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Chapter 5 • Theory & Derivations

Magnetic Field

Lorentz force on moving charges and currents, cyclotron motion and helical trajectories, magnetic torque on current loops, moving-coil galvanometers, the Hall effect, the Biot-Savart law with circular loop applications, and Ampere's circuital law applied to straight conductors, solenoids, and toroids.

§5.1 The Magnetic Field, Lorentz Force, and Charged Particle Trajectories

Magnetism originates from electric charges in motion. A magnetic field is established by moving charges or permanent magnetic dipoles and exerts forces exclusively on moving charges.

1. The Lorentz Force Law

A particle carrying electric charge $q$ moving with velocity $\vec{v}$ in a region containing both an electric field $\vec{E}$ and a magnetic field $\vec{B}$ experiences the unified Lorentz Force: $$\vec{F} = q \left( \vec{E} + \vec{v} \times \vec{B} \right)$$ The magnetic force component is: $$\vec{F}_B = q (\vec{v} \times \vec{B})$$ Magnitude: $F_B = |q| v B \sin\theta$, where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. Direction: Governed by the vector cross product (Right-Hand Rule). SI Unit: **Tesla (T)** ($1 \text{ Tesla} = 1 \text{ N}/(\text{A}\cdot\text{m}) = 10^4 \text{ Gauss}$).

2. Fundamental Properties of the Magnetic Force

  • Zero Work Property: Because $\vec{F}_B$ is everywhere perpendicular to velocity $\vec{v}$ ($\vec{F}_B \cdot \vec{v} = q (\vec{v} \times \vec{B}) \cdot \vec{v} \equiv 0$), the instantaneous power delivered by a magnetic field is identically zero: $$P = \vec{F}_B \cdot \vec{v} = 0 \implies dK = 0$$ A static magnetic field can NEVER change the kinetic energy or speed of a charged particle; it can only alter its direction of motion.

3. Cyclotron Motion and Helical Trajectories

Consider a particle of mass $m$ and charge $q$ injected into a uniform magnetic field $\vec{B} = B \hat{k}$ with initial velocity $\vec{v}_\perp = v_x \hat{i} + v_y \hat{j}$. The magnetic force acts as a pure centripetal force: $$q v_\perp B = \frac{m v_\perp^2}{r} \implies r_c = \frac{m v_\perp}{q B}$$ $r_c$ is the cyclotron (Larmor) radius. The period of circular revolution and the cyclotron angular frequency $\omega_c$ are: $$T = \frac{2\pi r_c}{v_\perp} = \frac{2\pi m}{q B}, \quad \omega_c = \frac{q B}{m}$$ Notice that $\omega_c$ and $T$ are **completely independent of the particle's speed or orbital radius** (isochronism of the cyclotron). If the particle possesses a parallel velocity component $v_\parallel = v_z \hat{k}$, it executes a **helical path** with pitch $p = v_\parallel T = \frac{2\pi m v_\parallel}{q B}$.

§5.2 Magnetic Force on a Current-Carrying Conductor and Torque on a Current Loop

Because electric current consists of an ensemble of moving charges, magnetic forces manifest macroscopically on current-carrying wires.

1. Magnetic Force on a Wire Element

Consider a differential segment of wire of cross-section $A$ and length $d\vec{l}$ carrying current $I = n q A v_d$. The number of charge carriers in the segment is $dN = n A dl$. The total magnetic force on the element is: $$d\vec{F} = dN \cdot q (\vec{v}_d \times \vec{B}) = (n A dl) q (\vec{v}_d \times \vec{B}) = (n q A v_d) (d\vec{l} \times \vec{B})$$ $$d\vec{F} = I (d\vec{l} \times \vec{B})$$ For a straight wire of finite length $\vec{L}$ in a uniform magnetic field: $$\vec{F} = I (\vec{L} \times \vec{B})$$

