Turing machine to compute $⌈\log_2n⌉$ with 1 tape and unary input/output
One simple approach is repeated halving.
Remove a single 1 from the input (if you fail, output N).
Then go over the tape in passes. Each pass deletes every other 1 encountered (starting with the first). The number of passes required to delete all 1s is $\lceil \log_2 n \rceil$. You can count the number of passes as you go.
For example, suppose that $n=6$:
111111       Delete the first 1
b11111       Delete every other 1
2b1b1b       Delete every other 1
22bb1b       Delete every other 1
222bbb       Convert back to 1
111bbb       Done

The number of passes is $3 = \lceil \log_2 n \rceil$.