Question

two long parallel wires separated by 1.20 cm experience a force per unit length of 1.49 n/m. they are carrying the same current. what is it? o 211 a o 598 a o 150 a o 299 a

Explanation:

Step1: Recall force - per - unit - length formula

The force per unit length between two parallel current - carrying wires is given by $F/L=\frac{\mu_0I_1I_2}{2\pi r}$. Since $I_1 = I_2=I$, the formula becomes $F/L=\frac{\mu_0I^{2}}{2\pi r}$.
We know that $\mu_0 = 4\pi\times10^{- 7}\ T\cdot m/A$, $r = 1.20\ cm=0.012\ m$ and $F/L = 1.49\ N/m$.

Step2: Rearrange the formula to solve for $I$

Starting from $F/L=\frac{\mu_0I^{2}}{2\pi r}$, we can solve for $I$. First, cross - multiply to get $(F/L)\times2\pi r=\mu_0I^{2}$. Then $I^{2}=\frac{2\pi r(F/L)}{\mu_0}$.
Substitute the values: $\mu_0 = 4\pi\times10^{-7}\ T\cdot m/A$, $r = 0.012\ m$ and $F/L = 1.49\ N/m$.
$I^{2}=\frac{2\pi\times0.012\ m\times1.49\ N/m}{4\pi\times10^{-7}\ T\cdot m/A}$.
The $\pi$ terms cancel out, and we have $I^{2}=\frac{2\times0.012\times1.49}{4\times10^{-7}}$.
$I^{2}=\frac{0.024\times1.49}{4\times10^{-7}}=\frac{0.03576}{4\times10^{-7}} = 89400$.

Step3: Calculate the value of $I$

Take the square - root of both sides: $I=\sqrt{89400}\approx299\ A$.

Answer:

299 A