Electricity and Magnetism

Electricity and magnetism are interconnected phenomena, unified under the field of electromagnetism, which governs much of modern technology and natural processes. Electricity deals with charges and their interactions, while magnetism arises from moving charges and magnetic materials. These principles, formalized by scientists like Coulomb, Faraday, and Maxwell, explain phenomena from lightning strikes to the operation of electric motors. This guide explores key concepts, provides all basic formulas (Coulomb’s Law, electric field, electric potential, Ohm’s Law, capacitance, magnetic field, magnetic force, and electromagnetic induction), and includes detailed examples and applications.

Coulomb’s Law (with Electric Field, Potential, Ohm’s Law, Capacitance)

Coulomb’s Law describes the force between two point charges, a fundamental principle in electrostatics:

\[ F = k \frac{|q_1 q_2|}{r^2} \]

Where:

  • \( F \): Force (in Newtons, N)
  • \( k \): Coulomb’s constant (\( 8.99 \times 10^9 \, \text{N·m}^2/\text{C}^2 \))
  • \( q_1, q_2 \): Charges (in Coulombs, C)
  • \( r \): Distance between charges (in meters, m)

Related Formulas:

  • Electric Field: \( E = \frac{F}{q} = k \frac{|q|}{r^2} \)
  • Electric Potential: \( V = k \frac{q}{r} \)
  • Ohm’s Law: \( V = I \cdot R \) (where \( I \) is current in Amperes, \( R \) is resistance in Ohms)
  • Capacitance: \( C = \frac{Q}{V} \), Energy in Capacitor: \( U = \frac{1}{2} C V^2 \)

Example 1: Coulomb’s Law

Two charges, \( q_1 = 2 \, \mu\text{C} \), \( q_2 = 3 \, \mu\text{C} \), 1 m apart:

\[ F = k \frac{|q_1 q_2|}{r^2} \] \[ = (8.99 \times 10^9) \frac{(2 \times 10^{-6})(3 \times 10^{-6})}{1^2} \] \[ = (8.99 \times 10^9) \frac{6 \times 10^{-12}}{1} \] \[ = 8.99 \times 6 \times 10^{-3} \] \[ = 0.05394 \, \text{N} \]

Example 2: Electric Field

Electric field due to a \( 5 \, \mu\text{C} \) charge at 2 m:

\[ E = k \frac{|q|}{r^2} \] \[ = (8.99 \times 10^9) \frac{5 \times 10^{-6}}{2^2} \] \[ = (8.99 \times 10^9) \frac{5 \times 10^{-6}}{4} \] \[ = 8.99 \times 1.25 \times 10^3 \] \[ = 11237.5 \, \text{N/C} \]

Example 3: Electric Potential

Potential due to a \( 4 \, \mu\text{C} \) charge at 0.5 m:

\[ V = k \frac{q}{r} \] \[ = (8.99 \times 10^9) \frac{4 \times 10^{-6}}{0.5} \] \[ = (8.99 \times 10^9) \times 8 \times 10^{-6} \] \[ = 71920 \, \text{V} \]

Example 4: Ohm’s Law

A resistor with \( R = 10 \, \Omega \) has a current \( I = 2 \, \text{A} \). Find the voltage:

\[ V = I \cdot R \] \[ = 2 \cdot 10 \] \[ = 20 \, \text{V} \]

Example 5: Capacitance

A capacitor with \( C = 2 \, \mu\text{F} \) is charged to \( V = 100 \, \text{V} \). Find the charge and energy stored:

\[ Q = C \cdot V \] \[ = (2 \times 10^{-6}) \cdot 100 \] \[ = 2 \times 10^{-4} \, \text{C} \] \[ U = \frac{1}{2} C V^2 \] \[ = \frac{1}{2} (2 \times 10^{-6}) (100)^2 \] \[ = \frac{1}{2} (2 \times 10^{-6}) \times 10000 \] \[ = 0.01 \, \text{J} \]

Magnetic Fields (with Magnetic Force, Induction)

A current-carrying wire generates a magnetic field. For a long straight wire, the Biot-Savart Law simplifies to:

\[ B = \frac{\mu_0 I}{2\pi r} \]

Where:

  • \( B \): Magnetic field (in Tesla, T)
  • \( \mu_0 \): Permeability of free space (\( 4\pi \times 10^{-7} \, \text{T·m/A} \))
  • \( I \): Current (in Amperes, A)
  • \( r \): Distance from the wire (in meters, m)

Related Formulas:

  • Magnetic Force on a Moving Charge: \( F = q v B \sin\theta \)
  • Magnetic Force on a Current-Carrying Wire: \( F = I L B \sin\theta \)
  • Faraday’s Law of Induction: \( \mathcal{E} = - \frac{d\Phi_B}{dt} \), where \( \Phi_B = B A \cos\theta \)

