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This does not seem like a realistic hope. When you say a lot of normal subgroups we can be more precise. A group is conjugacy separable if given two elements which are not conjugate then there is a finite quotient group such that the images of the two elements are not conjugate in this finite group. This is stronger than residually finite which is the special case where one of the group elements is the identity.

The point is I think the braid groups are conjugacy seperable, For $B_3$ we have the short exact sequence $$0 \rightarrow \mathbb{Z} \rightarrow B_3 \rightarrow \mathbb{Z}_2 * \mathbb{Z}_3 \rightarrow 0$$ and the free product of two cyclic groups is conjugacy seperable.

I don't know the argument for $B_n$.

Edit (in response to comments). The result that braid groups are residually finite is stated in Magnus (Property III) and Rolfsen (Theorem 2.5) . The argument is that free groups are residually finite and the automorphism group of a residually finite group is residually finite.

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This does not seem like a realistic hope. When you say a lot of normal subgroups we can be more precise. A group is conjugacy separable if given two elements which are not conjugate then there is a finite quotient group such that the images of the two elements are not conjugate in this finite group. This is stronger than residually finite which is the special case where one of the group elements is the identity.

The point is I think the braid groups are conjugacy seperable, For $B_3$ we have the short exact sequence $$0 \rightarrow \mathbb{Z} \rightarrow B_3 \rightarrow \mathbb{Z}_2 * \mathbb{Z}_3 \rightarrow 0$$ and the free product of two cyclic groups is conjugacy seperable.

I don't know the argument for $B_n$.