2. Torque on a Planar Current Loop and Magnetic Dipole Moment

Consider a closed rectangular loop of dimensions $a \times b$ (area $A = ab$) carrying current $I$ in a uniform magnetic field $\vec{B}$. The net translational force vanishes ($\vec{F}_{\text{net}} = 0$). However, forces on opposite arms form a couple, generating net torque: $$\vec{\tau} = \vec{\mu} \times \vec{B}$$ where $\vec{\mu}$ is the magnetic dipole moment vector: $$\vec{\mu} = N I \vec{A} = N I A \hat{n}$$ for a coil of $N$ turns, where $\hat{n}$ is the unit normal given by the right-hand grip rule. SI Unit: $\text{A}\cdot\text{m}^2 = \text{J/T}$. The potential energy of the magnetic dipole in field $\vec{B}$ is: $$U = -\vec{\mu} \cdot \vec{B} = -\mu B \cos\theta$$

3. The Moving-Coil Galvanometer

In a d'Arsonval galvanometer, a rectangular coil of $N$ turns is suspended in a cylindrical soft iron core that produces a radial magnetic field ($\vec{B} \parallel$ plane of coil always, $\sin\theta = 1$). Deflecting magnetic torque: $\tau_d = N I A B$. Restoring torsional torque of phosphor-bronze suspension fiber: $\tau_r = C \theta$. In equilibrium ($\tau_d = \tau_r$): $$\theta = \left( \frac{N A B}{C} \right) I$$ The angular deflection is strictly linear with current.
  • Current Sensitivity ($S_I$): $S_I = \frac{\theta}{I} = \frac{N A B}{C}$ (rad/A or div/$\mu$A).
  • Voltage Sensitivity ($S_V$): $S_V = \frac{\theta}{V} = \frac{N A B}{C R_g}$ (rad/V).

§5.3 The Hall Effect and Galvanomagnetic Measurement

Edwin Herbert Hall (1879) discovered that when a magnetic field is applied perpendicular to a current-carrying conducting strip, a transverse potential difference develops across the strip.

1. Physical Mechanism of the Hall Effect

Consider a flat conducting slab of width $w$ and thickness $t$ carrying current $I$ along $+x$. A uniform magnetic field $\vec{B} = B \hat{k}$ is applied along $+z$.
  1. Charge carriers moving with drift velocity $\vec{v}_d$ experience transverse Lorentz magnetic force: $$\vec{F}_B = q (\vec{v}_d \times \vec{B})$$
  2. If charge carriers are **negative electrons** ($q = -e, \vec{v}_d = -v_d \hat{i}$): $$\vec{F}_B = (-e) [(-v_d \hat{i}) \times (B \hat{k})] = -e v_d B \hat{j}$$ Electrons are deflected toward the bottom edge, charging it negative and leaving the top edge positive.
  3. If charge carriers are **positive holes** ($q = +e, \vec{v}_d = +v_d \hat{i}$): $$\vec{F}_B = (+e) [(+v_d \hat{i}) \times (B \hat{k})] = -e v_d B \hat{j}$$ Positive charges also deflect downward, charging the bottom edge positive!
Sign of Carriers: The polarity of the transverse **Hall Voltage ($V_H$)** immediately reveals the sign of the charge carriers (confirming that metals conduct via negative electrons, while p-type semiconductors conduct via positive holes).

2. Derivation of the Hall Voltage and Hall Coefficient

Accumulating transverse charge creates a transverse Hall electric field $\vec{E}_H$ pointing toward the negative edge. At steady state, the electrostatic force balances the magnetic force: $$q E_H = q v_d B \implies E_H = v_d B$$ The measured transverse potential difference is: $$V_H = E_H w = v_d B w$$ Using $I = n q A v_d = n q (w t) v_d \implies v_d = \frac{I}{n q w t}$: $$V_H = \left( \frac{I}{n q w t} \right) B w = \frac{I B}{n q t}$$ We define the Hall Coefficient ($R_H$): $$R_H = \frac{E_H}{J B} = \frac{1}{n q}$$ $$V_H = R_H \frac{I B}{t}$$ Applications: Hall effect sensors measure magnetic fields non-invasively (Gaussmeters), sense motor rotor position in brushless DC motors, and quantify carrier density $n$ in semiconductor wafer fabrication.

