Physics - Magnetism Question with Solution | TestHub

For a long straight solid wire of radius $a$ carrying a steady current $I$ uniformly distributed across its cross-section, we can find the magnetic field $B$ at a distance $r$ from the axis using Ampere's circuital law: $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{enclosed}$.

Case 1: Inside the wire ($r<a$)
The current enclosed by an Amperian loop of radius $r$ is $I_{enclosed}=I\left(\frac{\pi r^2}{\pi a^2}\right)=I\frac{r^2}{a^2}$.
Applying Ampere's law:
$B(2\pi r)={\mu }_0\left(I\frac{r^2}{a^2}\right)$
$B=\left(\frac{{\mu }_{0}I}{2\pi a^2}\right)r$
Thus, $B\propto r$. The graph is a straight line passing through the origin.

Case 2: Outside the wire ($r\ge a$)
The current enclosed by an Amperian loop of radius $r$ is the total current $I$.
Applying Ampere's law:
$B(2\pi r)={\mu }_{0}I$
$B=\frac{{\mu }_{0}I}{2\pi r}$
Thus, $B\propto \frac{1}{r}$. The graph is a rectangular hyperbola.

At the surface ($r=a$), the magnetic field is maximum: $B=\frac{{\mu }_{0}I}{2\pi a}$.

The plot that correctly represents this variation shows a linear increase from the origin up to $r=a$, followed by a $1/r$ decrease for $r>a$. This matches the first graph.

Answer: