Need to open file through VB.net on another directory - vb.net

Question

Through my form, I open a .exe which is located in a folder inside debug so:
\bin\Debug\folder\ .exe
The .exe opens and creates 3 different files then closes. It is meant to create them in the same folder as the .exe but instead creates it in the Debug folder when opened through VB.net using Process.Start().
Would anyone have a possible fix for this so I don't have to move the files?
EDIT
(the .exe itself creates the files, some things may be declared outside of this sub)
Private Sub btnRunExe_Click(sender As Object, e As EventArgs) Handles btnRunExe.Click
If AcptEULA.Checked = True Then
Localpath = Application.StartupPath() + "\MCserver" + "\minecraft_server." + txtVersion.Text + ".exe"
Downloadpath = "https://s3.amazonaws.com/Minecraft.Download/versions/" + txtVersion.Text + "/minecraft_server." + txtVersion.Text + ".exe"
LocalpathParent = Application.StartupPath() + "\MCserver"
Try
Dim dirs As String() = Directory.GetFiles(LocalpathParent, "minecraft_server*.exe")
Dim dir As String
For Each dir In dirs
Process.Start(dir)
Next
Catch
'Console.WriteLine("The process failed: {0}", e.ToString())
End Try
ElseIf AcptEULA.Checked = False Then
MsgBox("You must accept the Minecraft EULA before continuing")
End If
End Sub

Answer 1

Process.Start could be used with a ProcessStartInfo instance in which you could set the WorkingDirectory property.
Dim psi As ProcessStartInfo = New ProcessStartInfo()
psi.WorkingDirectory = LocalpathParent
For Each fileName In Directory.EnumerateFiles(LocalpathParent, "minecraft_server*.exe")
psi.FileName = fileName
Process.Start(psi)
Next
Notice that I have changed the Directory.GetFiles with Directory.EnumerateFiles that allows you to process a file while you loop over the folder files without loading all the filenames in memory inside an array. If you still want to use GetFiles then it is
Dim psi As ProcessStartInfo = New ProcessStartInfo()
psi.WorkingDirectory = LocalpathParent
Dim files as String() = Directory.GetFiles(LocalpathParent, "minecraft_server*.exe")
For Each fileName In files
psi.FileName = fileName
Process.Start(psi)
Next
If you look at the ProcessStartInfo documentation on MSDN you could find a lot of other useful properties to fine tune how your program executes.