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Archived Derivation of total magnetic field with changing distance in a coil?

  • Thread starter Violagirl
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1. Homework Statement

You are a member of a research team studying magnetotactic bacteria. Magnetotactic
bacteria from the southern hemisphere preferentially swim to the south along magnetic
field lines, while similar bacteria from the northern hemisphere preferentially swim to
the north along magnetic field lines. Your team wishes to quantify the behavior o f
magnetotactic bacteria in closely controlled magnetic fields.

You know from your physics class that a coil of wire can be used to produce a magnetic
field, which can be varied by changing the current through it. You set yourself the task
of calculating the magnetic field along the axis of the coil as a function of its current,
number of turns, radius, and the distance along the axis from the center of the coil. To
make sure you are correct, you decide to compare your calculation to measurements.

Calculate the magnitude of the magnetic field as a function of the position along the
central axis of a coil of known radius, the number of turns of wire, and the electric
current in the coil.

2. Homework Equations
Biel-Savart Law: B = µ0/4π ∫c Idl x r / r3


3. The Attempt at a Solution

I was going to see if someone would be able to check to see if I did the derivation correctly? Thanks so much!

See attached document for diagram of situation.

Derived equation:



∫ dBz = ∫ µ0 N I dl cos θ/4π (R2 + z2) (R/√R2 + z2)

µ0 N I R / 4π (R2 + z2)3/2 ∫ dl

2π µ0 I N R2 / 4π (R2 + z2)3/2

µ0 I N R2 / 2 (R2 + z2)3/2-final equation.
 

Attachments

BiGyElLoWhAt

Gold Member
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Man, that's really messy...
Reprint of the work, as is.
##\vec{B} = \frac{\mu_0}{4\pi}\int_C \frac{I d\vec{\ell}\times\vec{r}}{|r|^3}## -> note that ##\frac{\vec{r}}{|r|} = \hat{r}##
##\int dB_z = \int \mu_0 N I \frac{d\ell \text{cos}(\theta)}{4\pi (R^2 + z^2)}\frac{R}{R^2 + z^2}##
... to be edited later...
 

BiGyElLoWhAt

Gold Member
1,560
113
Man, that's really messy...
Reprint of the work, as is.
##\vec{B} = \frac{\mu_0}{4\pi}\int_C \frac{I d\vec{\ell}\times\vec{r}}{|r|^3}## -> note that ##\frac{\vec{r}}{|r|} = \hat{r}##
##\int dB_z = \int \mu_0 N I \frac{d\ell \text{cos}(\theta)}{4\pi (R^2 + z^2)}\frac{R}{R^2 + z^2}##
... to be edited later...
Continuing because it's too late to edit.
***
##\frac{\mu_0 NIR}{4\pi (R^2 + z^2)^{3/2}}\int_l d \ell## This is good, assuming that the point only lies on the z axis, about which the loop is centered (the z axis is normal to the area of the loop) and also that the wires have negligible width (all the loops are concentrated in a small range of z, call it A, such that A<<z). The limits (l) of the integral should be 1 full circle, so for example, 0 to 2pi r.
##\frac{\mu_0 NIR}{4\pi (R^2 + z^2)^{3/2}}(2\pi R)##
##\frac{\mu_0 NIR}{2 (R^2 + z^2)^{3/2}}## Given these assumptions, this is a valid solution.
 

BiGyElLoWhAt

Gold Member
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Originally, I wasn't thinking that this was simplified as such (see the assumptions). I'm working on a general form, lacking these assumptions, that would work for any point in R^3. However, I think I might have to essentially treat the coil as a cylindrical shell with current density J, and it keeps getting hairier and hairier. =D

I'll post back... eventually.
 

rude man

Homework Helper
Insights Author
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666
Originally, I wasn't thinking that this was simplified as such (see the assumptions). I'm working on a general form, lacking these assumptions, that would work for any point in R^3. However, I think I might have to essentially treat the coil as a cylindrical shell with current density J, and it keeps getting hairier and hairier. =D

I'll post back... eventually.
You will encounter hopelessly complex math, e.g. elliptic integrals plus the need to consider the radius of the wire.
Quit now while the getting is good! :smile:
 

BiGyElLoWhAt

Gold Member
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=D
Yea that's what I was finding. I actually looked up the solution to one of the integrals, and it was about 5 ridiculous substitutions... I still kind of want to work it out, but we'll see how it gets. Good looking out ;)
 

rude man

Homework Helper
Insights Author
Gold Member
7,510
666
=D
Yea that's what I was finding. I actually looked up the solution to one of the integrals, and it was about 5 ridiculous substitutions... I still kind of want to work it out, but we'll see how it gets. Good looking out ;)
Yes, and even then it will be an approximation to an arbitrary order.
 

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