Question

A wire carrying current $I$ is laid in shape of a curve which is represented in plane polar coordinate system as,
$r=b+\frac{c}{\pi }\theta for0\le \theta \le \frac{\pi }{2}$
Here $b$ and $c$ are positive constants. $\theta$ is the angle measured with respect to positive $X$ direction in anticlockwise sense and $r$ is the distance from origin (see figure). The magnetic field at the origin due to the wire is found to be $\frac{{\mu }_{o}I}{xc}ln(1+\frac{c}{yb})$
The value of xy is

Answer 1

Unit vectors along $\overrightarrow{r}$ and perpendicular to $\overrightarrow{r}$ are $\hat{r}$ and $\hat{\theta }$ respectively. A small element of length along the curve can be represented as,
$\overrightarrow{dl}=\left(dr\right)\hat{r}+\left(rd\theta \right)\hat{\theta }$
Field due to such an element at O is,
$\overrightarrow{dB}=\frac{{\mu }_0}{4\pi }\frac{I[\overrightarrow{dl}\times (-\hat{r}\left)\right]}{r^2}$
$\overrightarrow{dB}=\frac{{\mu }_{0}I}{4\pi r^2}[(\left(dr\right)\hat{r}+(rd\theta \left)\hat{\theta }\right)\times (-\hat{r}\left)\right]$
$\overrightarrow{dB}=\frac{{\mu }_{0}I}{4\pi r}d\theta \hat{k}$ $\odot$

$\therefore$ Magnetic field B
$B=\frac{{\mu }_{0}I}{4\pi }{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{b+\frac{c\theta }{\pi }}$ $\odot$
$B=\frac{{\mu }_{0}I}{4\pi }\frac{\pi }{c}{\left[ln(b+\frac{c\theta }{\pi })\right]}_0^{\frac{\pi }{2}}=\frac{{\mu }_{0}I}{4c}[ln(b+\frac{c}{2})-ln(b\left)\right]$

$B=\frac{{\mu }_{0}I}{4c}\left[ln(1+\frac{c}{2b})\right]$ $\odot$