§5.4 The Biot-Savart Law and Magnetic Fields of Current Geometries

Jean-Baptiste Biot and Félix Savart (1820) established the differential law governing the magnetic field generated by an infinitesimal current element.

1. The Biot-Savart Law

The magnetic induction $d\vec{B}$ at field point $P$ due to a differential current element $I d\vec{l}$ at source position $\vec{r}'$ is: $$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times \hat{r}}{r^2} = \frac{\mu_0}{4\pi} \frac{I d\vec{l} \times (\vec{r} - \vec{r}')}{|\vec{r} - \vec{r}'|^3}$$ where $\mu_0$ is the permeability of free space: $$\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A (exact by historical definition)} \approx 1.2566 \times 10^{-6} \text{ H/m}$$ For any closed circuit loop $C$: $$\vec{B}(\vec{r}) = \frac{\mu_0 I}{4\pi} \oint_C \frac{d\vec{l}' \times (\vec{r} - \vec{r}')}{|\vec{r} - \vec{r}'|^3}$$

2. Magnetic Field of a Long Straight Conductor

Integrating along an infinite straight wire carrying current $I$: At perpendicular distance $R$: $$B = \frac{\mu_0 I}{4\pi} \int_{-\infty}^\infty \frac{dx \sin\theta}{r^2} = \frac{\mu_0 I}{2\pi R}$$ Field lines form concentric circles centered on the wire.

3. Magnetic Field on the Axis of a Circular Current Loop

Consider a circular wire loop of radius $R$ carrying current $I$ lying in the yz-plane. At axial distance $x$ along the symmetry axis: By symmetry, components perpendicular to the axis cancel. The axial component is: $$B_x = \int dB \sin\alpha = \frac{\mu_0 I}{4\pi (x^2 + R^2)} (2\pi R) \left( \frac{R}{\sqrt{x^2 + R^2}} \right)$$ $$B(x) = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}$$
  • At the center of the loop ($x = 0$): $$B(0) = \frac{\mu_0 I}{2R}$$
  • Far from the loop ($x \gg R$): using dipole moment $\mu = I A = I (\pi R^2)$: $$B(x) \approx \frac{\mu_0 \mu}{2\pi x^3}$$ Decays as $1/x^3$, identical to an electric dipole.

4. Helmholtz Coils

A pair of identical coaxial coils of radius $R$, separated by distance equal to their radius ($d = R$), carrying identical current $I$ in the same direction. At the midpoint $x = R/2$: $$\frac{dB}{dx} = 0, \quad \frac{d^2 B}{dx^2} = 0$$ The second derivative vanishes, producing an exceptionally uniform magnetic field over a wide central volume.

§5.5 Ampere's Circuital Law, Solenoids, and Toroids

André-Marie Ampère (1826) formulated Ampere's circuital law, the magnetic analog of Gauss's law for high-symmetry current distributions.

1. Ampere's Circuital Law

The line integral of magnetic field $\vec{B}$ around any closed Amperian loop $C$ equals $\mu_0$ times the total net electric current enclosed by the loop: $$\oint_C \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$$ In differential form (applying Stokes' Theorem): $$\nabla \times \vec{B} = \mu_0 \vec{J}$$

2. The Ideal Long Solenoid

A helical coil of length $L$ and $N$ closely spaced turns carrying current $I$ ($n = N/L$ turns per unit meter). Inside an infinitely long solenoid, the magnetic field is uniform and parallel to the axis; outside, it is zero. Construct a rectangular Amperian loop of length $h$ with one side inside and one outside: $$\oint_C \vec{B} \cdot d\vec{l} = B h + 0 + 0 + 0 = B h$$ Enclosed current: $I_{\text{encl}} = n h I$. $$B h = \mu_0 (n h I) \implies B = \mu_0 n I$$ The field depends exclusively on turn density $n$ and current $I$, independent of solenoid cross-sectional diameter or position.

