Applications of Ampere law:-
- Infinite straight current-carrying wire:-
\(\begin{array}{l}\oint{\overrightarrow{B}\centerdot \overrightarrow{\delta}}={{mu }_{o}}I\end{array} \)
\(\begin{array}{l}B\oint{\overrightarrow{\delta }={{mu }_{o}}I},\Rightarrow B( 2\pi r)={{mu }_{o}}I\end{array} \)
\(\begin{array}{l}B=\frac{{{mu }_{o}}I}{2\pi r}\end{array} \)
\(\begin{array}{l}\Rightarrow \overrightarrow{B}=\frac{{{mu }_{o}}I}{2\pi r}\hat{\phi }\end{array} \)
(In cylindrical co-ordinates)
In this, the current is longitudinal (source is longitudinal) but the field produced is circumferential. The magnetic field is tangential to every point of the observed path.
- M.F of cylindrical current-carrying wire:-
A cylindrical wire having radius ‘R’, carrying a uniform current density
\(\begin{array}{l}\overrightarrow{J}\end{array} \)
. Once again the current here is longitudinal.
The magnetic field will be circumferential. We can choose a circular path as the integration path to calculate the magnetic field anywhere.
Inside:-
\(\begin{array}{l}\oint{\overrightarrow{B}}\cdot \overrightarrow{\delta}={{mu }_{o}}{{I}_{net}}\end{array} \)
\(\begin{array}{l}B\times 2\cancel{\pi }\cancel{r}=mu_{0}(J\cancel{pi }{{r}^{\cancel{2}}})\end{array} \)
\(\begin{array}{l}B=\frac{{{mu }_{o}}Jr}{2}\overrightarrow{B} (R)=\frac{{{mu }_{o}}(\overrightarrow{J}\times \overrightarrow{r})}{2}\end{array} \)
Outside:-
\(\begin{array}{l}B=2\cancel{\pi }r={{mu }_{o}}J(\cancel{pi }{{R}^{2}})\end{array} \)
\(\begin{array}{l}B=\frac{{{mu }_{o}}J{{R}^{2}}}{2r}\overrightarrow{B}( R)=\frac{{{mu }_{o}}{{R}^{2}}}{2r}(\overrightarrow{J}\times \hat{r})\end{array} \)
For outside points, the M.F is similar to the field attained in the case of an st. current-carrying wire. That means, for outside points, it is behaving as if the entire current is passing through the axis.
