91Ó°ÊÓ

Now, considering the Helmholtz Coils which consist of two circular coils with radius a, each wound with N turns of wire carrying a current I, circulating in the same direction in both coils. The coils are separated by a distance of a equal to their radii as shown in the following figure. First, we need to find the magnetic field due to the coils at point P. The magnetic field at a distance of x from a circular coil with a radius of a given by,

$B_x=\frac{{\mathrm{μ}}_{0}N/a^2}{2{\left(x^2+a^2\right)}^{3/2}}$

The left coils is at a distance of x + a / 2 from the point P, and the right coil is at a distance of x - a / 2 from the same point. Hence the changes are:

$\begin{array}{r}x→x+\frac{a}{2}\\ x→x−\frac{a}{2}\end{array}$

Now, for the left and the right coil respectively. Now we can write the magnetic fields due to these coils as:

${\mathrm{B}}_{\mathrm{L}}=\frac{{\mathrm{μ}}_0{\mathrm{Nia}}^2}{2{\left\{(\mathrm{x}+\mathrm{a}/2{)}^2+{\mathrm{a}}^2\right\}}^{3/2}}{\mathrm{B}}_{\mathrm{L}}=\frac{{\mathrm{μ}}_0{\mathrm{Nia}}^2}{2{\left\{(\mathrm{x}−\mathrm{a}/2{)}^2+{\mathrm{a}}^2\right\}}^{3/2}}$

Now, the total magnetic field at P is therefore:

$\begin{array}{r}\mathrm{B}={\mathrm{B}}_{\mathrm{L}}+{\mathrm{B}}_{\mathrm{R}}\\ =\frac{{\mathrm{μ}}_0\mathrm{N}\left({\mathrm{ai}}^2\right.}{2}\left(\frac{1}{{\left\{(\mathrm{x}+\mathrm{a}/2{)}^2+{\mathrm{a}}^2\right\}}^{3/2}}+\frac{1}{{\left\{(\mathrm{x}−\mathrm{a}/2{)}^2+{\mathrm{a}}^2\right\}}^{3/2}}\right)\end{array}$

Now, we need to the plot, however first we need to transform the equation to be in terms of the dimensionless variables,

$\begin{array}{r}\mathrm{x}→\mathrm{x}/\mathrm{a}\text{as}\\ \mathrm{B}=\frac{{\mathrm{μ}}_0{\mathrm{Nia}}^2}{2}\left(\frac{1}{{\left\{(\mathrm{x}+\mathrm{a}/2{)}^2+{\mathrm{a}}^2\right\}}^{3/2}}+\frac{1}{{\left\{(\mathrm{x}−\mathrm{a}/2{)}^2+{\mathrm{a}}^2\right\}}^{3/2}}\right)\\ =\frac{{\mathrm{μ}}_0{\mathrm{N}}^2{\mathrm{a}}^2}{2{\mathrm{a}}^3}\left(\frac{1}{{\left\{(\mathrm{x}/\mathrm{a}+1/2{)}^2+1\right\}}^{3/2}}+\frac{1}{\left\{(\mathrm{x}/\mathrm{a}−1/2{)}^2+1\right\}}\right)\end{array}$

Or,

$=\frac{\mathrm{B}}{{\mathrm{μ}}_0\mathrm{NI}/2\mathrm{a}}\left(\frac{1}{{\left\{(\mathrm{x}/\mathrm{a}+1/2{)}^2+1\right\}}^{3/2}}+\frac{1}{{\left\{(\mathrm{x}/\mathrm{a}−1/2{)}^2+1\right\}}^{3/2}}\right)$

Now, we can plot this equation from $\mathrm{x}=0\text{to}\mathrm{x}=\mathrm{a}/2,\text{or}\mathrm{x}/\mathrm{a}=0\text{to}\mathrm{x}/\mathrm{a}=1/2$. Now plotting this we get the graph:

To find the magnitude of the magnetic field at the point P(x=0), by substituting with x=0 into we get:

$\begin{array}{r}\mathrm{B}=\frac{{\mathrm{μ}}_0{\mathrm{Nia}}^2}{2}\frac{1}{{\left\{(\mathrm{a}/2{)}^2+{\mathrm{a}}^2\right\}}^{3/2}}+\frac{1}{{\left\{(−\mathrm{a}/2{)}^2+{\mathrm{a}}^2\right\}}^{3/2}}\\ =\frac{{\mathrm{μ}}_0{\mathrm{Nia}}^2}{{\left(5{\mathrm{a}}^2/4\right)}^{3/2}}\\ ={\left(\frac{4}{5}\right)}^{3/2}\frac{{\mathrm{μ}}_0\mathrm{NI}}{\mathrm{a}}\end{array}$

Now, to calculate the magnitude of the magnetic field at P (x = 0) given that N = 300 turns, I = 6.00 A and a = 8.00 cm as:

$\begin{array}{r}={\left(\frac{4}{5}\right)}^{3/2}\frac{\left(4\mathrm{Ï€}{10}^{−7}Tâ‹\dots mA\right)\left(300\right)\left(6.00A\right)}{0.080m)}\\ =0.0202T\end{array}$

Therefore, the magnitude is 0.0202T

Now, to derivative of Bat P (x) to check the stability of the magnetic field at this point as:

$\begin{array}{r}\frac{\mathrm{dB}}{\mathrm{dx}}=\frac{{\mathrm{μ}}_0{\mathrm{Nia}}^2}{2}\left(\frac{−3(\mathrm{x}+\mathrm{a}/2)}{{\left\{(\mathrm{x}+\mathrm{a}/2{)}^2+{\mathrm{a}}^2\right\}}^{5/2}}+\frac{{\mathrm{μ}}_0{\mathrm{Nia}}^2}{2}\frac{−3(\mathrm{x}−\mathrm{a}/2)}{{\left\{(\mathrm{x}+\mathrm{a}/2{)}^2+{\mathrm{a}}^2\right\}}^{5/2}}\right)\\ {\left.\frac{\mathrm{dB}}{\mathrm{dx}}\right|}_{\mathrm{x}=0}=0\end{array}$