Answer
a) $ \sum B= 2\times 10^{-7} \left[\dfrac{1}{ x}- \dfrac{1}{ 0.09-x }\right] $
b)
Work Step by Step
a)
The net magnetic field at point $x$ between the two wires is given by
$$\sum B=\sum B_y=B_{\rm by1}-B_{\rm by 2}$$
See the figure below to figure out the coming steps.
$$\sum B= \dfrac{\mu_0I}{2\pi x}- \dfrac{\mu_0I}{2\pi (d-x)}=\dfrac{\mu_0I}{2\pi }\left[\dfrac{1}{ x}- \dfrac{1}{ d-x }\right]$$
Plugging the known;
$$\sum B= \dfrac{4\pi\cdot 10^{-7}\cdot 1}{2\pi }\left[\dfrac{1}{ x}- \dfrac{1}{ 0.09-x }\right]$$
$$\boxed{\sum B= 2\times 10^{-7} \left[\dfrac{1}{ x}- \dfrac{1}{ 0.09-x }\right]}$$
b) See the figure below.
