A wire carrying a current $I$ is bent into the shape of an exponential spiral $r=e^{\theta }$ from $\theta =0$ to $\theta =2\pi$ as in Fig.

To complete a loop, the ends of the spiral are connected by a straight wire along the $x-$axis. Find the magnitude and direction of $\overrightarrow{B}$ at the origin.

The angle between radial line \& its tangent line at any point of the curve, $r=f\left(\theta \right)$ is related to the function in the following way

$\tan \beta =\frac{r}{\frac{dr}{d\theta }}$

Thus in this case $r=e^{\theta }, \tan \beta =1 \& \beta =\pi /4$

Therefore, the angle between $\overrightarrow{ds }\& \hat{r}$ is

$\pi -\beta =3\pi /4. \text{ Also}$

$ds=\frac{dr}{\sin \pi /4}=\sqrt{2}dr$

using Biosavart law :

$dB=\frac{{\mu }_{0}I\times ds\sin \beta }{4\pi r^2}=\frac{{\mu }_{0}Ids\times \sin \pi /4}{4\pi r^2};dB=\frac{{\mu }_{0}Idr}{4\pi r^2};B={\int }_{{\tau }_i}^{T_I}dB$

Where  $r_i=e^0=1$

after integration

$r_f=e^{2\pi }$

$B=\frac{{\mu }_{0}I}{4\pi }\left[1-{\text{e}}^{-2\pi }\right],\text{ out of page.}$