the process cannot access the file because it is being used by another process.

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C#

Hi,
I am Develop one application read csv files dynamically from given directory.after read that file move another folder.i am write code to read csv and move to another folder.csv file read perfect after file does not move to another folder given exception : the process cannot access the file because it is being used by another process. the process cannot access the file because it is being used by another process. . i am attach my code below Please solve this problem.
thank you
regards,
raju

code :

try
           {
               DialogResult result = folderBrowserDialog1.ShowDialog();

               if (result == DialogResult.OK)
               {
                   string abc = folderBrowserDialog1.SelectedPath;
                   DirectoryInfo selectedPath = new DirectoryInfo(abc);
                   textBox1.Text = abc;
                   //string[] fileList = System.IO.Directory.GetFiles(abc, "*.csv");
                   // using (System.IO.StreamReader sr = new System.IO.StreamReader("c:/Raj/abcd.txt"))
                   {
                       for (int i = 0; i != null; i++)
                       {
                           string[] fileList = System.IO.Directory.GetFiles(abc, "*.csv");

                           string abcd = fileList[i];

                           StreamReader sr = new System.IO.StreamReader(abcd);

                           string currentLine;
                           while ((currentLine = sr.ReadLine()) != null)
                           {
                               MessageBox.Show(currentLine);
                               textBox2.Text = currentLine;

                               string destinationPath = @"c:\Raj\log";
                               string fileToMove = abcd;
                               string moveTo = destinationPath;
                               File.Move(fileToMove, moveTo);
                               MessageBox.Show("file moved");

                           }

                       }
                   }
               }
           }
           catch (Exception e)
           {
               MessageBox.Show(e.Message);

           }