands along \(\hat{z}\); swapping the proton for an electron flips the sign but not the magnitude—mass-spectrometer separation in one stroke. ✓

Example 2 · Zero Force: the Parallel Case

An electron moves with \(\vec{v} = 5.0\times10^5\,\hat{z}\,\mathrm{m/s}\) in a field \(\vec{B} = 0.30\,\hat{z}\,\mathrm{T}\). Find the magnetic force.

Solution. Velocity is parallel to the field (\(\theta = 0\)): \[ \vec{F} = q\,(v\hat{z})\times(B\hat{z}) = qvB\,(\hat{z}\times\hat{z}) = \vec{0}. \] The force is exactly zero, regardless of speed, charge, or field strength. The common distractor \(qvB\) forgets the \(\sin\theta\) factor, which here is \(\sin0 = 0\). Check. With \(\vec{v}\parallel\vec{B}\) the cross product vanishes identically—no value of speed or field strength can revive it. The bare \(qvB\) silently drops the \(\sin\theta=0\). ✓

Example 3 · Proton in a Cyclotron Field

A proton enters a uniform field \(B = 0.80\,\mathrm{T}\) at speed \(v = 2.0\times10^6\,\mathrm{m/s}\), perpendicular to \(\vec{B}\). Find (a) the orbit radius, (b) the period, (c) the cyclotron frequency. (\(m_p = 1.67\times10^{-27}\,\mathrm{kg}\), \(e = 1.6\times10^{-19}\,\mathrm{C}\).)

Solution. (a) Radius. \[ r = \frac{m_p v}{eB} = \frac{(1.67\times10^{-27})(2.0\times10^6)}{(1.6\times10^{-19})(0.80)} \approx 2.6\times10^{-2}\,\mathrm{m} = 2.6\,\mathrm{cm}. \] (b) Period. \[ T = \frac{2\pi m_p}{eB} \approx 8.2\times10^{-8}\,\mathrm{s}. \] The period is independent of speed—the defining feature of cyclotron motion. (c) Frequency. \[ f = \frac{1}{T} \approx 1.2\times10^7\,\mathrm{Hz} \approx 12\,\mathrm{MHz}, \] in the radio-frequency band, which is why a fixed-frequency RF source can drive the acceleration. Check. The three results are mutually consistent: \(v=2\pi r/T=2\pi(0.026)/(8.2\times10^{-8}) \approx2.0\times10^6\,\mathrm{m/s}\), the given speed. Because \(T\) is speed-independent, a faster proton would trace a larger circle in the same time. ✓

Example 4 · Force on a Current-Carrying Wire

A straight horizontal wire of length \(0.60\,\mathrm{m}\) carries \(5.0\,\mathrm{A}\) due east. A uniform field \(B = 0.20\,\mathrm{T}\) points vertically upward. (a) Find the force on the wire. (b) What is the force if the current is instead directed vertically upward?

Solution. Take \(\hat{x}\) east, \(\hat{y}\) north, \(\hat{z}\) up. (a) Current east, field up. With \(\hat{x}\times\hat{z} = -\hat{y}\), \[ \vec{F} = I\vec{L}\times\vec{B} = 5.0\,(0.60\,\hat{x})\times(0.20\,\hat{z}) = -0.60\,\hat{y}\;\mathrm{N}. \] Magnitude \(0.60\,\mathrm{N}\), directed due south. (b) Current up, field up. \(\vec{L}\parallel\vec{B}\), so \(\vec{F} = \vec{0}\). A wire laid along the field feels no force. Check. In (a) the current (\(\hat{x}\)) crosses the field (\(\hat{z}\)) to give a southward force; in (b) aligning the wire with \(\vec{B}\) zeroes it. The magnitude \(0.60\,\mathrm{N}=ILB\) uses the full length only because current and field are perpendicular there. ✓

Example 5 · Force on a Tilted Wire

A wire of length \(0.80\,\mathrm{m}\) carries \(3.0\,\mathrm{A}\) at \(30^\circ\) to a field \(B = 0.50\,\mathrm{T}\). Find the force magnitude.

Solution. \[ F = ILB\sin\theta = (3.0)(0.80)(0.50)\sin30^\circ = 0.60\,\mathrm{N}. \] Equivalently, only the perpendicular projection \(L\sin30^\circ = 0.40\,\mathrm{m}\) contributes: \(F = IB(0.40) = 0.60\,\mathrm{N}\). Check. Using the full \(0.80\,\mathrm{m}\) would double the answer—a classic slip. The result \(0.60\,\mathrm{N}\) sits below the \(ILB=1.2\,\mathrm{N}\) maximum reached at \(90^\circ\). ✓

Moving Charges and Magnetism

Perplexing Questions

Why Magnetism Feels Different

Magnetic Force on a Moving Charge

Worked Examples

Motion of a Charged Particle in a Uniform Magnetic Field

Perpendicular Entry: Circular Motion

Oblique Entry: Helical Motion

Worked Examples

Solved examples

Problem bank

Chapter test

The chapter continues.