Sinusoidal Wire Magnetics (part 2)

Very nice problem:

Consider a point on the wire having a position vector:

$\vec{r}_p = x \hat{i} + \sin{x} \hat{j} + 0 \hat{k}$

The point of interest is:

$\vec{r}_c = 1 \hat{i} + 2 \hat{j} + 3 \hat{k}$

A vector joining these two points directed towards the point of interest is:

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

The current is assumed to flow rightwards along the sine curve. An arc length element vector along the curve is:

$d\vec{s} = \left(1 \hat{i} + \cos{x} \hat{j} + 0 \hat{k}\right)dx$

From here, Biot-Savart Law is applied, which is:

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

Substituting and simplifying gives the expressions:

$d\vec{B} = dB_x \hat{i} + dB_y \hat{j} + dB_z \hat{k}$

Where:

$dB_x = \frac{3\,\cos\left(x\right) \ dx}{{\left({\left(x-1\right)}^2+{\left(\sin\left(x\right)-2\right)}^2+9\right)}^{3/2}}$ $dB_y = -\frac{3 \ dx}{{\left({\left(x-1\right)}^2+{\left(\sin\left(x\right)-2\right)}^2+9\right)}^{3/2}}$ $dB_z = \frac{\left(\cos\left(x\right)\,\left(x-1\right)-\sin\left(x\right)+2\right) \ dx}{{\left({\left(x-1\right)}^2+{\left(\sin\left(x\right)-2\right)}^2+9\right)}^{3/2}}$

Finally:

$B_x = \int_{0}^{2 \pi} \frac{3\,\cos\left(x\right) \ dx}{{\left({\left(x-1\right)}^2+{\left(\sin\left(x\right)-2\right)}^2+9\right)}^{3/2}}$ $B_y = \int_{0}^{2 \pi} -\frac{3 \ dx}{{\left({\left(x-1\right)}^2+{\left(\sin\left(x\right)-2\right)}^2+9\right)}^{3/2}}$ $B_z = \int_{0}^{2 \pi} \frac{\left(\cos\left(x\right)\,\left(x-1\right)-\sin\left(x\right)+2\right) \ dx}{{\left({\left(x-1\right)}^2+{\left(\sin\left(x\right)-2\right)}^2+9\right)}^{3/2}}$

The net magnetic field at the point of interest is:

$\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$

And the required magnitude is:

$\boxed{\lvert \vec{B} \rvert = \sqrt{ B_x^2 + B_y^2 + B_z^2} \approx 0.3154}$