The current in a coil changes from 2 A to 10 A in 80 ms. As a result, an e.m.f of 5.0V is induced in the coil. Determine the magnitude of the self-inductance of the coil

The self-inductance of a coil can be calculated using Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a coil is equal to the rate of change of magnetic flux through the coil. This can be mathematically represented as:

E = -L * (ΔI/Δt)

where: E is the induced emf, L is the self-inductance, ΔI is the change in current, and Δt is the change in time.

We can rearrange this equation to solve for L:

L = -E / (ΔI/Δt)

Given in the problem: E = 5.0 V, ΔI = 10 A - 2 A = 8 A, and Δt = 80 ms = 0.08 s.

Substituting these values into the equation gives:

L = -5.0 V / (8 A / 0.08 s)

Solving this equation gives:

L = -5.0 V / (100 A/s)

L = -0.05 H

However, because self-inductance is always a positive quantity, we take the absolute value:

L = 0.05 H

So, the magnitude of the self-inductance of the coil is 0.05 Henrys.