Physics Notes on Moving Charges and Magnetism

Page 1: Introduction to Moving Charges and Magnetism

Electricity and Magnetism were recognized as distinct phenomena for over 2000 years. It was not until 1820 that Hans Christian Oersted (1777–1851), a Danish physicist, discovered their intimate relationship during a lecture. He observed that an electric current in a straight wire caused a compass needle to deflect.

Oersted's Observations:

• The magnetic needle aligns tangentially to an imaginary circle centered on the wire, with the wire sitting perpendicular to the circle's plane.

• The deflection is most evident when the current is high and the needle is close to the wire, allowing the observer to neglect the Earth's magnetic field.

• Reversing the current direction reverses the needle's orientation.

• The magnitude of deflection increases as the current increases or the distance between the wire and needle decreases.

• Iron filings spread around a current-carrying wire form concentric circular patterns centered on the wire.

Moving charges or currents generate magnetic fields in the space around them. In 1864, James Maxwell unified the laws of electricity and magnetism, identifying light as an electromagnetic wave. By the late 19th century, Hertz discovered radio waves, which were subsequently produced by J.C. Bose and G. Marconi.

Page 2: Electromagnetic Progress and Conventions

Significant advances in the 20th century were driven by understandings of electromagnetism and the invention of technologies to produce, amplify, transmit, and detect electromagnetic waves. Magnetism studies focus on how fields exert forces on moving charged particles (electrons, protons) and current-carrying wires.

Notation and Conventions:

• Dot ($⊙$): Represents a current or field (electric/magnetic) emerging out of the plane of the paper (like the tip of an arrow).

• Cross ($⊗$): Represents a current or field entering the plane of the paper (like the feathered tail of an arrow).

Magnetic Field Sources: Interactions between charges occur in two stages. First, a source charge $Q$ produce an electric field $E$:

$E = \frac{Q\hat{r}}{4\pi\epsilon_0 r^2}$

Second, a charge $q$ interacts with this field to experience a force $F = qE$. Unlike a mere mathematical artifact, the field carries energy and momentum, propagating over a finite time.

Page 3: The Lorentz Force and Field Properties

Fields can vary across space and time. Static charges produce electric fields, while moving charges or currents also produce magnetic fields, denoted as $B(r)$. Magnetic fields obey the principle of superposition, meaning the total field from multiple sources is the vector sum of their individual fields.

The Lorentz Force: A point charge $q$ moving with velocity $v$ at position $r$ in the presence of both electric field $E(r)$ and magnetic field $B(r)$ experiences a total force:

$F = q[E(r) + v \times B(r)] = F_{electric} + F_{magnetic}$

Features of Magnetic Force ($q[v \times B]$):

1. It depends on the charge $q$, velocity $v$, and magnetic field $B$. Force on a negative charge is opposite to that on a positive charge.

2. It vanishes if the velocity and magnetic field are parallel or anti-parallel ($\sin(0^\circ) = 0$ or $\sin(180^\circ) = 0$).

3. The force acts in a direction perpendicular to both velocity and the magnetic field, determined by the right-hand screw rule.

4. Static charges ($|v|=0$) experience no magnetic force.

Page 4: Units and Force on Conductors

Units of Magnetic Field: The magnitude of $B$ is defined in SI units as the Tesla (T), named after Nikola Tesla.

• $B = \frac{F}{qv \sin(\theta)}$

• One Tesla is the field where a unit charge ($1\,C$) moving at $1\,m/s$ perpendicular to the field experiences a force of $1\,Newton$.

• Gauss (G): A non-SI unit where $1\,G = 10^{-4}\,T$.

• Earth’s magnetic field is approximately $3.6 \times 10^{-5}\,T$.

Force on a Straight Current-Carrying Rod: For a rod with area $A$, length $l$, and mobile carrier density $n$, the total force is:

$F = (nlA)q v_d \times B$

Since current $I = nqv_d A$, the formula for a straight rod is:

$F = Il \times B$

where $l$ is a vector with magnitude equal to the rod length and direction matching the current flow. For wires of arbitrary shapes, the force is calculated by integrating the Lorentz force over infinitesimal strips $dl$.

