The magnetic field energy of the coil W = 12.8 J. Determine the self-induction flux linkage
| June 1, 2021 | education
The magnetic field energy of the coil W = 12.8 J. Determine the self-induction flux linkage and the inductance of the coil if the current in it is I = 6.4A.
Initial data: Wm (coil magnetic field energy) = 12.8 J; I (coil current) = 6.4 A.
1) Calculate the inductance of the coil from the formula:
Wm = L * I ^ 2/2.
L = 2 * Wm / I ^ 2 = 2 * 12.8 / 6.4 ^ 2 = 0.625 H.
2) Define the self-induction flux linkage:
ψ = L * I = 0.625 * 6.4 = 4 Wb.
Answer: The self-induction flux linkage of the coil is 4 Wb, the inductance of the coil is 0.625 H.
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