Question

Current flows through a straight cylindrical conductor of radius $r$. The current is distributed uniformly over its cross-section. The magnetic field at a distance $x$ from the axis of the conductor has magnitude $B$ :

• A. $B=0$ at the axis
• B. $B\propto x$ for $0\le x\le r$
• C. $B\propto \frac{1}{x}$ for $x>r$
• D. $B$ is maximum for $x=R$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $B=0$ at the axis
B $B\propto x$ for $0\le x\le r$
C $B\propto \frac{1}{x}$ for $x>r$
D $B$ is maximum for $x=R$

Inside the cylinder:
Consider the amperian loop shown as dotted circle-2 of radius $r$.
Applying Ampere's law
$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{i}_{enc}$
At every point on the circle, $\overrightarrow{B}\parallel \overrightarrow{dl}$
$\therefore B\left(2\pi r\right)={\mu }_{0}I\left(\frac{\pi r^2}{\pi R^2}\right)$ where $I$ is the current through the circular cross section of the cylinder of radius $R$. Since current is uniformly distributed on the surface, ratio of the areas is taken for $i_{enc}$.
$\therefore B_{inside}=\frac{{\mu }_{0}Ir}{2\pi R^2}$
Outside the cylinder:
Consider the loop-1.
Applying Ampere's law
$\oint \overrightarrow{B}.\overrightarrow{dl}={\mu }_0{i}_{enc}={\mu }_{0}I$
$B_{out}=\frac{{\mu }_{0}I}{2\pi r}$
On the surface of the cylinder:
we can use formula for inside point and put $r=R$
$B_{surface}=\frac{{\mu }_{0}I}{2\pi R}$
$B_{surface}$ is maximum since B increases until $r=R$ and then decreases.