Solenoid Magnetic Field Simulation
Interactive Solenoid
Simulation of the magnetic field inside a solenoid with length = 0.5 m, radius = 1 cm, 500 turns, and current = 5 A.
Example
Question:
A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5 A. What is the magnitude of the magnetic field inside the solenoid?
Solution:
Number of turns per unit length:
\[
n = \frac{500}{0.5} = 1000\,\text{turns/m}
\]
Length \(l = 0.5\,\text{m}\), radius \(r = 0.01\,\text{m}\), so \(l/a = 50\), i.e., long solenoid.
Magnetic field inside (using long solenoid formula):
\[
B = \mu_0 n I = 4\pi \times 10^{-7} \times 10^3 \times 5 = 6.28 \times 10^{-3}\,\text{T}
\]
Example Values
- Length: 0.5 m
- Radius: 1 cm
- Number of turns: 500
- Current: 5 A
- Turns per unit length (n): 1000 turns/m
- Calculated magnetic field (B): 6.28 × 10⁻³ T
Parameters
Length (l): 0.5 m
Radius (r): 0.01 m
Turns (N): 500
Current (I): 5 A
Calculations
Turns per unit length (n = N/l): 1000 turns/m
l/r ratio: 50
Approximation: Long solenoid (l ≫ r)
Magnetic Field
Formula: B = μ₀nI
Calculation: 4π×10⁻⁷ × 1000 × 5
Result: 6.28 × 10⁻³ T



