Question

Write Ampere's circuital law. A long straight wire of a circular cross-section (radius $a$) carrying steady current. Current is uniformly distributed in the wire. Calculate magnetic field inside the region $(r<a)$ in the wire.

Answer 1

According to Ampere's circuital law, integral of magnetic field along a closed curve equals the sum of all electric currents passing through the cross section of the closed curve times permeability.

$\oint \overrightarrow{I}.\overrightarrow{dl}={\mu }_o{I}_{enclosed}$

To find magnetic field inside a long straight wire of cross-sectional area $a$, refer the attached figure. Magnetic field lines in this configuration forms concentric circles centered at $O$, the axis of the cylindrical wire. Hence, it is the preferred path of the Amperian Loop.

Consider an Amperian Loop of radius $a<r$ as shown in the figure.

Current enclosed in a cross-section area $\pi a^2$ is $I$

Then, Current enclosed in a cross-section area $\pi r^2$ is $I_{enclosed}=\frac{r^2}{a^2}I$

Taking the magnetic field integral along the loop, we get

$\oint \overrightarrow{B}\cdot \overrightarrow{dl}=B\times 2\pi r$

Using Ampere's Law, we get

$B\times 2\pi r=\frac{{\mu }_{o}I{r}^2}{a^2}$

$B=\frac{{\mu }_{o}rI}{2\pi a^2}$

Direction of magnetic field can be found using Fleming's Right hand thumb rule.