Question

two long thin parallel wires 13.0 cm apart carry 29 - a currents in the same direction.
figure
< 1 of 1 >
part b
determine the direction of the magnetic - field vector at that point
express your answer using two significant figures.

Explanation:

Step1: Recall magnetic - field formula

The magnetic field due to a long - straight wire is given by $B = \frac{\mu_0I}{2\pi r}$, where $\mu_0=4\pi\times10^{- 7}\ T\cdot m/A$, $I$ is the current, and $r$ is the distance from the wire.

Step2: Calculate magnetic field due to each wire

Let the two wires be wire 1 and wire 2. For wire 1 with current $I = 29\ A$ and distance $r_1=6.0\ cm = 0.06\ m$, $B_1=\frac{\mu_0I}{2\pi r_1}=\frac{4\pi\times10^{-7}\times29}{2\pi\times0.06}=\frac{2\times10^{-7}\times29}{0.06}\ T$.
For wire 2 with current $I = 29\ A$ and distance $r_2 = 10.0\ cm=0.10\ m$, $B_2=\frac{\mu_0I}{2\pi r_2}=\frac{4\pi\times10^{-7}\times29}{2\pi\times0.10}=\frac{2\times10^{-7}\times29}{0.10}\ T$.

Step3: Use vector addition

Since the currents are in the same direction, we use the Pythagorean theorem for the magnetic - field vectors (because the magnetic - field vectors due to the two wires are perpendicular at point $P$). The magnitude of the net magnetic field $B=\sqrt{B_1^{2}+B_2^{2}}$.
$B_1=\frac{2\times10^{-7}\times29}{0.06}\approx9.67\times10^{-5}\ T$ and $B_2=\frac{2\times10^{-7}\times29}{0.10}=5.8\times10^{-5}\ T$.
$B=\sqrt{(9.67\times10^{-5})^{2}+(5.8\times10^{-5})^{2}}=\sqrt{9.35\times10^{-9}+3.364\times10^{-9}}=\sqrt{1.2714\times10^{-8}}\approx1.1\times10^{-4}\ T$.

Answer:

$1.1\times10^{-4}$