Sinusoidal Wire Magnetics

One can start by computing the resistance of the sinusoidal wire. This can be done by multiplying the resistance per unit length by the arc length of the sine wire. The expression for resistance is:

$R = \int_{0}^{2\pi}\sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx$ Which is: $R = \int_{0}^{2\pi}\sqrt{1 + \cos^2(x)}dx$

The current through the wire is computed using Ohm's law:

$I = \frac{5}{R}$

Now a general point P on the sinusoidal curve is parameterised as such:

$\vec{r}_p = (t,\sin(t),0)$

The test point is: $\vec{r}_t = (5,1,0)$

The vector joining the point P and the test point, directed away from P is:

$\vec{r} = \vec{r}_t - \vec{r}_p$

The arc length vector is tangential to the curve:

$d\vec{s} = (1,\cos(t),0)$

Finally, the use of Biot-Savart law is made:

$d\vec{B} = \frac{\mu_o}{4\pi}I\frac{d\vec{s}\times\vec{r}}{\mid\vec{r}\mid^3}$

The vector is directed along the Z-axis only. This leads to:

$\mid d\vec{B} \mid = \frac{I}{4\pi}\frac{\left(\cos\left(t\right)\,\left(t-5\right)-\sin\left(t\right)+1\right)}{{\left({\left(t-5\right)}^2+{\left(\sin\left(t\right)-1\right)}^2\right)}^{3/2}}dt$

$\mid \vec{B} \mid = \int_{0}^{2\pi}\left(\frac{I}{4\pi}\frac{\left(\cos\left(t\right)\,\left(t-5\right)-\sin\left(t\right)+1\right)}{{\left({\left(t-5\right)}^2+{\left(\sin\left(t\right)-1\right)}^2\right)}^{3/2}}\right)dt$

Integrating numerically gives the answer: $\boxed{\mid \vec{B} \mid = 0.06345}$

As always, it is a pleasure to solve your problems. Thank you for another good one.