Page 5: Applied Examples of Magnetic Force

Example 4.1: Wire Suspension A wire with mass $200\,g$ ($0.2\,kg$) and length $1.5\,m$ carries a current of $2\,A$. To suspend it in mid-air, the magnetic force $IlB$ must balance gravity $mg$:

$B = \frac{mg}{Il} = \frac{0.2 \times 9.8}{2 \times 1.5} \approx 0.65\,T$

Example 4.2: Direction of Lorentz Force A field exists along the positive y-axis ($B = B\hat{j}$) and a particle moves along the positive x-axis ($v = v\hat{i}$). The vector product $\hat{i} \times \hat{j} = \hat{k}$.

• (a) For an electron (negative), the force is along the $-z$ axis.

• (b) For a proton (positive), the force is along the $+z$ axis.

Page 6: Motion in a Uniform Magnetic Field

Magnetic force is always perpendicular to velocity ($F \perp v$); therefore, the work done is zero ($W = \int F \cdot dl = 0$). The speed and kinetic energy remain constant, but the direction of motion changes.

Circular Motion: When velocity is perpendicular to the uniform magnetic field, the force acts as a centripetal force:

$\frac{mv^2}{r} = qvB \implies r = \frac{mv}{qB}$

Cyclotron Frequency: The angular frequency $\omega$ and frequency $\nu$ are independent of the particle's velocity or energy:

$\omega = 2\pi\nu = \frac{qB}{m}$

Helical Motion: If the velocity has a component parallel to the field ($v_{∥}$), the particle moves along the field while circling, creating a helix.

Page 7: Path Characteristics and Biot-Savart Law

Pitch ($p$): The distance the particle moves along the magnetic field direction during one full rotation:

$p = v_{∥} T = \frac{2\pi m v_{∥}}{qB}$

Example 4.3: Electron Path Properties An electron ($m=9 \times 10^{-31}\,kg, q=1.6 \times 10^{-19}\,C$) moving at $3 \times 10^7\,m/s$ in a field of $6 \times 10^{-4}\,T$:

• Radius $r = \frac{mv}{qB} = 0.28\,m = 28\,cm$

• Frequency $\nu = \frac{v}{2\pi r} = 17\,MHz$

• Energy $E = \frac{1}{2}mv^2 ≈ 2.5\,keV$

Biot-Savart Law: Relates current to its magnetic field ($dB$) at distance $r$ for an element $dl$:

$dB = \frac{\mu_0}{4\pi} \frac{I dl \times r}{r^3}$

Magnitude: $dB = \frac{\mu_0}{4\pi} \frac{I dl \sin(\theta)}{r^2}$

Page 8: Constants and Comparisons

Permeability of Free Space ($\mu_0$): $\frac{\mu_0}{4\pi} = 10^{-7}\,Tm/A$

Comparison: Biot-Savart vs. Coulomb's Law:

1. Both are long-range and inverse-square laws; both obey superposition.

2. Electrostatic sources are scalar (charge), while magnetic sources are vectors ($I dl$).

3. Electrostatic fields are along the displacement vector; magnetic fields are perpendicular to the plane of the displacement vector and the source element.

4. Biot-Savart law has an angle dependence ($\sin(\theta)$) absent in Coulomb’s law.

Light Speed Relation: $c = \frac{1}{\sqrt{\mu_0\epsilon_0}} ≈ 3 \times 10^8\,m/s$

Page 9: Applications of Biot-Savart Law

Example 4.4: Field from a Current Element A $1\,cm$ element $dl = \Delta x \hat{i}$ at the origin with $I = 10\,A$. At $y = 0.5\,m$:

$dB = \frac{\mu_0}{4\pi} \frac{I dl \sin(90^\circ)}{r^2} = 10^{-7} \frac{10 \times 10^{-2}}{0.5^2} = 4 \times 10^{-8}\,T$

Direction is along $+\hat{k}$ ($z$-axis).

Field on the Axis of a Circular Loop: For a loop in the y-z plane with radius $R$, the field at an axial point $P$ at distance $x$ is:

$B = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}} \hat{i}$

Page 10: Loop Field and Right-Hand Rule

At the center of a circular loop ($x=0$):

$B = \frac{\mu_0 I}{2R}$

Right-Hand Thumb Rule (Loops): Curl fingers in the direction of the current; the thumb indicates the direction of the magnetic field along the axis.

