I've made benzyl bromide this way (actually, I just mix benzyl alcohol with straight 48% HBr. no chance for elimination, here, but I thought I'd throw in my 2 cents.
That is a great procedure. It also works for benzyl chloride as well. Unfortunately, both of those are extremely irritating lachrymating agents.
In this case, it's a simple protonation of the alcohol by the sulfuric and then the bromide comes along.
There is also possibility of rearrangement and having the Br on the second carbon.
The sulfuric acid is not in concentration sufficient to cause dehydration to an alkene. These reactions work well and in high yields for alkyl chlorides and bromides. (ref. Vogel) It's a pity there is not an easy way to make iodobutane or iodoethane--the sulfuric acid will simple oxidize your halide salt to iodine and nothing will happen...
Also, the base, sodium carbonate removes any formed bromine in addition to neutralizing any acid traces.