First, calculate the area of the loop: $$A = (0.040 \text{ m})^2 = 0.0016 \text{ m}^2$$. Next, evaluate the magnetic field at $$t=2.0$$ s: $$B(2.0) = 3.0(2.0) - 1.5(2.0)^2 = 6.0 - 1.5(4.0) = 6.0 - 6.0 = 0$$ T. Since the magnetic field is zero at this instant, the magnetic flux $$\Phi_B = BA$$ is also zero.

The magnetic flux through the entire coil is $$N$$ times the flux through a single turn, $$\Phi_{coil} = N(BA \cos\theta)$$. Here, $$A = s^2$$. In the final position, the axis of the coil is perpendicular to the field, meaning the angle $$\theta$$ between the normal vector and the field is $$90^\circ$$. Since $$\cos(90^\circ) = 0$$, the magnetic flux through the coil is zero.

According to Gauss's law for magnetism, the total magnetic flux through any closed surface is zero. Consider the closed surface formed by the hemisphere and its flat circular base. $$\Phi_{closed} = \Phi_{curved} + \Phi_{base} = 0$$. Therefore, $$\Phi_{curved} = -\Phi_{base}$$. The flux through the flat base of area $$A=\pi R^2$$ in the xy-plane is $$\Phi_{base} = \vec{B} \cdot \vec{A}{base}$$. With the outward normal for the base pointing in the $$-z$$ direction, $$\vec{A}{base} = -\pi R^2 \hat{k}$$, so $$\Phi_{base} = (B_0 \hat{k}) \cdot(-\pi R^2 \hat{k}) = -B_0 \pi R^2$$. The flux through the curved surface is thus $$\Phi_{curved} = -(-B_0 \pi R^2) = B_0 \pi R^2$$.

The magnetic flux is $$\Phi_B = BA \cos\theta$$, where $$\theta$$ is the angle between the field and the normal to the loop's plane. The magnitude is maximum when $$|\cos\theta|$$ is maximum, which occurs at $$\theta = 0^\circ$$ or $$180^\circ$$. In this case, the normal vector is parallel to the field, meaning the plane of the loop is perpendicular to the field.

Magnetic flux ($$\Phi_B$$) is conceptually a measure of the amount of magnetic field passing through a surface. It is quantified by counting the net number of magnetic field lines that pierce the surface.

Magnetic flux is given by $$\Phi_B = \vec{B} \cdot \vec{A} = BA \cos\theta$$, where $$\theta$$ is the angle between the magnetic field vector and the area vector (which is normal to the loop's plane). With $$\theta = 60^\circ$$, $$\cos(60^\circ) = 1/2$$, so the flux is $$\frac{1}{2} BA$$.

This is a statement of Gauss's law for magnetism, $$\oint \vec{B} \cdot d\vec{A} = 0$$. Since there are no magnetic monopoles, magnetic field lines always form closed loops. Therefore, any field line that enters a closed surface must also exit it, resulting in a net magnetic flux of zero.

The fundamental definition of magnetic flux for a uniform field is $$\Phi_B = BA \cos\theta$$. The unit of magnetic field ($$B$$) is the tesla (T) and the unit of area ($$A$$) is the meter squared ($$\text{m}^2$$). Therefore, the unit of magnetic flux is tesla-meter squared ($$\text{T}\cdot\text{m}^2$$), which is defined as the weber (Wb). While a volt-second is also equivalent to a weber, it is derived from Faraday's law of induction, not the direct definition of flux.

The area of the square is $$A = (2.0 \text{ m})^2 = 4.0 \text{ m}^2$$. Since the surface lies in the yz-plane, its area vector is perpendicular to that plane, meaning it points in the x-direction. We can write $$\vec{A} = 4.0 \hat{i} \text{ m}^2$$. The magnetic flux is the dot product $$\Phi_B = \vec{B} \cdot \vec{A} = ((4.0 \hat{i} - 3.0 \hat{j}) \text{ T}) \cdot(4.0 \hat{i} \text{ m}^2) = (4.0)(4.0) = 16$$ Wb.

The magnetic field is not uniform over the area of the loop; it depends on the distance $$r$$ from the wire. To find the total flux, we must integrate. Consider a thin strip of the loop of length $$L$$ and width $$dr$$ at a distance $$r$$ from the wire. The area of this strip is $$dA = L dr$$. The flux through this strip is $$d\Phi_B = B dA = (\frac{\mu_0 I}{2\pi r}) (L dr)$$. To find the total flux, we integrate this expression over the width of the loop, from $$r=a$$ to $$r=a+w$$.