Configuration:
Magnetic Field (B)
0.00 T
In Gauss
0.00 G
In mT
0.00 mT
In μT
0.00 μT

The Physics of Magnetic Fields

A magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. It is the fundamental force behind motors, generators, and transformers. The SI unit for magnetic field strength (magnetic flux density) is the Tesla (T).

Key Formulas for Common Conductors

1. Long Straight Wire

The magnetic field lines form concentric circles around a straight wire. The field strength decreases as you move further from the wire:

B = (μ₀ × I) / (2πr)

Where r is the distance from the wire and I is the current.

2. Circular Coil (at the Center)

For a flat circular coil with N turns and radius R, the field is strongest and most uniform at the exact center:

B = (μ₀ × N × I) / (2R)

3. Ideal Solenoid (Inside)

A solenoid is a long coil of wire. Inside a long solenoid, the magnetic field is very strong and uniform (parallel lines):

B = μ₀ × n × I = μ₀ × (N/L) × I

Where n is the number of turns per unit length (N/L).

Physical Constants:
• μ₀ (Permeability of Free Space): 4π × 10⁻⁷ T·m/A (approx. 1.257 × 10⁻⁶ T·m/A).

Laminated Cores and Permeability

In practical applications, magnetic fields are often amplified by placing a ferromagnetic material (like iron) inside the coil. This changes the permeability from μ₀ to μ = μ₀ × μᵣ, where μᵣ is the relative permeability (often 1000 or more for soft iron).

Deep Dive: Worked Examples

✅ Example 1: Solenoid Calculation

A student wounds a solenoid with 500 turns over a 20 cm length. If a current of 2.0 A flows through it, calculate the B-field inside.

Given: N = 500, L = 0.20 m, I = 2.0 A
Turns per meter (n): n = 500 / 0.20 = 2500 turns/m
Formula: B = μ₀nI
Solution:
B = (4π × 10⁻⁷) × 2500 × 2.0
B = 1.257 × 10⁻⁶ × 5000 = 0.00628 T = 6.28 mT

✅ Example 2: Field Near a Power Line

Calculate the magnetic field strength at a distance of 5.0 meters from a high-voltage DC power line carrying 1000 A.

Given: I = 1000 A, r = 5.0 m
Formula: B = (μ₀I) / (2πr)
Solution:
B = (4π × 10⁻⁷ × 1000) / (2π × 5.0)
B = (2 × 10⁻⁷ × 1000) / 5.0 = 2 × 10⁻⁴ / 5.0
B = 4 × 10⁻⁵ T = 40 μT (similar to Earth's magnetic field)

✅ Example 3: Circular Coil for a Helmholtz Setup

A circular coil with a radius of 10 cm has 100 turns. What current is required to produce a peak field of 1.0 mT at the center?

Given: B = 1.0 × 10⁻³ T, R = 0.10 m, N = 100
Formula: I = (2RB) / (μ₀N)
Solution:
I = (2 × 0.10 × 1.0 × 10⁻³) / (4π × 10⁻⁷ × 100)
I = 2 × 10⁻⁴ / 1.257 × 10⁻⁴ = 1.59 A

✅ Example 4: Comparing Solenoids

Solenoid A is 10 cm long with 100 turns. Solenoid B is 20 cm long with 400 turns. Which produces a stronger field for the same current?

nA: 100 / 0.1 = 1000 turns/m
nB: 400 / 0.2 = 2000 turns/m
Conclusion: Solenoid B has twice the turn density, so it will produce double the magnetic field strength of Solenoid A.

✅ Example 5: Force on a Charge in the Field

A proton (q = 1.6 × 10⁻¹⁹ C) moves at 3.0 × 10⁶ m/s perpendicular to a 0.5 T magnetic field. Calculate the magnetic force on it.

Given: q = 1.6 × 10⁻¹⁹ C, v = 3.0 × 10⁶ m/s, B = 0.5 T
Formula: F = qvB sin θ (here sin 90° = 1)
Solution:
F = (1.6 × 10⁻¹⁹) × (3.0 × 10⁶) × 0.5
F = 4.8 × 10⁻¹³ × 0.5 = 2.4 × 10⁻¹³ N