Question

Determine the magnetic field B at the point (0,0,2) produced by a circular wire of radius r carrying a current I. Assume the wire is positioned in the x-y plane and centred at (0,0,0).
Determine the magnitude and direction of the magnetic field for r = 1 cm and I=1 A

Answer 1

The magnetic field caused by a element of circular wire $dl$ can be found as following

$\displaystyle dB = \frac{\mu_0}{4 \pi} \frac{I dl}{r^2}$

To find B(z) we need to integrate dB for the whole circle. Vector dB has 2 components relative to z-axis: parallel and perpendicular. Because of circular symmetry, all $dB_\perp$ compensate each other. So field B at any point of z-axis is directed along this axis.

$\displaystyle dB_z = |dB| \cos \beta = |dB| \frac{R}{R^2+z^2} = \frac{\mu I_0}{4 \pi} \frac{R}{(R^2+z^2)^{3/2}} dl$

After integration we get

$\displaystyle B = \frac{\mu I_0}{2 } \frac{R^2}{(R^2+z^2)^{3/2}}$

For R=1cm, z=2cm and I = 1 A, the value of B is

$\displaystyle B = \frac{1.26 \cdot 10^{-6}}{2 } \frac{10^{-4}}{(5 \cdot 10^{-4})^{3/2}} = \frac{1.26 \cdot 10^{-6}}{2 } \frac{10^{-4}}{1.11 \cdot 10^{-5}}=0.57 \cdot 10^{-5} = 5.7 \; \mu T$