Question

A long straight wire of radius $a$ carries a steady current $I$. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at distance $\frac{a}{2}$ and $2a$ from axis of wire is

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

Answer 1

The correct option is D $1$

Here, current is uniformly distributed across the cross-section of the wire, thus current per unit cross-sectional area is

$\frac{i}{A}=\frac{I}{\pi a^2}$

Case 1: $B$ at a distance $r=\frac{a}{2}$

The current enclosed by loop $1$ is,


$\Rightarrow i_1=\frac{i}{A}\times \pi {\left(\frac{a}{2}\right)}^2$

or, $i_1=\frac{I}{\pi a^2}\times \frac{\pi a^2}{4}$

or, $i_1=\frac{I}{4}$

Using Ampere circuital law on loop $1$ we have,

$\oint \overrightarrow{B_1}.\overrightarrow{dl}={\mu }_0{I}_{en}$

$B_1\oint dl={\mu }_0\frac{I}{4}(\because \theta =0,I_{enc}=\frac{I}{4})$

$\Rightarrow B_1\left(2\pi \frac{a}{2}\right)={\mu }_0\left(\frac{I}{4}\right)$

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

Case 2: $B$ at a distance $r=\frac{a}{2}$

Similarly, for loop $2,r=2a$

$\oint \overrightarrow{B_2}.\overrightarrow{dl}={\mu }_0{I}_{en}$

$B_2(2\pi \times 2a)={\mu }_{0}I(\because I_{en}=I)$

$B_2=\frac{{\mu }_{0}I}{4\pi a}$

$\therefore \frac{B_1}{B_2}=1$

Why this Question ? Key point- Magnetic field calculation with the help of Ampere circuital law is easily possible only under highly symmetric conditions. Ampere circuital law is given by, $\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{I}_{en}$