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
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
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
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 $+z$.- Charge carriers moving with drift velocity $\vec{v}_d$ experience transverse Lorentz magnetic force: $$\vec{F}_B = q (\vec{v}_d \times \vec{B})$$
- 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.
- 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!
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
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
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.Rigorous Analytical & Numerical Solved Problems
Comprehensive step-by-step mathematical proofs, dimensional evaluations, and calculations matching B.Sc. Honors university examinations.
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).
The RF oscillator driving the dees must oscillate at 11.44 MHz.
The deuterons emerge at 12% the speed of light.
The cyclotron delivers a beam of 13.5 MeV deuterons.
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.
The negative sign of $R_H$ confirms that majority charge carriers are electrons.
Carrier density in this doped semiconductor is $1.35 \times 10^{22}$ m⁻³.
In semiconductors, because carrier density is low, drift velocity is much faster (3.85 m/s) than in copper.
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.
At the center of the 250-turn coil, field magnitude is 3.77 mT.
The field drops to one-eighth of its peak value at $x = R\sqrt{3} = 17.3$ cm.
The magnetic dipole moment of the coil is 18.85 A·m².