3. The Toroid (Toroidal Solenoid)

A solenoid bent into a closed donut-shaped ring of inner radius $a$ and outer radius $b$ with $N$ total turns. Construct a circular Amperian loop of radius $r$ inside the core ($a < r < b$): $$\oint \vec{B} \cdot d\vec{l} = B (2\pi r) = \mu_0 (N I) \implies B(r) = \frac{\mu_0 N I}{2\pi r}$$ Outside the toroid ($r < a$ or $r > b$), enclosed current is zero, so $\vec{B} = 0$ everywhere. Toroids have zero external magnetic leakage.
Standard University Exam Examination Problems

Rigorous Analytical & Numerical Solved Problems

Comprehensive step-by-step mathematical proofs, dimensional evaluations, and calculations matching B.Sc. Honors university examinations.

Honors Cyclotron Dynamics Exam Standard Example 5.1: Cyclotron Resonant Frequency and Relativistic Energy

A medical cyclotron accelerates deuterons (mass $m = 3.344 \times 10^{-27}\text{ kg}$, charge $q = +1.602 \times 10^{-19}\text{ C}$) in a uniform magnetic field $B = 1.50\text{ Tesla}$. The outer radius of the dees is $R_{\max} = 0.500\text{ m}$.\n(a) Calculate the cyclotron resonant frequency $f_c$ in Megahertz (MHz),\n(b) Determine the maximum exit speed $v_{\max}$ of the deuterons, and\n(c) Find the maximum kinetic energy $K_{\max}$ in both Joules and Mega-electron-volts (MeV).

Step 1: Compute cyclotron resonance frequency
$$\omega_c = \frac{q B}{m} = \frac{(1.6022 \times 10^{-19} \text{ C}) \times 1.50 \text{ T}}{3.344 \times 10^{-27} \text{ kg}} = \frac{2.4033 \times 10^{-19}}{3.344 \times 10^{-27}} = 7.1869 \times 10^7 \text{ rad/s}$$ $$f_c = \frac{\omega_c}{2\pi} = \frac{7.1869 \times 10^7}{2\pi} = 1.1438 \times 10^7 \text{ Hz} = 11.44 \text{ MHz}$$

The RF oscillator driving the dees must oscillate at 11.44 MHz.

Step 2: Maximum speed at outer radius
$$v_{\max} = \omega_c R_{\max} = (7.1869 \times 10^7 \text{ rad/s}) \times 0.500 \text{ m} = 3.5935 \times 10^7 \text{ m/s}$$ $$\frac{v_{\max}}{c} = \frac{3.5935 \times 10^7}{3.00 \times 10^8} = 0.120 = 12.0\% \text{ of light speed}$$

The deuterons emerge at 12% the speed of light.

Step 3: Maximum kinetic energy
$$K_{\max} = \frac{1}{2} m v_{\max}^2 = \frac{1}{2} \times (3.344 \times 10^{-27} \text{ kg}) \times (3.5935 \times 10^7 \text{ m/s})^2$$ $$K_{\max} = 0.5 \times 3.344 \times 10^{-27} \times 1.2913 \times 10^{15} = 2.159 \times 10^{-12} \text{ Joules}$$ $$K_{\max} = \frac{2.159 \times 10^{-12} \text{ J}}{1.6022 \times 10^{-13} \text{ J/MeV}} = 13.48 \text{ MeV}$$

The cyclotron delivers a beam of 13.5 MeV deuterons.

Experimental Solid-State Physics Standard Example 5.2: Hall Effect Carrier Density and Hall Coefficient

A thin rectangular strip of n-type germanium has width $w = 6.00\text{ mm}$ and thickness $t = 0.500\text{ mm}$. A longitudinal current $I = 25.0\text{ mA}$ is driven through the strip in a perpendicular magnetic field $B = 0.800\text{ Tesla}$. A digital millivoltmeter across the width records a transverse Hall voltage $V_H = -18.5\text{ mV}$.\n(a) Determine the Hall coefficient $R_H$ of the germanium sample,\n(b) Calculate the conduction electron number density $n$, and\n(c) Find the electron drift speed $v_d$ under these operating conditions.

