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Calculator of H along axis of round "thin" solenoid

Magnetic field strength H of a circular “thin” solenoid along its axis
$$ H(t,x) = \frac{I(t) · N}{2·L}· \left( \frac{L + 2·x}{\sqrt{ d^2 + (L+2·x)^2}} + \frac{L - 2·x}{\sqrt{ d^2 + (L-2·x)^2}} \right) $$ (A/m)
where: $I(t)$ - current (A) at time $t$ (s), $N$ - total number of turns (unitless), $L$ - length of the solenoid (m), $d \approx D$ - diameter of the solenoid (m), $x$ - location (m) from the centre of the solenoid (the centre is located at the point x = 0)
“Thin” solenoid ($d \approx D$) with a circular cross-section
Current I =      Diameter d = D =

Length L =      Number of turns N = (unitless)

Position on axis x =

H =        B0) =

Notes: This equation is valid only for uniformly wound solenoid, with infinitely thin wire. The instantaneous values of H are directly proportional to the instantaneous values of I. The value of B(μ0) is for vacuum.

calculator/solenoid_thin.txt · Last modified: 2022/10/11 14:51 by stan_zurek

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