When the current changes from 2.5A to 14.5A in the solenoid

When the current changes from 2.5A to 14.5A in an 800-turn solenoid, the magnetic flux flowing through its turns increases by 2.4 mWb. If the current changes in a time of 0.15 s, then what is the average DC self-induction occurring in the solenoid? It is also necessary to determine the inductance of the solenoid and the change in magnetic field energy.

To solve this problem, we will use formulas relating the change in magnetic flux, electric current and DC self-induction. According to Faraday's law, a change in the magnetic flux in a conductor creates an electromotive force (ΔMF) in it, which is proportional to the rate of change of the magnetic flux. The average DS self-induction that occurs when the current in the solenoid changes can be calculated using the formula:

L = ΔΦ/ΔI

where L is the inductance of the solenoid, ΔΦ is the change in magnetic flux, ΔI is the change in current.

From the conditions of the problem, ΔΦ and ΔI are known, so the inductance of the solenoid can be found:

L = ΔΦ/ΔI = 2.4 mWb / (14.5 A - 2.5 A) = 0.2 H

To find the change in magnetic field energy, we use the formula:

ΔW = 1/2 * L * ΔI²

where ΔW is the change in magnetic field energy, L is the inductance of the solenoid, ΔI is the change in current.

Substituting the known values, we get:

ΔW = 1/2 * 0.2 H * (14.5 A - 2.5 A)² = 4.2 J

Thus, the average DS self-induction that occurs in the solenoid when the current changes from 2.5A to 14.5A in a time of 0.15 s is 0.2 H. The inductance of the solenoid is 0.2 H, and the change in magnetic field energy is 4.2 J.

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There is no product description in your request. However, the condition of the problem is given for solving the average DS of self-induction, solenoid inductance and changes in magnetic field energy when the current changes from 2.5 A to 14.5 A in the solenoid in a time of 0.15 s.

Solution tasks:

From the conditions of the problem we know:

• number of solenoid turns: N = 800
• magnetic flux change: ΔΦ = 2.4 mWb
• current change time: Δt = 0.15 s
• initial current: I1 = 2.5 A
• final current: I2 = 14.5 A

To solve the problem, it is necessary to use the formula for the average DC self-induction:

ΔW = (1/2) * L * ΔI^2,

where ΔW is the change in magnetic field energy, L is the inductance of the solenoid, ΔI is the change in current.

Current change:

ΔI = I2 - I1 = 14.5 A - 2.5 A = 12 A.

Average DS of self-induction:

ΔW = (1/2) * L * ΔI^2

L = ΔW / [(1/2) * ΔI^2] = ΔW / (6 * 10^2) J/A^2

Magnetic field energy change value:

ΔW = ΔΦ * I2 = 2.4 * 10^-3 Wb * 14.5 A = 34.8 * 10^-3 J

Then:

L = (34.8 * 10^-3 J) / [(1/2) * (12 A)^2] = 2.9 * 10^-3 H

Thus, the average DS of self-induction arising when the current changes from 2.5 A to 14.5 A in a solenoid containing 800 turns of magnetic current, through which of its turns has increased by 2.4 mWb and the time of change of which is 0.15 s, equal to 2.9 * 10^-3 H. The solenoid inductance L = 2.9 * 10^-3 H and the change in magnetic field energy ΔW = 34.8 * 10^-3 J were also determined.

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