Self Inductance of a Coil: Why Not 1/2?

[jaejoon89]

Homework Statement

How will a coil's self inductance change if you unwind it and then re-wind half of the length of wire into a coil with the same diameter but with half of the number of turns?

The answer is that it will decrease by 1/4, but I don't understand why it isn't 1/2 instead.

Homework Equations

phi (magnetic flux) = LI (where L is inductance, I is current)
n = N/l where "l" is length
B = mew naught * n * I

The Attempt at a Solution

original: L = phi/I = (NBAcostheta) / I = (N* mew naught *n*I*costheta) / I
Now the L with the proposed change: L' = (N/2)* mew naught * ((N/2) / (l/2))* costheta = 1/2 L

 

[Vuldoraq]

I think it's because you have confused the length of the solenoid, l, with the length of the wire, 2*pi*R*l, where R is the radius of the coil. The length of the solenoid hasn't changed, only the number of turns per unit length (and the wire's length) have dropped.

Edit:Woopsey, no solenoid mentioned, sorry I've been doing a lot of these and went tunnel vision. Your right the length is halved. Where does your N come from in the NBAcostheta/I?

 

[thomate1]

I would explain that the self inductance of a coil is determined by the number of turns and the geometry of the coil. When the coil is unwound and half of the length of wire is re-wound, the number of turns is halved and the geometry is changed. This results in a decrease in the self inductance by a factor of 1/4, not 1/2. This is because the magnetic field produced by each turn of the coil is not only dependent on the current, but also on the number of turns and the geometry of the coil. Therefore, changing these factors will affect the overall self inductance of the coil. Additionally, the formula for self inductance assumes that the current is evenly distributed throughout the coil, which may not be the case with a halved coil.