Trace of a function from $W^{1,p}(\mathbb{R}^N)$ is zero?

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Assume that $u \in W^{1,p}(\mathbb{R}^N)$, where $p>1$ and $N \geq 2$. We know only that $u \equiv 0$ in an open ball $B \subset \mathbb{R}^N$. Is it true that the trace of $u$ equals zero on the boundary of $B$? In other words, is it true that $$ Tu = 0 \quad \text{on} \quad \partial B \,? $$

Intuitively, the answer seems to be affirmative, but I'm afraid that $u$ could have some "jumps" on $\partial B$ which prevent $Tu=0$.

As an approach, we definitely know that $Tu = 0$ on each sphere $\partial B_1$ compactly contained inside $B$. Then we can expand $B_1$ and look what happens with $Tu$. So, my initial question can be reformulated as follows: Is the trace operator continuous with respect to a smooth change of (smooth) boundary of a subdomain of $\mathbb{R}^N$?