Question

Two identical, flat, circular coils of wire each have 100 turns and a radius of 0.500 m. The coils are arranged as a set of Helmholtz coils, parallel and with separation 0.500 m Each coil carries a current of 10.0 A. Determine the magnitude of the magnetic field at a point on the common axis of the coils and halfway between them.

Answer 1

Answer:

The magnitude of the magnetic field at the point of interest is B=1.8·10^-3T

Explanation:

To solve this problem we can calculate the expression of the magnetic field for one of the coils and using the symmetry of the problem we can use twice this expression to obtain the expression of the total field as:

$B_{T}=B_{1}+B_{2}$

For the magnetic field $B{1}$ we use the Biot-Savart law:

$d\vec{B}=\frac{\mu_{0}i\vec{dl}}{4\pi}\times \frac{\vec{r}-\vec{r'}}{|\vec{r}-\vec{r'}|^3}$

Using cylindrical coordinates, the expression of the magnetic field in a circular coil is:

$B_{i}=\frac{\mu_{0}ir^2}{2(\sqrt{z^2+r^2})^3}$

The expression of the magnetic field in a circular coil with N turns is:

$B_{i}=\frac{\mu_{0}iNr^2}{2(\sqrt{z^2+r^2})^3}$

Because we have a point of symmetry in the middle between the 2 coils, the expression of the field in that point is:

$B_{T}=B_{1}+B_{2}=\frac{\mu_{0}iNr^2}{(\sqrt{z^2+r^2})^3}$

with r=0.5m, Z=0.25m, i=10A and N=100