Example 1: Magnetic Field from a Wire

Magnetic field at 0.1 m from a wire with \( I = 5 \, \text{A} \):

\[ B = \frac{\mu_0 I}{2\pi r} \] \[ = \frac{(4\pi \times 10^{-7})(5)}{2\pi (0.1)} \] \[ = \frac{4 \times 5 \times 10^{-7}}{2 \times 0.1} \] \[ = \frac{20 \times 10^{-7}}{0.2} \] \[ = 1 \times 10^{-5} \, \text{T} \]

Example 2: Magnetic Force on a Charge

Force on a charge \( q = 2 \, \mu\text{C} \) moving at \( v = 300 \, \text{m/s} \) in a \( B = 0.5 \, \text{T} \) field (\( \theta = 90^\circ \)):

\[ F = q v B \sin\theta \] \[ = (2 \times 10^{-6}) \cdot 300 \cdot 0.5 \cdot \sin(90^\circ) \] \[ = (2 \times 10^{-6}) \cdot 300 \cdot 0.5 \cdot 1 \] \[ = 3 \times 10^{-4} \, \text{N} \]

Example 3: Magnetic Force on a Wire

Force on a 0.2 m wire with \( I = 10 \, \text{A} \) in a \( B = 0.3 \, \text{T} \) field (\( \theta = 90^\circ \)):

\[ F = I L B \sin\theta \] \[ = 10 \cdot 0.2 \cdot 0.3 \cdot \sin(90^\circ) \] \[ = 10 \cdot 0.2 \cdot 0.3 \cdot 1 \] \[ = 0.6 \, \text{N} \]

Example 4: Electromagnetic Induction

A loop with area \( A = 0.05 \, \text{m}^2 \) in a field \( B \) changing from 0.2 T to 0.1 T in 0.01 s (\( \theta = 0^\circ \)):

\[ \Phi_B = B A \cos\theta \] \[ \Delta\Phi_B = (0.1 - 0.2) \cdot 0.05 \cdot \cos(0^\circ) \] \[ = (-0.1) \cdot 0.05 \cdot 1 \] \[ = -0.005 \, \text{Wb} \] \[ \mathcal{E} = - \frac{\Delta\Phi_B}{\Delta t} \] \[ = - \frac{-0.005}{0.01} \] \[ = 0.5 \, \text{V} \]

Applications

Electromagnetism is integral to technology and science. Here are detailed examples with calculations:

Example 1: Electric Motor

A motor with a 0.1 m wire carrying 15 A in a 0.4 T field (\( \theta = 90^\circ \)):

\[ F = I L B \sin\theta \] \[ = 15 \cdot 0.1 \cdot 0.4 \cdot 1 \] \[ = 0.6 \, \text{N} \]

Force produces torque for rotation.

Example 2: Transformer

A transformer with a secondary coil experiencing \( \mathcal{E} = 120 \, \text{V} \), \( R = 60 \, \Omega \). Find current:

\[ V = I \cdot R \] \[ 120 = I \cdot 60 \] \[ I = \frac{120}{60} \] \[ I = 2 \, \text{A} \]

Example 3: MRI Machine

Magnetic field at 0.05 m from a wire with \( I = 20 \, \text{A} \):

\[ B = \frac{\mu_0 I}{2\pi r} \] \[ = \frac{(4\pi \times 10^{-7})(20)}{2\pi (0.05)} \] \[ = \frac{4 \times 20 \times 10^{-7}}{0.1} \] \[ = 8 \times 10^{-5} \, \text{T} \]

Example 4: Wireless Charging

Induced EMF in a coil with \( \Delta\Phi_B = 0.02 \, \text{Wb} \) over 0.04 s:

\[ \mathcal{E} = - \frac{\Delta\Phi_B}{\Delta t} \] \[ = - \frac{0.02}{0.04} \] \[ = 0.5 \, \text{V} \]

Example 5: Capacitor in a Circuit

A \( 5 \, \mu\text{F} \) capacitor charged to 50 V:

\[ U = \frac{1}{2} C V^2 \] \[ = \frac{1}{2} (5 \times 10^{-6}) (50)^2 \] \[ = \frac{1}{2} (5 \times 10^{-6}) \times 2500 \] \[ = 0.00625 \, \text{J} \]

Example 6: Electric Field in a Defibrillator

Electric field between plates 0.02 m apart with \( V = 3000 \, \text{V} \), \( E = \frac{V}{d} \):

\[ E = \frac{V}{d} \] \[ = \frac{3000}{0.02} \] \[ = 150000 \, \text{V/m} \]