How can I solve this?

Question

How many turns of wire would be required to make a 160-mH inductor out of a 32.0-cm-long air-filled solenoid with a diameter of 6.3 cm ?

N = ______ turns

Answer 1

How many turns of wire would be required to make a 160-mH inductor out of a 32.0-cm-long air-filled solenoid with a diameter of 6.3 cm ?

Self inductance L is given as

L = µ₀ N² A/l

µ₀ = 4pi * 10 ^-7 for air

L = 160* 10 ^-3 H

l = 32*10 ^-2 m

A = pi * r² , r = radius = 3.15 * 10 ^-2 m

N=?

N = sqrt ( L*l/ µ₀ *pi * r² )

= sqrt (160 *32*10 ⁶/ 4*9.86*3.15 *3.15) = 3610 turns

Pl check calculations

Answer 2

The short version of this, if you don't need any derivation, is to use the solenoid inductance formula,

L = mu * N^2 *A / length

Where L is the inductance- in your case, the target inductance.

mu is the permeability of the core (in this case, we'll use the permeability of free space, which is close enough to air's permeability that the two are usually treated as identical), a constant: 4π * 10^−7 H/m

N is the number of turns in your solenoid,

A is its cross-sectional area,

and all that is divided by the length of the wire when coiled.

Solving this equation for N should allow you to find your desired term by plug-and-chug.

As for the reason this is true, I recommend looking into

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html#c2 (Hyperphysics derivation of solenoid field)

and

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indsol.html#c1 (Inductance formula for a solenoid)

Let me know if I can be of any additional help, or if you'd like to schedule a video session about this or other topics. Thanks!

-Miles