Parallel conductors

Let $\vec{B_{a,b}}$ the magnetic field that conductor $a$ produces on point $b$ and let $r_{a,b}$ the distance from conductor $a$ to point $b$ . We have:

$B_{1,3}=\dfrac{\mu_0 I_1}{2 \pi r_{1,3}}=\dfrac{\left(4\pi \times 10^{-7}\frac{\text{N}}{\text{A}^2}\right)(5\text{ A})}{2\pi(0.1\text{ m})}=1\times 10^{-5}\text{ T}$

$B_{2,3}=\dfrac{\mu_0 I_2}{2 \pi r_{2,3}}=\dfrac{\left(4\pi \times 10^{-7}\frac{\text{N}}{\text{A}^2}\right)(10\text{ A})}{2\pi(0.1\text{ m})}=2\times 10^{-5}\text{ T}$

To determine the directions we use the right hand rule as shown above. By simple geometry we get:

$\hat{B_{1,3}}=\cos 30^\circ \hat{i}+\sin 30^\circ \hat{j}=\dfrac{\sqrt{3}}{2}\hat{i}+\dfrac{1}{2}\hat{j}$

$\hat{B_{2,3}}=\cos 270^\circ \hat{i}+\sin 270^\circ \hat{j}=-\hat{j}$

Then we find the total magnetic field on conductor 3:

$\vec{B_3}=\vec{B_{1,3}}+\vec{B_{2,3}}=1\times 10^{-5}\text{ T}\left(\dfrac{\sqrt{3}}{2}\hat{i}+\dfrac{1}{2}\hat{j}\right)+2\times 10^{-5}\text{ T}(-\hat{j})=\dfrac{\sqrt{3}}{2}\times 10^{-5}\text{ T}\hat{i}-\dfrac{3}{2}\times 10^{-5}\text{ T}\hat{j}$

Finally, we find the force per unit length on conductor 3, where $\hat{l}$ is the direction of the current:

$\dfrac{\vec{F_3}}{l}=I_3\hat{l}\times\vec{B_3}=(20\text{ A})(-\hat{k})\times\left(\dfrac{\sqrt{3}}{2}\times 10^{-5}\text{ T}\hat{i}-\dfrac{3}{2}\times 10^{-5}\text{ T}\hat{j}\right)=-3\times 10^{-4}\frac{\text{N}}{\text{m}}\hat{i}-\sqrt{3}\times 10^{-4}\frac{\text{N}}{\text{m}}\hat{j}$

We convert to polar form to get the answer:

$\dfrac{\vec{F_3}}{l}=2\sqrt{3}\times 10^{-4}\frac{\text{N}}{\text{m}}\angle 210°$

Hence $a=2,b=3,c=4,\theta=210°$ and $a+b+c+\theta=\boxed{219}$ .