Given,
Current carried by wire, I = 0.40 A,
Radius of circular wire, r = 8.0 cm = $8\times {10}^{-2}\mathrm{m}$
Number of turns in the coil, $\mathrm{n} = 100$

Using the formula, we get the magnetic field at the centre of coil as

$\boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{nI}}{2\mathrm{r}}} = \frac{4\mathrm{\pi }\times {10}^{-7}\times 100\times 0.4}{2\times 8.0\times {10}^{-2}}\mathrm{T}$
               $= 3.1 \times {10}^{-4}\mathrm{T}.$
                        

Given,
Current carried by conductor, I = 35 A
Distance, r = 20 cm = 0.2 m

Using Ampere's circuital law we get, 

 $$\begin{gathered} \boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}} \\ = \frac{4\mathrm{\pi }\times {10}^{-7}\times 35}{2\mathrm{\pi }\times 0.20}\mathrm{T} \\ = 3.5 \times {10}^{-5}\mathrm{T}. \end{gathered}$$

Given,
Current carried by wire, I= 50 A
Distance from the wire, r = 2.5 m
Magnetic field at point r,  B = ?

Using ampere's circuital law
                       $$\begin{gathered} \boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}} \\ = \frac{4\mathrm{\pi }\times {10}^{-7}\times 50}{2\mathrm{\pi }\times 2.5} \\ = 4 \times {10}^{-6}\mathrm{T} \end{gathered}$$ 

The direction of magnetic field can be found out by applying right hand thumb rule. As the direction of current is from north to south represented by thumb, the direction iof magnetic field is vertically upwards in east direction of wire.

Given, current, I= 90 A
distance ,r = 1.5 m
Thus,
the magnitude of magnetic field is given by Ampere's circuital law 

               $$\begin{gathered} \boxed{\mathrm{B} = \frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}} \\ = \frac{4\mathrm{\pi }\times {10}^{-7}\times 90}{2\mathrm{\pi }\times 1.5} \\ = 1.2 \times {10}^{-5} T \end{gathered}$$  

Applying the right-hand thumb rule, we find that the magnetic field at the observation point is directed towards south.