Page 11: Arc Fields and Ampere's Law

Example 4.5: Semicircular Arc

• Straight segments pointing toward the center contribute zero field because $dl \parallel r$.

• A semicircular arc provides half the field of a full loop: $B_{arc} = \frac{\mu_0 I}{4R}$.

Example 4.6: Tightly Wound Coil A coil with $100$ turns, $R=10\,cm$, $I=1\,A$:

$B = 100 \times \frac{\mu_0 I}{2R} = \frac{100 \times 4\pi \times 10^{-7} \times 1}{2 \times 0.1} ≈ 6.28 \times 10^{-4}\,T$

Ampere’s Circuital Law: $\oint B \cdot dl = \mu_0 I$

Page 12: Applications of Ampere's Law

Ampere’s Law is the magnetic equivalent of Gauss’s Law. For a straight infinite current-carrying wire, we choose a circular Amperian loop of radius $r$:

$B(2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$

Andre Marie Ampere (1775–1836): A French prodigy who founded electrodynamics, hypothesized that all magnetism stems from circulating currents, and explored the relationship through rigorous experimentation.

Page 13: Symmetry and Long Wires

Infinite Wire Characteristics:

• Field magnitude depends only on distance $r$ (cylindrical symmetry).

• Field lines are concentric circles.

• Field is directly proportional to current and inversely to distance.

Right-Hand Rule (Straight Wires): Grasp the wire with the thumb pointing in current direction; fingers curl in the field direction.

Page 14: Field Inside and Outside a Thick Wire

Example 4.7: Thick Wire of Radius $a$

• Outside ($r > a$): Current enclosed is $I$. $B = \frac{\mu_0 I}{2\pi r}$. ($B \propto 1/r$).

• Inside ($r < a$): Assuming uniform current density, current enclosed $I_e = I(r^2/a^2)$.

$B(2\pi r) = \mu_0 I \frac{r^2}{a^2} \implies B = \frac{\mu_0 I r}{2\pi a^2}$. ($B \propto r$).

Page 15: The Solenoid

A solenoid is a wire wound into a long helix. When turns are closely spaced, the field inside is uniform and axial, while the field outside is nearly zero.

Page 16: Solenoid Calculations

Using a rectangular Amperian loop inside a long solenoid with $n$ turns per unit length:

$Bh = \mu_0 (nh I) \implies B = \mu_0 n I$

Example 4.8: Solenoid Magnitude Length $0.5\,m$, $500$ turns, $I = 5\,A$.

$n = 500 / 0.5 = 1000\,turns/m$

$B = (4\pi \times 10^{-7}) \times 1000 \times 5 ≈ 6.28 \times 10^{-3}\,T$

Page 17: Force Between Parallel Currents

Two parallel conductors $a$ and $b$ separated by distance $d$ carry currents $I_a$ and $I_b$. Conductor 'a' produces a field $B_a$ at 'b'. The force per unit length $f_{ba}$ is:

$f_{ba} = \frac{\mu_0 I_a I_b}{2\pi d}$

• Rule: Parallel currents attract; antiparallel currents repel.

• Newton’s third law holds for these magnetic forces in steady current conditions.

Page 18: The Definition of the Ampere

Definition established in 1946: The Ampere is the steady current which, when maintained in two infinite parallel conductors $1\,m$ apart in vacuum, produces a force of $2 \times 10^{-7}\,N/m$ on each conductor.

Coulomb: The charge that flows through a cross-section in $1\,s$ when a current of $1\,A$ is maintained.

Page 19: Torque on a Rectangular Current Loop

A current loop in a uniform field experiences torque but no net force.

Case 1: Field in the plane of the loop Force on arm AB is $F_1 = IbB$ (into plane). Force on arm CD is $F_2 = IbB$ (out of plane). Torque $\tau = IAB$, where $A$ is the area ($a \times b$).

Page 20: General Torque and Magnetic Moment

Case 2: Loop normal makes angle $\theta$ with field $\tau = IAB \sin(\theta)$

Magnetic Moment ($m$): $m = I A$ In vector notation: $\tau = m \times B$

• Dimensions: $[L^2 A]$. Units: $Am^2$.

• Stable equilibrium: $m \parallel B$ (torque vanishes).

