JEE Main & Advanced Physics Magnetic Effects of Current / करंट का चुंबकीय प्रभाव Magnetic Field Due to a Cylindrical Wire

Magnetic field due to a cylindrical wire is obtained by the application of Ampere's law (1) Outside the cylinder

In all above cases magnetic field outside the wire at \[P\,\,\oint{\overrightarrow{B}.\overrightarrow{dl}={{\mu }_{0}}i}\] \[\Rightarrow \] \[B\int{dl}={{\mu }_{0}}i\]\[\Rightarrow \] \[B\times 2\pi r={{\mu }_{0}}i\]\[\Rightarrow \]\[{{B}_{out}}=\frac{{{\mu }_{0}}i}{2\pi r}\]

In all the above cases \[{{B}_{surface}}=\frac{{{\mu }_{0}}i}{2\pi R}\]

(2) Inside the hollow cylinder : Magnetic field inside the hollow cylinder is zero.

(3) Inside the solid cylinder : Current enclosed by loop \[(i')\] is lesser then the total current \[(i)\]

Current density is uniform i.e. \[l=l'\Rightarrow i'=i\times \frac{A'}{A}=i\,\left( \frac{{{r}^{2}}}{{{R}^{2}}} \right)\]

Hence at inside point \[\oint{\overrightarrow{{{B}_{in}}}.\,d\overrightarrow{l\,\,}={{\mu }_{0}}i'}\] \[\Rightarrow \]\[B=\frac{{{\mu }_{0}}}{2\pi }.\frac{ir}{{{R}^{2}}}\]

(4) Inside the thick portion of hollow cylinder : Current enclosed by loop is given as \[i'=i\times \frac{A'}{A}=i\times \frac{({{r}^{2}}-R_{1}^{2})}{(R_{2}^{2}-R_{1}^{2})}\]

Hence at point Q   \[\oint{\overrightarrow{B}.\,d\overrightarrow{l\,\,}={{\mu }_{0}}i'}\]\[\Rightarrow \]\[B=\frac{{{\mu }_{0}}i}{2\pi r}.\frac{({{r}^{2}}-R_{1}^{2})}{(R_{2}^{2}-R_{1}^{2})}\]