An inductor in the form of an air-core solenoid contains 191 turns, is of length 17.9 cm, and has a cross-sectional area of 1.2 cm^2 . The permeability of free space is 1.25664 × 10^−6 N/A^2 . What is the magnitude of the uniform rate of change in current through the inductor that induces an emf of 205 µV? Answer in units of A/s.

Question

contestada

Respuesta :

zoialaraib zoialaraib · Answer 1 · 2019-12-15T12:59:10+01:00

Answer:

The magnitude of the uniform rate of change in current through the inductor is 13.66 A/s.

Explanation:

To find the magnitude of the uniform rate of change in current through the inductor we will use the formula:

ΔI/Δt = e×l / μ0×N×A

Now, put all the given values in this formula and then you will get your answer,

ΔI/Δt = 205×10^-6×19.2 / 1.2566×10^-6×191×1.2

ΔI/Δt = 13.66 A/s.

okpalawalter8 okpalawalter8 · Answer 2 · 2022-03-24T12:26:06+01:00

The magnitude of the uniform rate of change in current is mathematically given as

dI/dt = 13.66 A/s.

Question Parameters:

the form of an air-core solenoid contains 191 turns

of length 17.9 cm

The permeability of free space is 1.25664 × 10^−6 N/A^2

Generally, the equation for the rate of change in current is mathematically given as

dI/dt = e×l / μ0*N*A

Therefore

dI/dt = 205×10^-6×19.2 / 1.2566×10^-6×191×1.2

dI/dt = 13.66 A/s.

In conclusion, magnitude of the uniform rate of change in current

dI/dt = 13.66 A/s.