Step 1: Calculate Hall coefficient RH
$$V_H = R_H \frac{I B}{t} \implies R_H = \frac{V_H t}{I B}$$ $$R_H = \frac{(-18.5 \times 10^{-3} \text{ V}) \times (0.500 \times 10^{-3} \text{ m})}{(25.0 \times 10^{-3} \text{ A}) \times (0.800 \text{ T})} = \frac{-9.25 \times 10^{-6}}{0.0200} = -4.625 \times 10^{-4} \text{ m}^3/\text{C}$$

The negative sign of $R_H$ confirms that majority charge carriers are electrons.

Step 2: Calculate electron carrier density n
$$R_H = -\frac{1}{n e} \implies n = \frac{1}{|R_H| e}$$ $$n = \frac{1}{(4.625 \times 10^{-4}) \times (1.6022 \times 10^{-19})} = \frac{1}{7.410 \times 10^{-23}} = 1.3495 \times 10^{22} \text{ electrons/m}^3$$

Carrier density in this doped semiconductor is $1.35 \times 10^{22}$ m⁻³.

Step 3: Electron drift speed
$$v_d = \frac{E_H}{B} = \frac{|V_H| / w}{B} = \frac{18.5 \times 10^{-3} \text{ V} / (6.00 \times 10^{-3} \text{ m})}{0.800 \text{ T}} = \frac{3.0833 \text{ V/m}}{0.800 \text{ T}} = 3.854 \text{ m/s}$$

In semiconductors, because carrier density is low, drift velocity is much faster (3.85 m/s) than in copper.

Undergraduate Classical Exam Problem Example 5.3: Biot-Savart Law for a Circular Coil and Axial Magnetic Field

A circular flat coil of radius $R = 10.0\text{ cm}$ contains $N = 250$ closely wound turns and carries current $I = 2.40\text{ A}$.\n(a) Calculate the magnetic field at the center of the coil $B(0)$,\n(b) Find the axial distance $x$ where the magnetic field drops to $1/8$ of its value at the center, and\n(c) Calculate the magnetic dipole moment $\mu$ of the coil.

Step 1: Compute magnetic field at coil center
$$B(0) = \frac{\mu_0 N I}{2R} = \frac{(4\pi \times 10^{-7}) \times 250 \times 2.40}{2 \times 0.100}$$ $$B(0) = \frac{(1.2566 \times 10^{-6}) \times 600}{0.200} = \frac{7.5398 \times 10^{-4}}{0.200} = 3.770 \times 10^{-3} \text{ Tesla} = 3.77 \text{ mT}$$

At the center of the 250-turn coil, field magnitude is 3.77 mT.

Step 2: Find axial distance for 1/8 field strength
$$B(x) = \frac{\mu_0 N I R^2}{2 (R^2 + x^2)^{3/2}} = \frac{B(0) R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{8} B(0)$$ $$\frac{R^3}{(R^2 + x^2)^{3/2}} = \frac{1}{8} \implies \frac{(R^2 + x^2)^{3/2}}{R^3} = 8$$ $$\left( \frac{R^2 + x^2}{R^2} \right)^{3/2} = 8 = 2^3 \implies \frac{R^2 + x^2}{R^2} = (2^3)^{2/3} = 2^2 = 4$$ $$1 + \frac{x^2}{R^2} = 4 \implies \frac{x^2}{R^2} = 3 \implies x = R \sqrt{3}$$ $$x = 10.0 \text{ cm} \times \sqrt{3} = 17.32 \text{ cm}$$

The field drops to one-eighth of its peak value at $x = R\sqrt{3} = 17.3$ cm.

Step 3: Magnetic dipole moment
$$\mu = N I A = N I (\pi R^2) = 250 \times 2.40 \times \pi (0.100)^2$$ $$\mu = 600 \times \pi \times 0.0100 = 6.00 \pi = 18.85 \text{ A}\cdot\text{m}^2 = 18.85 \text{ J/T}$$

The magnetic dipole moment of the coil is 18.85 A·m².