Chapter 4: Moving Charges and Magnetism

A current-carrying wire and a magnet exert force on each other — this is the discovery of Oersted (1820) that united electricity and magnetism into a single subject, electromagnetism. This chapter develops:

1. The magnetic force $\vec{F} = q\vec{v}\times\vec{B}$ acting on moving charges.
2. The sources of magnetic fields — currents — via the Biot–Savart law and Ampère's law.
3. Devices built on these principles: the cyclotron, solenoid, moving-coil galvanometer, and conversions to ammeter/voltmeter.

The mathematics is more vector-heavy than electrostatics: dot products and especially cross products are everywhere.

Concept Map

• 4.1 Magnetic field — Oersted's experiment
• 4.2 Magnetic force on a moving charge — Lorentz force
• 4.3 Motion in a magnetic field — circular and helical
• 4.4 Cyclotron
• 4.5 Magnetic force on a current-carrying conductor
• 4.6 Biot–Savart law and applications
• 4.7 Ampère's circuital law: wire, solenoid, toroid
• 4.8 Force between parallel currents; definition of ampere
• 4.9 Torque on a current loop; moving-coil galvanometer
• 4.10 Galvanometer $\to$ ammeter and voltmeter

4.1 Magnetic Field — Oersted's Experiment

Observation

A compass needle placed near a current-carrying wire deflects, showing that the wire creates a magnetic field. Reversing the current reverses the deflection. Thus a magnetic field can be produced by moving charges (currents) as well as by permanent magnets.

Magnetic Field $\vec{B}$

The magnetic field $\vec{B}$ is a vector field. It is sometimes called the magnetic flux density.

SI unit. Tesla ($\text{T}$): $1\text{ T} = 1\text{ N}/(\text{A m}) = 1\text{ Wb/m}^2$. The CGS unit is the gauss ($1\text{ G} = 10^{-4}\text{ T}$).

Typical values: Earth's surface $\sim 5\times 10^{-5}\text{ T}$; small bar magnet $\sim 10^{-2}\text{ T}$; MRI machine $\sim 1\text{–}3\text{ T}$.

Right-Hand Rule for Wire

For a straight wire, $\vec{B}$ encircles the wire; curl your right hand with thumb along the current — your fingers curl in the direction of $\vec{B}$.

4.2 Magnetic Force on a Moving Charge — Lorentz Force

Definition

A charge $q$ moving with velocity $\vec{v}$ in a magnetic field $\vec{B}$ experiences the magnetic force

Magnitude: $F = qvB\sin\theta$ where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$.

Direction. Perpendicular to both $\vec{v}$ and $\vec{B}$ — right-hand rule.

Properties

1. Force is zero when $\vec{v}\parallel\vec{B}$ ($\theta=0$ or $\pi$).
2. Force is maximum ($qvB$) when $\vec{v}\perp\vec{B}$.
3. The magnetic force is always perpendicular to $\vec{v}$, so it does no work: $\vec{F}\cdot\vec{v}=0$. Kinetic energy of the charge is conserved.

Lorentz Force (combined with electric field)

If both $\vec{E}$ and $\vec{B}$ are present:

Worked Example

An electron ($q = -1.6\times 10^{-19}\text{ C}$) moves at $\vec{v} = 2\times 10^6\,\hat{i}\text{ m/s}$ in $\vec{B} = 0.5\,\hat{j}\text{ T}$. Find the force.

$\vec{v}\times\vec{B} = 2\times 10^6\cdot 0.5\,\hat{k} = 10^6\,\hat{k}$. $\vec{F} = q\vec{v}\times\vec{B} = -(1.6\times 10^{-19})(10^6)\hat{k} = -1.6\times 10^{-13}\,\hat{k}\text{ N}$.

Pitfalls

• The magnetic force changes the direction of $\vec{v}$, not its magnitude.
• Sign of charge: for an electron, $\vec{F}$ is in the opposite direction to $\vec{v}\times\vec{B}$.
• Don't confuse $\vec{F} = q\vec{v}\times\vec{B}$ (force on a charge) with $\vec{F} = I\vec{L}\times\vec{B}$ (force on a current).

