question_answer

A long straight wire of a circular cross-section of radius a, carries a steady current I. The current is uniformly distributed across the cross-section of the wire. Use Ampere's circuital law to show that the magnetic field, due to this wire, in the region inside the wire, increases in direct proportional to the distance of the field point from the axis of the wire.

Write the value of this magnetic field on the surface of the wire.

Answer:

The current is distributed uniformly across the cross-section of radius a. \[\therefore \] Current passes per unit cross-section \[=\frac{\text{I}}{\pi {{\text{a}}^{\text{2}}}}\] \[\therefore \] Current passes through the cross-section of radius r is \[\text{I }\!\!'\!\!\text{ =}\left( \frac{\text{I}}{\pi {{\text{a}}^{2}}}\times \text{ }\!\!\pi\!\!\text{ }{{\text{r}}^{2}} \right)=\frac{\text{I}{{\text{r}}^{2}}}{{{\text{a}}^{2}}}\]      ? (i) (i) Consider a loop of radius r whose centre lies at the axis of wire where, \[r<a\]as shown in figure inside the wire. Applying Ampere?s circuital law,                         \[\oint{B\cdot dI={{\mu }_{0}}I'}\] \[\oint{Bdl\cos 0{}^\circ ={{\mu }_{0}}\left( \frac{I{{r}^{2}}}{{{a}^{2}}} \right)}\]                [From eqn. (i)]                         \[B\,\,\oint{\,dl={{\mu }_{0}}\frac{I{{r}^{2}}}{{{a}^{2}}}}\] \[B\times 2\pi r=\frac{{{\mu }_{0}}I{{r}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \]   \[B=\frac{{{\mu }_{0}}Ir}{2\pi {{a}^{2}}}\] \[\Rightarrow \]   \[B\propto r\] Now, the value of magnetic field on the surface of wire, i.e. \[(\pi =a),\]                         \[B=\frac{{{\mu }_{0}}I}{2\pi {{a}^{2}}}\times a=\frac{{{\mu }_{0}}I}{2\pi a}\]