Question

A long straight wire of radius a carries a steady current $I.$ The current is uniformly distributed over its cross-section. The ratio o the magnetic fields $B$ and $B^{\prime },$ at radial distances $\frac{a}{2}$ and $2a$ respectively, from the axis of the wire is

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $1$
• D. $4$

Answer 1

Answer: (C)

$1$

Consider two amperian loops of radius $a/2$ and $2a$ as shown in the diagram$.$

Applying ampere's circuital law for these loops$,$ we get

$\oint B.dL={\mu }_0{I}_{enclosed}$

For the smaller loop$,$

$\Rightarrow B\times 2\pi \frac{a}{2}={\mu }_0\times \frac{1}{\pi a^2}\times {\left(\frac{a}{2}\right)}^2$

$={\mu }_{0}I\times \frac{1}{4}=\frac{{\mu }_{0}I}{4}$

$\Rightarrow B_1=\frac{{\mu }_{0}I}{4\pi a}$

$B^{\prime }\times 2\pi \left(2a\right)={\mu }_{0}I$

$\frac{B}{B^{\prime }}=\frac{{\mu }_{0}I}{4\pi a}\times \frac{4\pi a}{{\mu }_{0}I}=1$

Hence,

option $\left(C\right)$ is correct answer.