Question

A long, straight wire of diameter 5.0 mm carries a uniformly distributed current of
15 A. At what distance from the axis of the wire will the magnitude of B be maximum? Justify your answer.

Answer 1

$R=5.0 mm$

$I=15A$

According to Ampere's circuital law

$\int{Bdl=\mu_0I}$

If $r<R$

$B\cdot 2\pi r=\mu_0\cdot \frac{I}{4\pi R^2}\cdot 4\pi r^2\to B=\frac{\mu_0I}{2\pi R^2}\cdot r$

If $r>R$

$B\cdot 2\pi r=\mu_0\cdot I\to B=\frac{\mu_0I}{2\pi r}$

So, if $r=R=5.0 mm$ then $B=B_{max}$

$B_{max}=\frac{\mu_0I}{2\pi R}=\frac{4\cdot \pi\cdot 10^{-7}\cdot15}{2\cdot\pi \cdot5\cdot 10^{-3}}=6\cdot 10^{-4}T$