Hemholtz Coils

What is the easiest way to create a uniform magnetic field?

Historically, this has been a crucial piece equipment in our cyclotrons, early television sets, mass spectrometers, etc. This is something you'll likely find covered quite extensively in your undergraduate physics textbook.

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Mathematical Derivation

Imagine a current loop in the xyxy-plane:

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dB=μ04πIdl×r^r2Biosavart Law\begin{align*} \textbf{dB} &= \frac{\mu_0}{4 \pi} \frac{I \textbf{dl} \times \hat{\textbf{r}}}{r^2} \quad \text{Biosavart Law} \end{align*}

Note that this is only for a infinitesimal part of the current loop! Off the bat, we can make the simplication about the magnitude:

dB=μ04πIdlr2Always dl⊥r^\begin{align*} dB &= \frac{\mu_0}{4 \pi} \frac{I dl }{r^2} \quad \text{Always } \textbf{dl} \perp \hat{\textbf{r}} \end{align*}

Now we just need to integrate for all the infinitesimal parts of the ring. To do this, we can realize that the x^,y^\hat{\textbf{x}}, \hat{\textbf{y}} components cancel and only the z^\hat{\textbf{z}} component will remain. Hence, we can just add the z^\hat{\textbf{z}} components.

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dBz=μ04πIdlr2cos⁡θ Just adding the z^ components=μ04πIdlr2Rr=μ04πIdl(R2+z2)RR2+z2=μ0I4πRdl(R2+z2)3/2\begin{align*} dB_z &= \frac{\mu_0}{4 \pi} \frac{I dl }{r^2} \cos{\theta} \quad \text{ Just adding the } \hat{\textbf{z}} \text{ components}\\ &= \frac{\mu_0}{4 \pi} \frac{I dl }{r^2} \frac{R}{r} = \frac{\mu_0}{4 \pi} \frac{I dl }{(R^2 + z^2)} \frac{R}{\sqrt{R^2 + z^2}} \\ &= \frac{\mu_0 I}{4 \pi} \frac{R dl}{(R^2 + z^2)^{3/2}} \end{align*}

Now just integrate over loop!

B=∫dBz=∫02πRμ0I4πRdl(R2+z2)3/2=μ0I2R2(R2+z2)3/2[Note that: B(center)=μ0I2R]\begin{align*} B &= \int dB_z = \int_0^{2\pi R} \frac{\mu_0 I}{4 \pi} \frac{R dl}{(R^2 + z^2)^{3/2}}\\ &= \frac{\mu_0 I}{2} \frac{R^2}{(R^2 + z^2)^{3/2}} \\ &\left[ \text{Note that: } B(\text{center}) = \frac{\mu_0 I}{2R} \right] \end{align*}

Now say we have two of these single loops coils at 2D2D distance apart:

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B(z)=B1+B2=μ0I2R2(R2+(D−z)2)3/2+μ0I2R2(R2+(D+z)2)3/2=μ0IR22[1(R2+(D−z)2)3/2+1(R2+(D+z)2)3/2]\begin{align*} B(z) &= B_1 + B_2\\ &= \frac{\mu_0 I}{2} \frac{R^2}{(R^2 + (D - z)^2)^{3/2}} + \frac{\mu_0 I}{2} \frac{R^2}{(R^2 + (D+z)^2)^{3/2}} \\ &= \frac{\mu_0 I R^2}{2} \left[ \frac{1}{(R^2 + (D - z)^2)^{3/2}} + \frac{1}{(R^2 + (D+z)^2)^{3/2}} \right] \\ \end{align*}

Note that at the center, the magnetic field is simply:

B(z)=μ0IR2(R2+D2)3/2\begin{align*} B(z) &= \frac{\mu_0 I R^2}{(R^2 + D^2)^{3/2}} \\ \end{align*}

Where II is simply total induced current.

Small experiments

Guitar Coils work in a very similar way but the other way. Disturbances in the surrounding magnetic field cause a current in the coil that we process as sound. As a result, when I use my Hemholtz Coil to manually disturb these guitar coils, I can manually change the induced current and sound produced by the guitar.