4.3 Motion in a Magnetic Field

(a) Charge Moving Perpendicular to $\vec{B}$

If $\vec{v}\perp\vec{B}$, the magnetic force always remains perpendicular to $\vec{v}$ — it acts as a centripetal force. The charge moves in a circle.

Step 1. $\vert qvB\vert = mv^2/r$.

Step 2. Radius:

Step 3. Angular speed: $\omega = v/r = qB/m$. Period:

$T$ is independent of $v$ and $r$ — the basis of the cyclotron.

(b) Helical Motion

If $\vec{v}$ has a component along $\vec{B}$, decompose: $v_\parallel$ (unaffected) and $v_\perp$ (gives circular motion). Result: a helix with:

• Radius $r = mv_\perp/(qB)$.
• Pitch (distance per turn along axis) $p = v_\parallel T = 2\pi m v_\parallel/(qB)$.

Worked Example

A proton ($m = 1.67\times 10^{-27}$ kg) enters $\vec{B} = 0.1\text{ T}$ at $v = 10^6\text{ m/s}$ perpendicular to $\vec{B}$.

$r = mv/(qB) = (1.67\times 10^{-27})(10^6)/((1.6\times 10^{-19})(0.1)) = 0.104\text{ m}$. $T = 2\pi r/v = 6.55\times 10^{-7}\text{ s}$ ($\nu \approx 1.53\text{ MHz}$).

Pitfalls

• Cyclotron period is independent of $v$ only in the non-relativistic regime. At very high $v$, $m$ grows and the period drifts.
• For an electron the circle is in the opposite sense compared to a proton in the same $\vec{B}$.

4.4 Cyclotron

Principle

The cyclotron exploits the fact that the period of circular motion is independent of $v$. Two semicircular "dees" sit in a uniform magnetic field. An alternating voltage between them, with frequency matching $\nu_c = qB/(2\pi m)$, accelerates the particle each time it crosses the gap — the radius grows but the timing stays in sync.

Resonance Condition

Maximum Kinetic Energy

When the particle reaches the maximum radius $R$:

Limitations

1. Relativistic mass increase — at high $v$, $m$ rises, $\nu_c$ drops, particle goes out of sync.
2. Cannot accelerate uncharged particles (no Lorentz force).
3. Cannot accelerate electrons efficiently — they go relativistic too quickly.
4. Limited to positive ions of moderate energy ($\sim$ tens of MeV).

Improvements: synchrocyclotron (variable frequency), synchrotron (variable $B$).

Worked Example

Cyclotron with $B = 1\text{ T}$ and $R = 0.5\text{ m}$, accelerating protons.

$K_{\max} = (1.6\times 10^{-19})^2(1)^2(0.5)^2/(2\cdot 1.67\times 10^{-27}) = 1.92\times 10^{-12}\text{ J} = 12\text{ MeV}$.

Pitfalls

• The dees themselves do not accelerate the particle — they shield it from the AC field; acceleration only occurs in the gap.
• The frequency depends on $q/m$ ratio; protons and deuterons need different cyclotron frequencies.

4.5 Magnetic Force on a Current-Carrying Wire

Derivation

A wire of length $L$, cross-section $A$, current $I$ in field $\vec{B}$ contains $n A L$ free electrons drifting at $\vec{v}_d$. Total force:

But $I = nAev_d$ and the current direction is opposite to $\vec{v}_d$ (electron flow), so $I\vec{L} = -nAe v_d \vec{L} = -nAe\vec{v}_d L$... working out carefully:

where $\vec{L}$ is the length vector in the direction of conventional current.

For a curved wire of arbitrary shape:

In a uniform $\vec{B}$, the integral $\int d\vec{l}$ depends only on the endpoints — a closed loop in uniform $\vec{B}$ experiences zero net force (but possibly a torque).

Worked Example

A horizontal wire of length $0.5\text{ m}$ carries $I = 4\text{ A}$ northward. $\vec{B} = 0.2\text{ T}$ vertically downward. Force on the wire:

$F = ILB\sin 90° = 4\cdot 0.5\cdot 0.2 = 0.4\text{ N}$, directed eastward (right-hand rule).

