Magnetostatics

Solenoid Field

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Parameters

Current (I)20 A
Turns (N)10
Radius (R)1.5 m
Length (L)6.0 m
Probe Position
Probe B-Field (μT)
Bx
0.00
By
0.48
Bz
37.59
Derived Quantities
B ideal (μ₀nI)
41.89 μT
|B| numerical
37.59 μT
Self-Inductance L
148.044 μH
Energy U_B
0.0296 J
Flux Φ_B
296.09 μWb
Turns density n
1.67 /m
B=37.59μT|\vec{B}| = 37.59\,\mu\text{T}
3D B-Field Vector
xyz
θ = 0.7°
φ = 90.0°
|B| = 37.59 μT
Live Telemetry

Awaiting Telemetry

Governing Dynamics

A solenoid produces a nearly uniform magnetic field inside its core. The ideal infinite solenoid result:

Binside=μ0nIB_{\text{inside}} = \mu_0 \, n \, In=NL(turns per unit length)n = \frac{N}{L} \quad \text{(turns per unit length)}

The magnetic flux through the solenoid cross-section and its self-inductance:

ΦB=BA=μ0nIπR2\Phi_B = B \cdot A = \mu_0 n I \cdot \pi R^2L=μ0n2A=μ0N2LπR2L = \mu_0 n^2 \ell A = \mu_0 \frac{N^2}{L} \pi R^2

Energy stored in the magnetic field:

UB=12LI2=B22μ0VU_B = \frac{1}{2} L I^2 = \frac{B^2}{2\mu_0} \cdot V

Outside an ideal infinite solenoid B=0B = 0. For a finite solenoid, fringe fields are computed numerically via Biot-Savart over the helical path.