Helmholtz coils are two identical circular coils having the same radius R and th

Question

Helmholtz coils are two identical circular coils having the same radius R and the same number of turns N, separated by a distance equal to the radius R and carrying the same current I in the same direction. (a) Determine the magnetic field B at points x along the line joining their centers. Let x = 0 at the center of one coil and x = R at the center of the other coil. (b) What is the magnetic field at the midpoint between the coils, where x = R/2? Helmholtz coils are often used in science experiments that require a nearly uniform magnetic field. (c) Demonstrate how uniform the field is at the midpoint between the coils by showing (by hand!) that dB/dx = 0. If you are feeling adventurous, you can show that d^2B/dx^2 = 0 is also zero at the midpoint between the coils! A function that has both the first and second derivatives zero has very little dependence on the argument. The field is very uniform.

Explanation / Answer

1. At the centre line
B1(x) = mu*nIR^2/2[R^2 + x^2]^3/2 [field due to coil 1 at the axis]
B2(x) = mu*nIR^2/2[R^2 + (R-x)^2]^3/2
Net B = mu*nIR^2[1/2(R^2 + x^2)^3/2 + 1/2(R^2 + R^2 + x^2 - 2Rx)^3/2]
2. At midpoint
B(R/2) = mu*nIR^2[R^2 + R^2/4]^3/2 = mu*nIR^2/(5/4)^3/2*R^3 = 8mu*nI/5sqroot(5)*R
3. dB/dx = mu*nIR^2[-6x/4(R^2 + x^2)^5/2 - 3(2x-2R)/4(2R^2 + x^2 - 2Rx)^5/2] = 0
=> -x/(R^2 + x^2)^3/2 -(x-R)/(2R^2 + x^2 - 2Rx)^3/2 = 0
=> -x(2R^2 + x^2 - 2Rx)^3/2 = (x-R)(R^2 + x^2)^3/2
=> x^2(2R^2 + x^2 - 2Rx)^3 = (x-R)^2(R^2 + x^2)^3
for x = R/2
=> R^2/4 (2R^2 + R^2/4 - R^2)^3 = R^2/4 * (5R^2/4)^3 = 125 R^8 / 256 -> LHS
=> (R/2 - R)^2(R^2 + R^2/4)^3 = R^2/4(5R^2/4)^3 = 125 R^8 / 256 --> (RHS)
Hence dB/dx = 0 for x = R/2

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