Helmholtz coil

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A Helmholtz coil
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A Helmholtz coil

The term Helmholtz coils refers to a device for producing a region of nearly uniform magnetic field. It is named in honor of the German physicist Hermann von Helmholtz.

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[edit] Description

A Helmholtz pair consists of two identical circular magnetic coils that are placed symmetrically one on each side of the experimental area along a common axis, and separated by a distance equal to the radius of the coil. Actually, a slightly larger separation improves the field uniformity. Each coil carries an equal electrical current flowing in the same direction. A cylindrical region extending between the centers of the two coils and approximately 1/5 of their diameter will have a nearly spatially uniform magnetic field.

[edit] Mathematics

The calculation of the exact magnetic field has mathematical complexities and involves the study of Bessel functions. An approximate calculation gives the correct value at the center point. If the radius is R, the number of turns in each coil is n and the current flowing through the coils is I, then the magnetic flux density, B at the midpoint between the coils will be given by

B = {\left ( \frac{4}{5} \right )}^{3/2} \frac{\mu_0 n I}{R}

μ0 is the permeability constant (1.26 \times 10^{-6} Tm/A, 1.26 \times 10^{-4} Tcm/A or 4.95 \times 10^{-5} Tin/A, for coils measured in meters, centimeters and inches, respectively)

[edit] Derivation

Start with the formula for the on-axis field due to a single wire loop [1] (which is itself derived from the Biot-Savart law [2]):

B = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}
Where:
\mu_0\; = the permeability constant = 4\pi \times 10^{-7} Tm/A = 1.26 \times 10^{-6} Tm/A
I\; = coil current, in amperes
R\; = coil radius, in meters
x\; = coil distance, on axis, to point, in meters

Since the field strength is proportional to the number of loops, we can add that to the formula:

B = \frac{\mu_0 n I R^2}{2(R^2+x^2)^{3/2}}
Where:
n\; = number of wire loops in coil

In a Helmholtz coil, a point halfway between the two loops has an x value equal to R/2, so let's perform that substitution:

B = \frac{\mu_0 n I R^2}{2(R^2+(R/2)^2)^{3/2}}

There are also two coils instead of one, so let's multiply the formula by 2, then reduce the formula:

B = \frac{2\mu_0 n I R^2}{2(R^2+(R/2)^2)^{3/2}}


B = \frac{\mu_0 n I R^2}{(R^2+(R/2)^2)^{3/2}}


B = \frac{\mu_0 n I R^2}{(R^2+(R^2/4))^{3/2}}


B = \frac{\mu_0 n I R^2}{((5/4)R^2)^{3/2}}


B = \frac{\mu_0 n I R^2}{(5/4)^{3/2}R^3}


B = \frac{\mu_0 n I }{(5/4)^{3/2}R}


B = {\left ( \frac{4}{5} \right )}^{3/2} \frac{\mu_0 n I}{R}

[edit] Quadrupole Magnetic Field

This is a Magnetic Field produced by two electric coils; see for example: [3]

[edit] See also

[edit] External links

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