• For $N$ turns: $m = NIA$.

Page 21: Torque Dynamics and Orientations

Example 4.10: Coil Rotation A $100$-turn coil, $R = 10\,cm, I = 3.2\,A, B = 2\,T$:

• Magnetic moment $m = 10\,Am^2$.

• Torque at $90^\circ$: $\tau = mB = 20\,Nm$.

Example 4.11: Orientation and Equilibrium

• A loop in a vertical field cannot rotate about a vertical axis because the torque always lies in the plane of the loop.

• Stable equilibrium occurs when the area vector coincides with the field (maximizing total flux).

• Flexible loops become circular to maximize enclosed area for a given perimeter.

Page 22: Loop as a Magnetic Dipole

At large axial distance ($x >> R$):

$B ≈ \frac{\mu_0 2m}{4\pi x^3}$

This behavior is identical to an electric dipole where:

• $B \to E$

• $m \to p_e$

• $\mu_0 \to 1/\epsilon_0$

Page 23: The Moving Coil Galvanometer (MCG)

MCG Principle: A coil rotates in a uniform radial magnetic field. A soft iron core concentrates the field.

Torque Balance: $Magnetic Torque = Restoring Torque$ $NIAB = k\phi$

Where $k$ is the torsional constant of the spring and $\phi$ is the angular deflection.

$\phi = \left(\frac{NAB}{k}\right) I$

Page 24: Galvanometer Sensitivity and Ammeter Conversion

Current Sensitivity: Deflection per unit current = $\frac{NAB}{k}$.

Conversion to Ammeter: To measure high currents and avoid circuit resistance changes, a small shunt resistance ($r_s$) is connected in parallel. Total resistance: $R_{eff} = \frac{R_G r_s}{R_G + r_s} ≈ r_s$.

Page 25: Voltmeter Conversion and Sensitivity Dynamics

Conversion to Voltmeter: A large resistance $R$ is connected in series with the galvanometer. Total resistance: $R_{eff} = R_G + R ≈ R$.

Voltage Sensitivity: Deflection per unit voltage = $\frac{NAB}{kR}$. Note: Doubling turns to increase current sensitivity might double resistance, leaving voltage sensitivity unchanged.

Page 26: Practical Examples and Summary

Example 4.12: Ammeter Accuracy Resistance R = 3\, ̦, V = 3\,V.

• (a) With MCG (R_G = 60\, ̦): $I = 3 / 63 = 0.048\,A$.

• (b) With Shunt (r_s = 0.02\, ̦): $I ≈ 3 / 3.02 ≈ 0.99\,A$.

• (c) Ideal Ammeter (0\, ̦): $I = 3 / 3 = 1.00\,A$.

Page 27: Comprehensive Summary and Formulas

• Lorentz Force: $F = q(v \times B + E)$.

• Centripetal Motion: Particles execute circular orbits with cyclotron frequency $\nu_c = \frac{qB}{2\pi m}$.

• Biot-Savart: $dB = \frac{\mu_0}{4\pi} \frac{I dl \times r}{r^3}$.

• Wire Field: $B = \frac{\mu_0 I}{2\pi R}$.

• Solenoid Field: $B = \mu_0 n I$.

• Torque: $\tau = NIA B \sin(\theta)$.

Page 28: Physical Quantities Table and Exercises

Points to Ponder:

• Magnetic field lines always form closed loops.

• Newton’s third law requires taking the momentum of electromagnetic fields into account for time-varying currents.

Page 29: Exercises and Problems

4.5: Wire $I = 8\,A$, $\theta = 30^\circ$, $B = 0.15\,T$. Force per length $f = IB \sin(\theta) = (8)(0.15)(0.5) = 0.6\,N/m$. 4.7: Two wires $A (8\,A)$ and $B (5\,A)$, $4\,cm$ apart. Force on $10\,cm$ section: $F = \frac{\mu_0 I_a I_b L}{2\pi d} = \frac{(2 \times 10^{-7})(8)(5)(0.1)}{0.04} = 2 \times 10^{-5}\,N$. 4.11: Electron $B = 6.5\, G$, $v = 4.8 \times 10^6\,m/s$. Path is circular because force is perpendicular to velocity. Radius $r = \frac{mv}{eB} = 4.2\,cm$.