Pitfalls

• Closed loop in uniform $\vec{B}$: $\vec{F}_{\text{net}} = 0$.
• Force depends on the wire's orientation, not where it physically is in space (as long as $\vec{B}$ is uniform).

4.6 Biot–Savart Law

Statement

A small element $d\vec{l}$ of a wire carrying current $I$ creates, at a point $P$ at displacement $\vec{r}$, a magnetic field

where $\mu_0 = 4\pi\times 10^{-7}\text{ T m/A}$ is the permeability of free space, and $\hat{r}$ is from $d\vec{l}$ to $P$.

Magnitude: $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$.

Comparison with Coulomb's Law

Application (a) — Straight Wire (Finite)

Long straight wire of length spanning angles $\theta_1$ to $\theta_2$ as seen from $P$ (perpendicular distance $a$).

Step 1. For an element $dl$ at angle $\theta$, $r = a/\sin\theta$, $dl = a\,d\theta/\sin^2\theta$ (using $l = -a\cot\theta$).

Step 2. All $d\vec{B}$ point in the same direction (out/into page).

Step 3.

(with appropriate sign convention; an alternative common form is $B = \dfrac{\mu_0 I}{4\pi a}(\sin\phi_1 + \sin\phi_2)$ where $\phi$ measured from perpendicular).

Special case — Infinite wire ($\theta_1=0, \theta_2=\pi$):

Semi-infinite wire (from $-\infty$ to the foot of perpendicular):

Application (b) — Circular Loop (Centre)

Loop of radius $R$ carrying $I$.

Step 1. For every element $d\vec{l}$, $d\vec{l}\perp\hat{r}$, so $\sin\theta = 1$. Also $r = R$.

Step 2. All elements produce $dB$ in the same direction along the axis at the centre.

Step 3.

For $N$ turns: $B = \mu_0 N I/(2R)$.

Application (c) — Circular Loop (Axis)

Point on axis at distance $z$ from centre.

Step 1. Each element gives $dB = \mu_0 I\,dl/(4\pi (R^2+z^2))$ (since $d\vec{l}\perp\hat{r}$).

Step 2. Components perpendicular to axis cancel; axial components add. Axial fraction: $\cos\alpha = R/\sqrt{R^2+z^2}$.

Step 3.

For $z\gg R$: $B_z \approx \mu_0 I R^2/(2z^3) = \mu_0(2 m)/(4\pi z^3)$ where $m = I\pi R^2$ is the magnetic moment — same $1/r^3$ dependence as an electric dipole on axis.

Pitfalls

• Biot–Savart is the magnetic analog of Coulomb's law — for currents, not charges.
• Always identify the direction of $d\vec{l}\times\hat{r}$ before applying magnitude formulas.
• For curved wires, use symmetry — most elements give components that cancel.

4.7 Ampère's Circuital Law

Statement

The line integral of $\vec{B}$ around any closed loop equals $\mu_0$ times the net current enclosed by the loop:

Sign convention: $I_{\text{enc}}$ is positive when current flows in the direction given by the right-hand rule from the orientation of the loop.

Like Gauss's law for $\vec{E}$, Ampère's law is always true, but useful only when symmetry lets us pull $B$ out of the integral.

Application (a) — Infinite Straight Wire

Circular Amperian loop of radius $r$ coaxial with the wire. By symmetry, $\vec{B}$ is tangent and constant in magnitude.

$\oint B\,dl = B\cdot 2\pi r = \mu_0 I$, so

(Same as Biot–Savart — but with much less work.)

Application (b) — Solenoid

A long solenoid: closely wound, $n$ turns per unit length, current $I$.

Inside (away from ends), $\vec{B}$ is uniform along the axis; outside, $B\to 0$ (for an ideal long solenoid).

Amperian loop: a rectangle of length $L$, one side parallel to axis inside the solenoid, the opposite side far outside, two short sides perpendicular to axis.

