Question

a semi - infinite straight wire is carrying a current of 14.0 a. the wire runs along the x - axis from the origin to x = +∞. at what point on the y - axis is the magnitude of the magnetic field 4.20×10^(-6) t?

Explanation:

Step1: Recall the formula for magnetic - field due to a semi - infinite wire

The magnetic - field due to a semi - infinite wire at a perpendicular distance $r$ from the wire is given by $B=\frac{\mu_0I}{4\pi r}$, where $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$ is the permeability of free space, $I$ is the current in the wire, and $r$ is the perpendicular distance from the wire.

Step2: Rearrange the formula to solve for $r$

We know that $B=\frac{\mu_0I}{4\pi r}$, so we can solve for $r$ as $r=\frac{\mu_0I}{4\pi B}$.
Given $I = 14.0\ A$ and $B=4.20\times10^{-6}\ T$.
Substitute $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$ into the formula:
$r=\frac{4\pi\times10^{-7}\ T\cdot m/A\times14.0\ A}{4\pi\times4.20\times10^{-6}\ T}$

Step3: Calculate the value of $r$

The $4\pi$ terms cancel out. Then $r=\frac{10^{-7}\times14.0}{4.20\times10^{-6}}$.
$r=\frac{14.0\times10^{-7}}{4.20\times10^{-6}}=\frac{14.0}{4.20}\times10^{-7 + 6}=\frac{14.0}{4.20}\times10^{-1}$.
$r=\frac{140}{42}\times10^{-1}=\frac{10}{3}\times10^{-1}=\frac{1}{3}\ m\approx0.333\ m$.

Answer:

$0.333\ m$