1. **State the problem:** We need to find the magnetic flux through a wire loop placed inside a solenoid's uniform magnetic field. 2. **Formula used:** Magnetic flux $\Phi$ through a loop is given by $$\Phi = B \cdot A \cdot \cos(\theta)$$ where: - $B$ is the magnetic field strength, - $A$ is the area of the loop, - $\theta$ is the angle between the magnetic field lines and the normal (perpendicular) to the loop's plane. 3. **Important rule:** Magnetic flux depends on the component of the magnetic field perpendicular to the loop. If the magnetic field is parallel to the plane of the loop, then $\theta = 90^\circ$ and $\cos(90^\circ) = 0$. 4. **Given data:** - $A = 2.5$ m$^2$ - $B = 6.0$ T - Magnetic field lines are in the same plane as the loop, so $\theta = 90^\circ$ 5. **Calculate magnetic flux:** $$\Phi = 6.0 \times 2.5 \times \cos(90^\circ)$$ $$\Phi = 15 \times 0 = 0$$ 6. **Interpretation:** Since the magnetic field is parallel to the loop's plane, no magnetic field lines pass perpendicularly through the loop, so the magnetic flux is zero. **Final answer:** $$\boxed{0 \text{ Wb}}$$