Question

A cylindrical wire of radius R carries a current directed out of the page as shown below. The current has a constant cross-sectional area density J. According to Ampere's Law, the current creates magnetic field both within the wire and outside the wire.
What is the magnetic field at point Q, which is at a position a distance r < R from the axis of the wire?

• A. $\frac{{\mu }_{0}Jr}{3}$
• B. $\frac{{\mu }_{0}J{r}^2}{2}$
• C. $\frac{{\mu }_{0}Jr}{2}$
• D. ${\mu }_{0}J$
• E. $\frac{{\mu }_{0}J{r}^2}{6}$

Answer 1

The correct option is C $\frac{{\mu }_{0}Jr}{2}$

Cross-sectional area of the loop of radius $r$, $A=\pi r^2$

$\therefore$ Current enclosed by the loop of radius $r<R$, $I_{enc}=JA=J\left(\pi r^2\right)=\pi J{r}^2$

Using Ampere circuital law : $\oint B.dl={\mu }_o{I}_{enc}$

$\therefore$ $B\times 2\pi r={\mu }_o\times \pi J{r}^2$ $⟹B=\frac{{\mu }_{o}Jr}{2}$ for $r<R$