For Faraday's law, why does the emf decrease as you increase the area of the loop? I've only recently started learning basic electrodynamics, but I don't understand why a loop of coil with a small area and a magnet falling through will produce a larger emf than a loop of coil with a larger area with the same magnetic field falling through. 
To clarify, lets say you have a loop of coil and you drop a magnet through the coil. This is will produce an emf according to Faraday's law. Now if you have a loop of coil with a larger radius, and you drop the same magnet through at the same speed, it will produce a smaller emf. Why is this?
 A: Faraday's Law: $\mathcal{E}=-\frac{d\Phi}{dt}$. For a bigger loop, the rate of change of magnetic flux through the loop must be smaller than for a small loop. To go any further/be less hand wavy we need vector calculus. The magnetic field of a dipole is: $$ \mathbf{B}(\mathbf{r})=\frac{\mu_0}{4\pi}\left( \frac{3\mathbf{r}(\mathbf{m}.\mathbf{r})}{r^5}-\frac{\mathbf{m}}{r^3}\right) $$ Whilst the flux through the loop can be calculated from this, if you can understand the integral you need for this $\int\mathbf{B}.d\mathbf{A}$ then you'll hopefully be happy with me just saying that there's another expression for EMF:$$\mathcal{E}=\oint\mathbf{F}.d\mathbf{l}$$
where $\mathbf{F}$ is the force producing the current we're concerned with. See Griffiths Introduction to Electrodynamics for a proof of equivalence, for example. I'm going to do it this way because most of the working is geometric which you should be able to follow if you haven't done vector calc. The magnetic force is $\mathbf{v}\wedge\mathbf{B}$. If we drop the magnet in "end on" then by symmetry the force at all points on the loop is the same magnitude, just in different direction. Then $\mathcal{E}=F(r)\oint\hat{\mathbf{F}}.d\mathbf{l}$. $\mathbf{v}$ is in the vertical direction, and $\mathbf{B}$ is coplanar with $\mathbf{v}$ and the axis of symmetry of the problem. To see why, make a quick sketch of the magnetic field lines of the dipole. Then $\hat{\mathbf{F}}$ is the direction of $d\mathbf{l}$ and the integral is just the length of the loop $2\pi R$.
$$\mathcal{E}=2\pi r*F(\mathbf{r})=2\pi R|\mathbf{v}\wedge\mathbf{B}|=2\pi RvB\sin\theta$$
I've thrown a lot of variables around so I'll just define what's left in this expression for clarity. $R$ is the radius of the loop. $v$ is the magnitude of the velocity of the loop in the magnet frame. $B$ is the magnitude of the magnetic field at at the loop. $\theta$ is the angle between the velocity vector and magnetic field vector.
Let's pick a particular point in time, with the dipole somewhere above the loop, and compare two loops of differing radius. Let us assume that the object falls with near constant velocity, then that is common along with $2\pi$. Now we find the modulus of $\mathbf{B}$
$$ B=\frac{\mu_0m}{4\pi r^3}|3\hat{r}(\hat{z}.\hat{r})-\hat{z}| $$ Note that $\hat{z}.\hat{r}=\frac{h}{r}$ where $h$ is the height of the magnet above the loop. Then
$$ B=\frac{\mu_0m}{4\pi r^3}\sqrt{\left(\frac{3h}{r}\hat{r}-\hat{m}\right).\left(\frac{3h}{r}\hat{r}-\hat{m}\right)}=\frac{\mu_0m}{4\pi r^4}\sqrt{3h^2+r^2} $$
so $$ \mathcal{E}\sim\frac{R}{r^4}\sqrt{3h^2+r^2}\sin\theta $$ To calculate $\sin\theta$, consider the scalar product of $\mathbf{B}$ with $\mathbf{v}$:
$$ \mathbf{B}.\mathbf{v}=|\mathbf{B}||\mathbf{v}|cos\theta $$
$$\cos\theta = \frac{r}{\sqrt{3h^2+r^2}}\left(\frac{3h}{r}\hat{r}-\hat{z} \right).\hat{z}=\frac{3h^2-r^2}{r\sqrt{3h^2+r^2}}  $$
then $$\sin\theta=\sqrt{1-\cos^2\theta}=\frac{3h}{r}\sqrt{\frac{r^2-h^2}{3h^2+r^2}}$$
Putting everything together
$$ \mathcal{E}\sim \frac{R}{r^4}\sqrt{3h^2+r^2}\frac{3h}{r}\sqrt{\frac{r^2-h^2}{3h^2+r^2}} \sim \frac{3hR}{r^5}\sqrt{r^2-h^2} $$
and using $r^2=h^2+R^2$
$$ \mathcal{E}\sim \frac{3hR^2}{(h^2+R^2)^{5/2}}$$
For a given $R$, where is $\mathcal{E}$ maximum?
$$\frac{\partial\mathcal{E}}{\partial h}=3R^2(h^2+R^2)^{-5/2}-15h^2R^2(h^2+R^2)^{-7/2}$$
which is $0$ when $h=R/2$. At $h=R/2$ the EMF is
$$ \mathcal{E}\sim \frac{3}{400R^2}$$
which is strictly increasing as you decrease $R$ i.e. maximum EMF felt by smaller loops is greater (as long as we drop from a height $>R/2$ for $R$ of the bigger loop.
Corrections most welcome. Image and tidying to follow.
A: Yes decreases because change in  magnetic field decreases due to increase in distance between coil and magnet
A: Assuming quite new to vector calc $\implies$ some experience with vector calc, I thought it'd be worth doing the flux calculation (also to check the form of the above answer). This method is much simpler.
$$\mathbf{B}=\frac{\mu_0m}{4\pi} \left(\frac{3\mathbf{r}(\mathbf{r}.\mathbf{\hat{z}})}{r^5}-\frac{\mathbf{\hat{z}}}{r^3}  \right) $$
Taking the integral $\int d\mathbf{A}.\mathbf{B}$ over the disc described by the ring such that $d\mathbf{A}=\mathbf{\hat{z}}dA$, and introducing cylindrical co-ordinates z (height above the disc), and $(\rho,\theta)$ parameterising the disc itself gives
$$ \phi \sim \int B_z\rho d\rho d\phi \sim \int_{\rho=0}^{\rho=R}\left(\frac{3(\mathbf{r}.\mathbf{\hat{z}})^2}{r^5}-\frac{\mathbf{\hat{z}}.\mathbf{{\hat{z}}}}{r^3}  \right)\rho d\rho\sim\int_{\rho=0}^{\rho=R}\left(\frac{3z^2}{r^5}-\frac{1}{r^3}  \right)\rho d\rho$$
Since $r^2=\rho^2+z^2$
$$ \phi\sim \int_{\rho=0}^{\rho=R}\left(\frac{3z^2}{(\rho^2+z^2)^{5/2}}-\frac{1}{(\rho^2+z^2)^{3/2}}  \right)\rho d\rho\sim\left[ -\frac{2z^2}{(\rho^2+z^2)^{3/2}} + \frac{2}{(\rho^2+z^2)^{1/2}}\right]_{\rho=0}^{\rho=R}$$ 
$$ \phi\sim\frac{R^2}{(R^2+z^2)^{3/2}} $$
Then
$$ \mathcal{E} \sim \frac{d\phi}{dt} \sim \frac{R^2z\frac{dz}{dt}}{(R^2+z^2)^{5/2}}$$
so it seemingly works out.
