Finding The Magnetic Field From A Helmholtz Coil 2n4173

Finding the Magnetic Field from a Helmholtz Coil Cody Goolsby∗ Department of Physics, University of Southern Maine 96 Falmouth Street, Portland, ME 04104-9300 (Dated: November 16, 2011)

One might find it of interest to know the magnetic field produced by a loop of wire carrying an electric current. Let’s first examine the most rudimentary case, the magnetic field along the central axis of a loop of wire.

~ a distance r from a loop of wire carrying a current, I. FIG. 1: Magnetic field, dB,

~ it produces a distance Fig. 1 gives a general picture of such a loop carrying a current, I, and the magnetic field, B, z from the center of the loop at point P. Using the Biot-Savart Law,

~ = dB

µ0 I d~l × ~rˆ , 4π r2

(1)

let’s find the magnetic field along the central axis of the loop. First, break down the field into horizontal and vertical components. As can be seen in Fig. 1 the horizontal components cancel each other out, since the horizontal component of the loop 180◦ opposite the loop has the same magnitude but opposite direction. This leaves us with only the vertical components which can simply be added together. Since we only care about the vertical component ~ of the field, let’s find out how it relates to the total field, dB,

cos θ =

dB~z . ~ dB

(2)

This leaves us with, ~ cos θ dB~z = dB

(3)

2 Thus our revamped Biot-Savart Equation is µ0 I d~l × ~rˆ cos θ, 4π r2

~z = dB

(4)

This leaves us with three unknowns: the cross product, r, and cos θ. The cross product is d~l sin φ. φ, in this instance, will always be 90◦ , leaving us with only d~l. r from Fig. 1 is evidently, r 2 = R2 + z 2 .

(5)

And finally, again from Fig. 1 we can see that cos θ is R R =√ 2 r R + z2

(6)

µ0 I R dl, 4π (R2 + z 2 )3/2

(7)

cos θ = Plugging these back into Eq. 5 we find, ~z = dB

Now integrate along dl or around the loop to obtain the magnetic field, Z

~z = dB

µ0 I R 4π (R2 + z 2 )3/2

Z dl,

(8)

which reduces to, R ~ z = µ0 I B (2πR), 2 4π (R + z 2 )3/2

~z = B

µ0 IR2 , 2(R2 + z 2 )3/2

(9)

(10)

What will the magnetic field look like at z=0? ~ z=0 = µ0 I B 2R

(11)

What will the field look like as z → ∞? µ0 IR2 z→∞ 2(R2 + z 2 )3/2

(12)

~ z→∞ = 0 B

(13)

~ z→∞ = lim B

Now that we know the field along the central axis, let’s next consider the field anywhere. To make our job easier, let’s set the point in space we’re intersted in, point P, along the y-axis. To determine the magnetic field, we will again use the Biot-Savart Law:

~ = dB

µ0 I d~l × ~rˆ . 4π r2

(14)

3

~ a distance r from a loop of wire carrying a current, I. FIG. 2: Magnetic field, dB,

Unlike in our previous example, the cross product, d~l × ~rˆ, does not nicely reduce down to something easily calculatable. Instead this time we must work out the cross product in full. First switch from the unit vector, rˆ, to the vector, ~r, that is ~r ~r = rˆ r =⇒ rˆ = . r

(15)

µ0 I d~l × ~r . 4π r3

(16)

Substituting this back into Eq. 232, we find

~ = dB

The cartesian coordinates of a point on the wire, W, are

x = R cos θ, y = R sin θ, z = 0,

(17)

dx = −R sin θ dθ, dy = R cos θ dθ, dz = 0.

(18)

their derivatives with respect to θ are

The coordinates at point P are (0, y, z), thus the distance from P to W are P (x, y, z) − W (x, y, z) =⇒ (−R cos θ, y − R sin θ, z).

(19)

These are the cooridnates for ~r. The coordinates for d~l are d~l = dxˆ x + dy yˆ + dz zˆ,

(20)

plugging in the differentials from Eq. 2343 we have, d~l = (−R sin θ dθ)ˆ x + (R cos θ dθ)ˆ y + (0)ˆ z. Thus, d~l × ~r is

(21)

4 x ˆ yˆ zˆ d~l × ~r = −R sin θ dθ R cos θ dθ 0 −R cos θ y − R sin θ z The components of the resulting cross product are d~l × ~r = (Rz cos θ dθ) x ˆ − (−Rz sin θ dθ) yˆ + [(−R sin θ dθ)(y − R sin θ) − (R cos θ dθ)(−R cos θ)]ˆ z

(22)

d~l × ~r = (Rz cos θ dθ) x ˆ + (Rz sin θ dθ) yˆ + (R2 − Ry sin θ)dθ zˆ

(23)

Now we must figure out the magnitude of ~r, but since we know the components from Eq. 2343 we simply square each component, add them together, and take the square root,

(24)

(−R cos θ)2 + (y − R sin θ)2 + z 2

(25)

R2 cos θ2 + y 2 − 2yR sin θ + R2 sin θ2 + z 2

(26)

|r| =

|r| =

p

q

rx2 + ry2 + rz2

|r| =

p

|r| =

p

R2 + y 2 + z 2 − 2yR sin θ

(27)

Finally, we can plug Eq. 343 and Eq. 343 back into Eq. 343 and solve for the magnetic field in of cartesian components, ~x = dB

µ0 I Rz cos θdθ . 4π (R2 + y 2 + z 2 − 2yR sin θ) 32

(28)

~y = dB

µ0 I Rz sin θdθ . 2 2 4π (R + y + z 2 − 2yR sin θ) 32

(29)

~z = dB

µ0 I R2 − Ry sin θdθ . 4π (R2 + y 2 + z 2 − 2yR sin θ) 23

(30)

As you can see the only integral that is ”easy” to do is B~x , ~ x = µ0 IRz B 4π

Z 0

2π

cos θ (R2

+

y2

3

+ z 2 − 2yR sin θ) 2

dθ.

(31)

Performing a u-substitution we find, ~ x = µ0 Iz [(R2 + y 2 + z 2 − 2yR sin 2π)−1/2 − (R2 + y 2 + z 2 − 2yR sin 0)−1/2 ]. B 4πy

(32)

But since the sin of 0 and 2π are both zero, B~x = 0. Leaving us with B~y and B~z to still calculate. In order to calculate B~y and B~z we must first become familiar with complete elliptic integrals of the first and second kind. The complete elliptic integral of the first kind is

5

Z

π/2

K(k) = 0

dθ p , 1 − k 2 sin2 θ

(33)

p 1 − k 2 sin2 θ dθ.

(34)

and of the second kind is Z E(k) =

π/2

0

Evidently we need to get B~y and B~z into the above forms. Plugging Eqs. 33 and 34 into Mathematica we find, ~ y = µ0 I B 4π



xz √ y(m + 2Ry) m − 2Ry



π 3π π 3π m E[ , k] + E[ , k] − (m + 2Ry) F[ , k] + F[ , k] 4 4 4 4

µ0 I π 3π π 3π x 2 ~ √ Bz = 2πxR − m E[ , k] + E[ , k] − (m + 2Ry) F[ , k] + F[ , k] 4π 4 4 4 4 (m + 2Ry) m − 2Ry m = R2 + y 2 + z 2

∗

Electronic address: [email protected]

k=

−4Ry m−2Ry

(35)

(36)