• Inside leg: $\int \vec{B}\cdot d\vec{l} = BL$.
• Outside leg: $\approx 0$.
• Short sides: $\vec{B}\perp d\vec{l}$, contribution zero.

$\oint = BL$. Enclosed current: $nLI$.

For a solenoid of finite length, near the ends $B$ drops to half: $B_{\text{end}} = \mu_0 n I/2$.

Application (c) — Toroid

A toroid is a solenoid bent into a doughnut. $N$ total turns, mean radius $r$. Amperian loop: a circle of radius $r$ inside the toroid.

$\oint B\,dl = B\cdot 2\pi r = \mu_0 N I$:

Outside the toroid: $B = 0$ (Amperian loops outside enclose zero net current).

Worked Example

A solenoid has $1000$ turns over $50\text{ cm}$ and carries $2\text{ A}$.

$n = 1000/0.5 = 2000\text{ /m}$. $B = (4\pi\times 10^{-7})(2000)(2) = 5\times 10^{-3}\text{ T}$.

Pitfalls

• Ampère's law works only for steady currents (in this chapter — Maxwell's correction comes later).
• Symmetry is essential; randomly drawn loops produce useless equations.
• The integral is over the loop, not the current.

4.8 Force Between Parallel Currents; Definition of the Ampere

Setup

Two long parallel wires, separation $d$, currents $I_1, I_2$.

Step 1. Field due to wire 1 at the location of wire 2: $B_1 = \mu_0 I_1/(2\pi d)$.

Step 2. Force per unit length on wire 2: $F/L = I_2 B_1 = \dfrac{\mu_0 I_1 I_2}{2\pi d}$.

Step 3.

Direction:

• Currents in same direction: wires attract.
• Currents in opposite directions: wires repel.

Definition of the Ampere

The historical SI definition (pre-2019): one ampere is the steady current which, when flowing in two parallel infinite wires $1\text{ m}$ apart in vacuum, produces a force per unit length of $2\times 10^{-7}\text{ N/m}$ between them.

(Since 2019, the ampere is defined in terms of the elementary charge $e$; but the parallel-wire formula remains the working test.)

Worked Example

Two parallel wires, $0.5\text{ m}$ apart, both carrying $10\text{ A}$ in the same direction. Force per meter:

Pitfalls

• "Same direction $\to$ attract" — counter-intuitive if you only know charges. The field/force geometry differs from Coulomb.
• The formula is for infinite parallel wires; for finite segments it's an approximation.

4.9 Torque on a Current Loop; Moving-Coil Galvanometer

Torque on a Rectangular Loop

Loop of sides $a$ (horizontal) and $b$ (vertical), current $I$, in uniform $\vec{B}$ horizontal. The loop's normal $\hat{n}$ makes angle $\theta$ with $\vec{B}$.

Two horizontal sides experience opposite forces along the loop axis — no torque (or cancellation).

Two vertical sides (length $b$) carry currents perpendicular to $\vec{B}$. Each experiences force $F = IbB$, equal and opposite, forming a couple.

Perpendicular distance between the forces: $a\sin\theta$.

where $A = ab$ is the loop area.

For $N$ turns: $\tau = NIAB\sin\theta$.

Magnetic Dipole Moment

Define $\vec{m} = NI\vec{A}$ (vector area, by right-hand rule); SI unit: $\text{A m}^2$. Then

(Direct analog of electric dipole: $\vec{p}\leftrightarrow\vec{m}$, $\vec{E}\leftrightarrow\vec{B}$.)

Moving-Coil Galvanometer

A coil of $N$ turns, area $A$, suspended in a radial magnetic field (produced by curved pole pieces). When current $I$ flows:

• Deflecting torque: $\tau_{\text{def}} = NIAB$ (because $\theta = 90°$ always — that's the point of the radial field).
• Restoring torque: $\tau_{\text{rest}} = k\phi$ (a torsion spring with constant $k$, $\phi$ is the deflection).

At equilibrium:

Linear scale: deflection $\propto$ current.

