We have three infinite straight conductors carrying current as shown above. Let \(I_n\) be the current on the \(n^\text{th}\) conductor. Find the force per unit length \(\frac{\vec{F_3}}{l}\) on conductor 3. If \(\frac{\vec{F_3}}{l}=a\sqrt{b}\times 10^{-c} \frac{\text{N}}{\text{m}} \angle \theta\), find \(a+b+c+\theta\).

**Details and assumptions**

- \(I_1=5\text{ A}\), \(I_2=10\text{ A}\), \(I_3=20\text{ A}\)
- \(a\), \(b\), \(c\) and \(\theta\) are positive integers, \(b\) is a square-free number, \(1\leq a \leq 9\) and \(0^\circ < \theta < 360^\circ\).
- \(\times\) (a cross) means that the current goes inside the screen, and \(\cdot\) (a dot) means that the current goes outside the screen.
- \(\vec{r}=r\angle\theta\) is a vector with magnitude \(r\) and angle \(\theta\) with respect to the positive x-axis.

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