In the inductance coil, within 0.2 s, the current strength decreased from 12 A to 4 A.

| March 4, 2021 | education

In the inductance coil, within 0.2 s, the current strength decreased from 12 A to 4 A. Determine the change in the magnetic field energy of the coil, if the self-induction EMF of 12 V occurs.

To calculate the change in the energy of the magnetic field of the indicated coil, we use the formula: ΔW = L * (Ik ^ 2 – In ^ 2) / 2 = εi * Δt * (Ik ^ 2 – In ^ 2) / (2 * (Ik – In) ) = εi * Δt * (Ik – In) * (Ik + In) / (2 * (Ik – In)) = εi * Δt * (Ik + In) / 2.

Variables: εi – emerging EMF of self-induction (εi = 12 V); Δt – time change (Δt = 0.2 s); Ik – final current in the coil (Ik = 4 A); In – initial current (In = 12 A).

Let’s make a calculation: ΔW = εi * Δt * (Ik + In) / 2 = 12 * 0.2 * (4 + 12) / 2 = 19.2 J.

Answer: The energy of the magnetic field of the indicated coil changed by 19.2 J.



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