# In a coil containing 400 turns wound on a cardboard cylinder with a radius of 2.0 cm and a length of 0.4 m

**In a coil containing 400 turns wound on a cardboard cylinder with a radius of 2.0 cm and a length of 0.4 m, the current strength changes according to the law i = 0.20t. Determine the energy of the magnetic field of the coil at the end of the tenth second and the EMF of the self-induction in the coil.**

Data: N (number of turns in the specified coil) = 400 pieces; r (radius of the cardboard cylinder) = 2 cm = 0.02 m; l (coil length) = 0.4 m; I (law of current change) = 0.2t; t (considered time) = 10 s.

Constants: μ0 (magnetic constant) = 1.26 * 10 ^ -6 H / m.

1) Current at the end of 10 s: I = 0.2t = 0.2 * 10 = 2 A.

2) Coil cross-section: S = Π * r ^ 2 = 3.14 * 0.02 ^ 2 = 1.256 * 10 ^ -3 m2.

3) Coil inductance: L = μ0 * N ^ 2 * S / l = 1.26 * 10 ^ -6 * 400 ^ 2 * 1.256 * 10 ^ -3 / 0.4 ≈ 633 * 10 ^ -6 H = 633 μH.

4) Energy of the magnetic field: W (10) = 0.5 * L * I ^ 2 = 0.5 * 633 * 10 ^ -6 * 2 ^ 2 = 1266 * 10 ^ -6 J ≈ 1.3 mJ.

5) EMF of self-induction: εс = L * (I – I0) / Δt = 633 * 10 ^ -6 * (2 – 0) / 10 = 126.6 * 10 ^ -6 V = 126.6 μV.