Sensitivity

• Current sensitivity: $\phi/I = NAB/k$.
• Voltage sensitivity: $\phi/V = NAB/(kR_g)$ where $R_g$ is the galvanometer's own resistance.

To increase current sensitivity: increase $N$, $A$, $B$; decrease $k$.

But increasing $N$ also increases $R_g$, so voltage sensitivity may not improve.

Pitfalls

• Galvanometer's pole pieces are curved precisely to keep $\theta = 90°$ — this is what makes the scale linear.
• "Sensitivity" and "accuracy" are not the same; a sensitive instrument can be precise but biased.

4.10 Conversion of a Galvanometer

To an Ammeter

An ammeter must measure large currents and have low resistance (so it doesn't disturb the circuit). Add a small shunt $S$ in parallel with the galvanometer.

Step 1. Let full-scale current of galvanometer be $I_g$ (with $R_g$). To read up to $I$, shunt must carry $I - I_g$.

Step 2. Same voltage across $G$ and $S$:

Step 3. Effective resistance of ammeter:

For $I\gg I_g$, $S\ll R_g$, so $R_A$ is very small — ideal.

To a Voltmeter

A voltmeter must measure large voltages without drawing current — high resistance. Add a large series resistance $R$.

Step 1. For full-scale voltage $V$:

Step 2. Effective resistance: $R_V = R + R_g \gg R_g$ — ideal voltmeter draws minimal current.

Worked Example

A galvanometer with $R_g = 50\,\Omega$, $I_g = 1\text{ mA}$:

• To read up to $1\text{ A}$ as an ammeter: $S = (10^{-3})(50)/(1 - 10^{-3}) \approx 0.05\,\Omega$.
• To read up to $10\text{ V}$ as a voltmeter: $R = 10/10^{-3} - 50 = 9950\,\Omega$.

Pitfalls

• An ammeter goes in series in the circuit; a voltmeter goes in parallel.
• Putting a voltmeter in series with a load gives essentially zero current — students often make this mistake.
• Ammeters are dangerous to apply directly across a battery — low resistance means huge currents.

Solved Problems

Problem 1 (Easy)

A wire carries $5\text{ A}$ in a horizontal direction. The horizontal component of Earth's magnetic field is $4\times 10^{-5}\text{ T}$. The force on a $1\text{ m}$ length of the wire (perpendicular to $\vec{B}$):

Problem 2 (Easy)

A proton enters a $\vec{B} = 0.2\text{ T}$ field at $v = 2\times 10^7\text{ m/s}$ perpendicular to $\vec{B}$. Find the radius.

Problem 3 (Medium)

A circular coil of radius $0.1\text{ m}$, $100$ turns, carries $2\text{ A}$. Find $\vec{B}$ at its centre and at $0.1\text{ m}$ on the axis.

Centre: $B = \mu_0 N I/(2R) = (4\pi\times 10^{-7})(100)(2)/(2\cdot 0.1) = 1.26\times 10^{-3}\text{ T}$.

On axis ($z = 0.1$): $B = \mu_0 N I R^2/(2(R^2+z^2)^{3/2}) =$ with $R^2+z^2 = 0.02$, $(R^2+z^2)^{3/2} = 2.83\times 10^{-3}$. $B = (4\pi\times 10^{-7})(100)(2)(0.01)/(2\cdot 2.83\times 10^{-3}) = 4.44\times 10^{-4}\text{ T}$.

Problem 4 (Medium)

A solenoid of length $0.5\text{ m}$ and $500$ turns carries current $I$. Determine $I$ such that $B$ inside is $0.01\text{ T}$.

Problem 5 (Medium) — Parallel Wires

Two long parallel wires, $5\text{ cm}$ apart, carry $4\text{ A}$ and $6\text{ A}$ in opposite directions. Find the force per meter.

Problem 6 (Hard) — Cyclotron Design

A cyclotron of dee radius $0.4\text{ m}$ accelerates protons to $10\text{ MeV}$. Find the magnetic field needed and the required AC frequency.

