The field of a straight wire

Three long wires carry $6\ \text{A}$, $15\ \text{A}$ and $25\ \text{A}$. You will make the program find the field of each at $3\ \text{cm}$, the distance at which the $15\ \text{A}$ wire's field equals the Earth's $50\ \mu\text{T}$, and how many times stronger one wire's field is at $2\ \text{cm}$ than at $8\ \text{cm}$. Then it draws the field against the distance, with the Earth's field as a dotted line.

The idea. $B = \dfrac{\mu_0 I}{2\pi r} = 2 \times 10^{-7}\,\dfrac{I}{r}$, in tesla, with $r$ in metres. $1\ \text{T} = 10^{6}\ \mu\text{T}$.

Your tasks.

1. Inside the loop, work out B3, the field in microtesla $3\ \text{cm}$ from the wire with current I.
2. Find r_earth, the distance in centimetres at which the $15\ \text{A}$ wire's field is $50\ \mu\text{T}$.
3. Find ratio, the field at $2\ \text{cm}$ divided by the field at $8\ \text{cm}$, for the same wire.