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 $a/2$ and $2a$ is:

• A. $1/2$
• B. $1/4$
• C. $4$
• D. $1$

Answer 1

The correct option is D $1$

Here, current is uniformly distributed across the cross-section of the wire, therefore, current enclosed in the amperean path formed at a distance $r_1=\left(\frac{a}{2}\right)$
$I_{in}=\left(\frac{\pi r_1^2}{\pi a^2}\right)\times I$, where $I$ is total current.

$\therefore$ Magnetic field at
$\left(B_1\right)=\frac{{\mu }_0\times currentinclosed}{path}$

$=\frac{{\mu }_0\times \left(\frac{\pi r_1^2}{\pi a^2}\right)\times I}{2\pi r_1}=\frac{{\mu }_0\times I{r}_1}{2\pi a^2}$

Now, magnetic filed at point
$\left(B_2\right)=\frac{{\mu }_0}{2\pi }\cdot \frac{I}{2a}=\frac{{\mu }_{0}I}{4\pi a}$

$\therefore$ Required Ratio $=\frac{B_1}{B_2}=\frac{{\mu }_{0}I{r}_1}{2\pi a^2}\times \frac{4\pi a}{{\mu }_{0}I}$
$=\frac{2r_1}{a}=\frac{2\times \frac{a}{2}}{a}=1$