Question

two long parallel wires separated by 5.70 cm experience a force per unit length measured in n/m. they are carrying the same current, 270 a. what is the force per unit length experienced by the wires? o 0.256 n/m o 0.128 n/m o 0.512 n/m o 0.181 n/m

Explanation:

Step1: Recall the formula for force per unit length

The formula for the force per - unit length between two parallel current - carrying wires is $\frac{F}{L}=\frac{\mu_0I_1I_2}{2\pi r}$. Since $I_1 = I_2=I = 270\ A$ and $\mu_0 = 4\pi\times10^{- 7}\ T\cdot m/A$, and $r=5.70\ cm=0.057\ m$.

Step2: Substitute the values into the formula

$\frac{F}{L}=\frac{\mu_0I^2}{2\pi r}=\frac{4\pi\times10^{-7}\times(270)^2}{2\pi\times0.057}$.
First, simplify the expression:
$\frac{F}{L}=\frac{2\times10^{-7}\times(270)^2}{0.057}$.
$(270)^2 = 72900$, then $2\times10^{-7}\times72900=\ 0.01458$.
$\frac{F}{L}=\frac{0.01458}{0.057}=0.256\ N/m$.

Answer:

0.256 N/m