a) Step 1: The magnetic field $B$ at a point on the axis of a circular current-carrying conductor at a distance $x$ from its center, with radius $a$ and current $I$, is given by: $$B = \frac{\mu_0 I a^2}{2(a^2 + x^2)^{3/2}}$$ where $\mu_0$ is the permeability of free space. The direction of the magnetic field is along the axis of the loop. b) Step 1: Magnetic field at the center of the loop. At the center, $x=0$. Substitute $x=0$ into the formula: $$B_{center} = \frac{\mu_0 I a^2}{2(a^2 + 0^2)^{3/2}} = \frac{\mu_0 I a^2}{2(a^2)^{3/2}} = \frac{\mu_0 I a^2}{2a^3} = \frac{\mu_0 I}{2a}$$ The magnetic field at the center is $\boxed{\frac{\mu_0 I}{2a}}$. Step 2: Magnetic field at an axial distance $x = a/2$ from the center. Substitute $x = a/2$ into the formula: $$B_{a/2} = \frac{\mu_0 I a^2}{2(a^2 + (a/2)^2)^{3/2}}$$ $$B_{a/2} = \frac{\mu_0 I a^2}{2(a^2 + a^2/4)^{3/2}}$$ $$B_{a/2} = \frac{\mu_0 I a^2}{2(\frac{5a^2}{4})^{3/2}}$$ $$B_{a/2} = \frac{\mu_0 I a^2}{2 \left(\frac{\sqrt{5}a}{2}\right)^3}$$ $$B_{a/2} = \frac{\mu_0 I a^2}{2 \frac{5\sqrt{5}a^3}{8}}$$ $$B_{a/2} = \frac{\mu_0 I a^2}{\frac{5\sqrt{5}a^3}{4}}$$ $$B_{a/2} = \frac{4 \mu_0 I a^2}{5\sqrt{5}a^3}$$ $$B_{a/2} = \frac{4 \mu_0 I}{5\sqrt{5}a}$$ The magnetic field at $x=a/2$ is $\boxed{\frac{4 \mu_0 I}{5\sqrt{5}a}}$. c) Due to the symmetry of the circular loop, for every current element producing a magnetic field component perpendicular to the axis, there is another diametrically opposite current element that produces an equal and opposite perpendicular component. These perpendicular components cancel each other out, leaving only the components parallel to the axis to contribute to the net magnetic field. 18. a) Einstein's photoelectric equation is given by: $$K_{max} = h\nu - \phi$$ where $K_{max}$ is the maximum kinetic energy of the emitted electron, $h\nu$ is the energy of the incident photon, and $\phi$ is the work function of the metal. This equation demonstrates the conservation of energy. The energy of the incident photon ($h\nu$) is used to overcome the work function ($\phi$) and the remaining energy is converted into the maximum kinetic energy ($K_{max}$) of the emitted electron. b) Alkali metals are important in solar energy generation because they have very low work functions. This means that less energy is required to eject electrons from their surface when light strikes them. This property makes them highly efficient at converting light energy into electrical energy, which is crucial for the operation of photovoltaic cells and other photoemissive devices used in solar energy technologies.