Question

A long straight wire of a radius $a$ carries a steady current $i$. The current is uniformly distributed across its cross-section. Then,

• A. The magnetic field at radial distance $2a$ is $\frac{{\mu }_{0}I}{4\pi a}$.
• B. The magnetic field at radial distance $2a$ is $\frac{{\mu }_{0}I}{\pi a}$.
• C. The magnetic field at radial distance $\frac{a}{2}$ is $\frac{{\mu }_{0}I}{4\pi a}$.
• D. The magnetic field at radial distance $\frac{a}{2}$ is $\frac{{\mu }_{0}I}{\pi a}$.

Answer 1

Answer: (A), (C)

The correct options are
A The magnetic field at radial distance $2a$ is $\frac{{\mu }_{0}I}{4\pi a}$.
C The magnetic field at radial distance $\frac{a}{2}$ is $\frac{{\mu }_{0}I}{4\pi a}$.



$\text{Field at any outside point :}$ Considering Amperian loop of radius $r=2a$ and using Ampere's law, we get

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_0{I}_{net}$

$\overrightarrow{B}\oint \overrightarrow{dl}={\mu }_0{I}_{net}$

$\Rightarrow B\left(2\pi r\right)={\mu }_0{I}_{net}[\because r=2a;I_{net}=I]$

$\Rightarrow B_{out}=\frac{{\mu }_{0}I}{\left(4\pi a\right)}$

$\text{Field at any inside point :}$

For inside net current enclosed is,

$I_{net}=JdA=\frac{I}{\pi (a{)}^2}\times \pi {\left(\frac{a}{2}\right)}^2=\frac{I}{4}$

Now, using Ampere's law,
$\overrightarrow{B}\oint \overrightarrow{dl}={\mu }_0{I}_{net}$

$\Rightarrow B\left(2\pi r\right)={\mu }_0{I}_{net}[\because r=\frac{a}{2};I_{net}=\frac{I}{4}]$

$\Rightarrow B_{in}=\frac{{\mu }_{0}I}{\left(4\pi a\right)}$

Hence, $\left(A\right)$ and $\left(C\right)$ are the correct answers.

Why this question ? To understand the concept of Ampere's circuital law and its application.