Isolate $z$ in equation

Question

This is an equation to find the field strength of a cylindrical magnet at a given distance. I'd like to reverse it and find a distance for a given field strength along the polar axis.

z is length

D is thickness

R is radius

h is remanence field

b is field strength in teslas

Can someone walk me through the steps to isolate z from this equation?

$$ b = \frac{h}{2} \left(\frac{D+z}{\sqrt{(D+z)^2+R^2}}-\frac{z}{\sqrt{ R^2+z^2}}\right) $$

I have a known b,h,D,R but I have an unknown z

example with numbers.

$$ 0.008801 = \frac{1.48}{2} \left(\frac{0.09375+0.438}{\sqrt{(0.09375+0.438)^2+0.125^2}}-\frac{0.438}{\sqrt{ 0.125^2+0.438^2}}\right) $$

Edit: I edited my equation. Sorry I messed it up when converting to mark up.

Equation is from this website

Answer 1

Let's suppose that $\;z=R\,\sinh(t-a)\,$ and $\;D+z=R\,\sinh(t+a)\;$ then \begin{align} \frac{2b}h&= \frac{D+z}{\sqrt{(D+z)^2+R^2}}-\frac{z}{\sqrt{ R^2+z^2}}\\ &=\tanh(t+a)-\tanh(t-a)\\ &=\frac{\sinh(t+a)}{\cosh(t+a)}-\frac{\sinh(t-a)}{\cosh(t-a)}\\ &=\frac{\sinh((t+a)-(t-a))}{(\cosh(t)\cosh(a))^2-(\sinh(t)\sinh(a))^2}\\ &=\frac{\sinh(2a)}{\cosh(t)^2+\cosh(a)^2-1}\\ \end{align} That we will rewrite as : $$\tag{1}\cosh(t)^2+\sinh(a)^2=\frac{h\,\sinh(2a)}{2b}$$ We have too $\;D=R\,(\sinh(t+a)-\sinh(t-a))\;$ so that $$\tag{2}D=2\,R\,\cosh(t)\sinh(a)$$ Let's note $A:=\sinh(a)^2\,$ and use the expression of $\;\cosh(t)$ from $(2)$ : $$\tag{3}\cosh(t)=\frac{D}{2R\,\sqrt{A}}$$ and replace it in $(1)$ to get : $$\frac {D^2}{4R^2\,A}+A=\frac{h\,\sinh(a)\sqrt{1+\sinh(a)^2}}{b}=\frac{h\sqrt{A(1+A)}}{b}$$ Multiplying by $\,A$ and squaring we get : $$\tag{4}\left(\frac {D^2}{4R^2}+A^2\right)^2=\left(\frac hb\right)^2\,A^3(1+A)$$ Solving the quartic you may

• obtain $A$ (the smallest positive real solution when it exists I think)
• deduce $t$ (or rather use directely $\cosh(t)$)

• and finally obtain the wished $z$.

  This is not really straightforward but could be the wished solution (all this should be verified of course).