Haskell Concurrent.Channel: What is difference between this two codes?

Why this code

import Control.Monad
import Control.Concurrent
import Control.Concurrent.STM
import Control.Concurrent.STM.TChan

main = do
  ch <- newTChanIO
  forkIO $ consumer ch 0
  forkIO $ consumer ch 1
  ----------------------
  forkIO $ producer ch
  ----------------------
  return ()


producer ch = loop 0
  where
    loop n = do
      atomically $ writeTChan ch n
      threadDelay (10^5 :: Int)
      loop (n+1)


consumer ch n = forever $ do
  v <- atomically $ readTChan ch
  print (n, v)

doesn't behave the same way as

import Control.Monad
import Control.Concurrent
import Control.Concurrent.STM
import Control.Concurrent.STM.TChan

main = do
  ch <- newTChanIO
  forkIO $ consumer ch 0
  forkIO $ consumer ch 1
  ----------------------
  producer ch
  ----------------------
  return ()


producer ch = loop 0
  where
    loop n = do
      atomically $ writeTChan ch n
      threadDelay (10^5 :: Int)
      loop (n+1)


consumer ch n = forever $ do
  v <- atomically $ readTChan ch
  print (n, v)

The latter code's output is

(0,0)
(1,1)
(1,2)
(1,3)
(1,4)
(0,5)
(0,6)
(1,7)
...

But, the former one's output is only

(0,0)

I intended that producer indefinitely add a value to the channel queue ch, and consumer indefinitely take a value from ch.

The latter code acts as well as my intension, but the former one doesn't.

In the eventlog (ghc-events), block happens on an MVar in producer thread

4775372: cap 0: stopping thread 8 (blocked on an MVar)

Why the former code (forkIO $ producer ch) doesn't indefinitely run.