A long, straight wire of radius 'a',  caries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wires at distance $\frac{\mathrm{a}}{3}$ and 2a. respectively from the axis of the wire is:

Given data: 

ratio of magnetic fields distance $\frac{\mathrm{a}}{3}\text{ and }2\mathrm{a}$

Concept used: Bio-savant law

Detailed solution:

${\mathrm{r}}_{\mathrm{p}}=\frac{\mathrm{d}}{3}$

${\mathrm{r}}_{\mathrm{Q}}=2\mathrm{a}$

At P, from Ampere's law

$\int {\mathrm{B}}_{\mathrm{p}}·\mathrm{dl}={\mu }_0{\mathrm{I}}_{\text{in }}$

${\mathrm{B}}_{\mathrm{p}}\overrightarrow{\mathrm{d}l}\cos \theta ={\mu }_0\sigma \pi (\mathrm{a}/3{)}^2$

${\mathrm{B}}_{\mathrm{p}}\times 2\times \pi \frac{\mathrm{a}}{3}=\mu .\sigma \pi {\left(\frac{\mathrm{a}}{3}\right)}^2$

${ \mathrm{B}}_{\mathrm{p}}=\frac{\mu .\sigma \mathrm{a}}{6}$

 $\text{ At }\mathrm{Q}$

$\int \mathrm{Bdl}=\mu .{\mathrm{I}}_{\text{in }}$

$\int {\mathrm{B}}_{\mathrm{Q}}\mathrm{dlcos}\theta =\mathrm{r}·\sigma \times \pi {\mathrm{a}}^2$

${ \mathrm{B}}_{\mathrm{Q}}2\pi \left(2\mathrm{a}\right)=\mu \sigma \pi {\mathrm{a}}^2$

${\mathrm{B}}_{\mathrm{Q}}=\frac{\mu \sigma \mathrm{a}}{4}$

$\text{ Ratio }=\frac{{\mathrm{B}}_P}{{ \mathrm{B}}_{\mathrm{Q}}}=\frac{\mu \sigma \mathrm{a}/6}{\mu \sigma \mathrm{a}/4}=\frac{4}{6}$

$\frac{{\mathrm{B}}_{\mathrm{P}}}{{\mathrm{B}}_{\mathrm{Q}}}=\frac{2}{3}$