$K = q^2B^2R^2/(2m)$. Solving:

$2mK = 2\cdot 1.67\times 10^{-27}\cdot 1.6\times 10^{-12} = 5.34\times 10^{-39}$. $\sqrt{5.34\times 10^{-39}/(2.56\times 10^{-38})} = \sqrt{0.2086} = 0.457\text{ ... in units T m}$. $B = 0.457/0.4 = 1.14\text{ T}$.

Frequency: $\nu = qB/(2\pi m) = (1.6\times 10^{-19}\cdot 1.14)/(2\pi\cdot 1.67\times 10^{-27}) = 1.74\times 10^7\text{ Hz} = 17.4\text{ MHz}$.

Problem 7 (Hard) — Galvanometer Conversion

A galvanometer ($R_g = 100\,\Omega$, $I_g = 10\,\mu\text{A}$) is to be converted (a) to an ammeter of range $0\text{–}1\text{ A}$, and (b) to a voltmeter of range $0\text{–}100\text{ V}$.

(a) $S = I_g R_g/(I - I_g) = 10^{-5}\cdot 100/(1 - 10^{-5}) \approx 10^{-3}\,\Omega = 1\text{ m}\Omega$.

(b) $R = V/I_g - R_g = 100/10^{-5} - 100 = 10^7 - 100 \approx 10\text{ M}\Omega$.

JEE/NEET Edge Cases

1. Velocity selector: Crossed $\vec{E}$ and $\vec{B}$ select particles with $v = E/B$. Used in mass spectrometers.

2. Cyclotron loses sync for high-energy particles due to relativistic mass increase; synchrocyclotrons modulate frequency.

3. Force on a closed loop in uniform $\vec{B}$ is zero, but torque generally non-zero.

4. Net force on a current loop in a non-uniform field: Use $\vec{F} = \nabla(\vec{m}\cdot\vec{B})$. This is how a magnet attracts a current loop.

5. Helical pitch ratio: for a charge entering at angle $\theta$ to $\vec{B}$, pitch/circumference $= v_\parallel/v_\perp = \cot\theta$.

6. Solenoid end vs middle: $B_{\text{end}} = B_{\text{mid}}/2$ for a long solenoid.

7. Current loop near a long wire: the loop experiences a net force (one side closer to the wire than the other).

8. Toroid outside is zero: but only for an ideal toroid; real-world toroids leak a little.

9. Galvanometer sensitivity vs robustness: increasing $N$ and $A$ makes coil heavier; spring constant $k$ has practical lower bound.

10. Ammeter has lowest resistance, voltmeter has highest — swap them and you'll either short the battery or read essentially nothing.

11. Magnetic field of a finite straight wire at perpendicular foot: $B = (\mu_0 I/4\pi a)(\sin\phi_1+\sin\phi_2)$ where $\phi_i$ are angles to the ends from the perpendicular. Don't forget this — the infinite formula doesn't always apply.

Quick Recap

• $\vec{F} = q(\vec{E} + \vec{v}\times\vec{B})$; magnetic force does no work.
• Circle in $\vec{B}$: $r = mv/qB$; $T = 2\pi m/qB$.
• Cyclotron: $\nu = qB/(2\pi m)$; $K_{\max} = q^2B^2R^2/(2m)$.
• Wire force: $\vec{F} = I\vec{L}\times\vec{B}$.
• Biot–Savart: $d\vec{B} = (\mu_0/4\pi)(Id\vec{l}\times\hat{r})/r^2$.
• Wire: $B = \mu_0 I/(2\pi a)$. Loop centre: $\mu_0 I/(2R)$. Solenoid: $\mu_0 n I$. Toroid: $\mu_0 N I/(2\pi r)$.
• Parallel wires: $F/L = \mu_0 I_1 I_2/(2\pi d)$; same direction $\to$ attract.
• Loop torque: $\vec{\tau} = \vec{m}\times\vec{B}$; $\vec{m} = NI\vec{A}$.
• Galvanometer: $\phi = NABI/k$.
• Ammeter shunt: $S = I_gR_g/(I-I_g)$; Voltmeter series: $R = V/I_g - R_g$.

Formula Sheet