Why Does Rearranging an Inductor Yield Different Magnetic Fields?

AI Thread Summary
Rearranging an inductor can yield different magnetic fields due to variations in inductance and geometry. Two methods of calculating the magnetic field produced by the inductor yielded different results, highlighting the complexity of the problem. The first method correctly relates the magnetic fields to the inductance and dimensions of the inductors, while the second method introduces additional factors that complicate the comparison. The discussion reveals confusion regarding the assumptions made about the wire's layers and the stability of current during the rearrangement. Ultimately, understanding the relationship between inductance, geometry, and current is crucial for accurate calculations.
matej1408
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I found two ways to solve this problem, but I get two different solutions, it's confusing because I can't see the flaw in wrong solution.
1. Homework Statement

Long cylindrical inductor of diameter D1 and inductance L1 is connected to battery and creates magnetic field B1. Inductor is then rearranged to new inductor of diameter D2 and inductance L2. Find magnetic field B2 which is created when "new" inductor is connected to the same battery as before. Assume that length of wire is much larger then length on inductor.

Homework Equations


L=μN2S/l
B=μNi/l
i-current, l-length of coil

The Attempt at a Solution


1st way(correct):[/B]
B1=μN1i/l1 => μN1/l1=B1/i
L1=μN12S1/l1
L1=N1S1B1/i
current is same in both case so:
L2=N2S2B2/i
dividing and rearranging_
L1=N1S1B1/i
L2=N2S2B2/i
B2=B1N1S1L2/(N2S2L1)
wire is same length in both case so:
h- thickness of wire
N1hD1π=N2hD2π
N1/N2=D2/D1
and S∝D2
B2=B1D1L2/(D2L1)
2nd way:
Bn=μNni/ln => B2= B1N2l1/(l2N1)
Ln=μNn2Sn/ln
Nn=√(Lnln /(μSn))
so N2/N1=√(L2l2S1/(L1l1S2))
substituting: B2= B1√(L2l1S1/(L1l2S2))
wire is same length so: Dπl= constant => l1/l2=D2/D1
and and S∝D2 =>
B2=B1√(D1L2/(D2L1))
 
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Hello, and welcome to PF!

matej1408 said:
1st way(correct):
wire is same length in both case so:
h- thickness of wire
N1hD1π=N2hD2π
I'm not quite sure why you have a factor of h on each side of the above equation. In any case, it does lead to the following
N1/N2=D2/D1
which I believe is correct.

Keep in mind that either solenoid can have more than one layer of wraps of wire.

2nd way:
wire is same length so: Dπl= constant => l1/l2=D2/D1

How is Dπl related to the length of wire?
 
When you connect battery to an inductor the current does not stabilize but ramps up at a constant rate di/dt = V/L. So how a re you comparing currents? the problem also implies that the inductor is changed 'on the fly' while current is flowing, which itself constitutes a V = i dL/dt problem in addition to the usual V = L di/dt one.
Bottom line, the problem is hard to understand, at least for me.
 
I see where i was wrong, I assumed that there is just one layer but there isn't, thank you TSny.
Coil has some resistance and when is connected to DC there is no voltage change=> current is constant
 
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