In the diagram, the circle has radius $\sqrt 7$ and and centre $O.$ Points $A, B$ and $C$ are on the circle. If $\angle BOC=120^\circ$ and $AC = AB + 1,$ determine the length of $AB.$
[asy]
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draw(circle((2.34,2.4),2.01),qqttff); draw((2.34,2.4)--(1.09,0.82),fftttt); draw((2.34,2.4)--(4.1,1.41),fftttt); draw((1.09,0.82)--(1.4,4.18),fftttt); draw((4.1,1.41)--(1.4,4.18),fftttt);
dot((2.34,2.4),ds); label("$O$", (2.1,2.66),NE*lsf); dot((1.09,0.82),ds); label("$B$", (0.86,0.46),NE*lsf); dot((4.1,1.41),ds); label("$C$", (4.2,1.08),NE*lsf); dot((1.4,4.18),ds); label("$A$", (1.22,4.48),NE*lsf); clip((-4.34,-10.94)--(-4.34,6.3)--(16.14,6.3)--(16.14,-10.94)--cycle);
[/asy]
1. Given that the circle has a radius of $\sqrt{7}$ and center $O$, and points $A$, $B$, and $C$ are on the circle. We know $\angle BOC = 120^\circ$ and $AC = AB + 1$. We need to determine the length of $AB$.

2. Since $O$ is the center of the circle, $OB = OC = \sqrt{7}$. The angle $\angle BOC = 120^\circ$ is given.

3. Using the Law of Cosines in $\triangle BOC$:
   \[
   BC^2 = OB^2 + OC^2 - 2 \cdot OB \cdot OC \cdot \cos(\angle BOC)
   \]
   Substituting the known values:
   \[
   BC^2 = (\sqrt{7})^2 + (\sqrt{7})^2 - 2 \cdot \sqrt{7} \cdot \sqrt{7} \cdot \cos(120^\circ)
   \]
   \[
   BC^2 = 7 + 7 - 2 \cdot 7 \cdot \left(-\frac{1}{2}\right)
   \]
   \[
   BC^2 = 14 + 7 = 21
   \]
   Therefore,
   \[
   BC = \sqrt{21}
   \]

4. Now, consider $\triangle ABC$. We know $AC = AB + 1$ and $\angle BAC = 60^\circ$ (since $\angle BOC = 120^\circ$ and $\angle BAC$ is the external angle to $\angle BOC$).

5. Using the Law of Cosines in $\triangle ABC$:
   \[
   BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cdot \cos(\angle BAC)
   \]
   Substituting the known values:
   \[
   21 = AB^2 + (AB + 1)^2 - 2 \cdot AB \cdot (AB + 1) \cdot \cos(60^\circ)
   \]
   \[
   21 = AB^2 + (AB^2 + 2AB + 1) - 2 \cdot AB \cdot (AB + 1) \cdot \frac{1}{2}
   \]
   \[
   21 = AB^2 + AB^2 + 2AB + 1 - AB^2 - AB
   \]
   \[
   21 = AB^2 + AB + 1
   \]

6. Simplifying the equation:
   \[
   AB^2 + AB + 1 - 21 = 0
   \]
   \[
   AB^2 + AB - 20 = 0
   \]

7. Solving the quadratic equation $AB^2 + AB - 20 = 0$ using the quadratic formula $AB = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
   \[
   AB = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-20)}}{2 \cdot 1}
   \]
   \[
   AB = \frac{-1 \pm \sqrt{1 + 80}}{2}
   \]
   \[
   AB = \frac{-1 \pm \sqrt{81}}{2}
   \]
   \[
   AB = \frac{-1 \pm 9}{2}
   \]
   \[
   AB = \frac{8}{2} \quad \text{or} \quad AB = \frac{-10}{2}
   \]
   \[
   AB = 4 \quad \text{or} \quad AB = -5
   \]

8. Since the length cannot be negative, we have:
   \[
   AB = 4
   \]

The final answer is $\boxed{4}$