1. **Problem statement:** Find the point on the second quadrant of a unit circle whose x-coordinate is $-\frac{1}{2}$. The unit circle is defined by the equation $$x^2 + y^2 = 1.$$ 2. **Formula and rules:** For any point $(x,y)$ on the unit circle, the coordinates satisfy $$x^2 + y^2 = 1.$$ Since $x = -\frac{1}{2}$, substitute this into the equation to find $y$. 3. **Substitute and solve:** $$\left(-\frac{1}{2}\right)^2 + y^2 = 1$$ $$\frac{1}{4} + y^2 = 1$$ $$y^2 = 1 - \frac{1}{4} = \frac{3}{4}$$ $$y = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}$$ 4. **Determine the correct sign of $y$:** The second quadrant is where $x$ is negative and $y$ is positive. Since $x = -\frac{1}{2}$ is negative, $y$ must be positive. Therefore, $$y = \frac{\sqrt{3}}{2}.$$ 5. **Final answer:** The point on the unit circle in the second quadrant with $x = -\frac{1}{2}$ is $$\boxed{\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)}.$$ This corresponds to option B.