Does a turing machine halt with input left on tape?

is the string aabb in the language of this Turing machine ?? as upon reading aa it goes to the final state but upon b there is no transition defined so won't it result in a dead configuration?

• Define transition function properly. $\delta(q_2,a) = ?$ Feb 3 '18 at 9:13
• @Complexity the transition function is not given for q2 on a there is no move defined upon q2 for any input Feb 3 '18 at 9:17
• in PDA with final state if final state is reached there can be inputs on the stack but in TM i dont think so but in wikipedia - " If δ is not defined on the current state and the current tape symbol, then the machine halts; is given in wiki so once on reaching the final state doesnt it consider the rest of the input ? Feb 3 '18 at 9:30

When a Turing machine enters an accepting state, it immediately stops running and accepts whatever the original input string was (regardless of the state of the tape). So given the input aabb, the Turing machine will read the two as and enter the accepting state $q_2$, at which point it will halt and accept aabb.
In this case it doesn't matter that there's no transition for reading a b from $q_2$ because we've already accepted (though you are correct in general that whenever we're running a Turing machine and we arrive at a state with no transition defined for the character we're reading, we implicitly go to a dead state and reject).