An inductor in the form of a solenoid contains 410 turns and is 15.4 cm in length. A uniform rate of decrease of current through the inductor of 0.421 A/s induces an emf of 175 µV. What is the radius of the solenoid?

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+3 votes

Best answer

Being with the formula for inductance for a solenoid, which is

**L = μ _{0 }* N^2 * A/l **(*where L=inductance, μ

Area = pi*r^2,

plugging this into the formula above we get:

**L = μ _{0 }* N^2 * (pi*r^2)/l **

Now solve for r,

**r = squareroot[(L*l)μ _{0 }* N^2 * (pi)]**

we are given the following values,

l = 15.4 cm = 15.4*10^-2 m

N = 410 turns

dI/dt = 0.421 A/s

Emf = 175 µV = 175*10^-6 V

Since we know the emf, we can calculate late L from the formula for emf:

**Emf = -L*dI/dt**

**L = Emf / ****dI/dt**

L = 175*10^-6 V / 0.421 A/s = 4.157*10^-4 H

Now we can plug this back into the formula for r and solve

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