Magnetic field of a solenoid at the poles? It's known that at the center of an electromagnet/solenoid the magnetic field $B$ is strongest there. At the pole/edge of the solenoid/ or electromagnet is the magnetic field $B$ strong? Let's assume $B$ = $1$ tesla at the center, how different is $B$ at the edges? 

(source: gsu.edu) 
In general, is it simple to create a $1$ tesla magnetic field or higher? 
MRI's use field's above $4$ tesla yet it seems very complex and requires massive power & cooling. 
 A: The magnetic field of a circular current thread lying in the y,z-plane with center at the origin is given as $B = \frac{\mu I}{2} \frac{R^2}{(x^2+R^2)^{\frac{3}{2}}}$ on the x-axis at the coordinate value $x$. The B-field is aligned to the x-axis. The direction of rotation of the current flow and the direction of the B-field build a right-handed screw.
We integrate this to find the B-field on the x-axis at the coordinate $x$ for a solenoid with $n$ windings of current $I$, and with the x-axis as its symmetry axis.
The current flow is modelled as uniform current distribution of strength $\frac{nI}{L}$ on a cylinder surface with radius $R$ starting at $x_{\rm coil}=0$ and ending at $L$.
The result of the integration is:
$$B = \frac{\mu n I}{2L} \left[\frac{x}{\sqrt{x^2+R^2}} + \frac{(L-x)}{\sqrt{(L-x)^2+R^2}}\right].$$
Now at the ends the field is simply 
$$B_{\rm end} = \frac{\mu n I}{2\sqrt{L^2+R^2}}$$
At the middle of the coil the field on the x-axis is
$$
B_{\rm middle} = \frac{\mu n I}{2\sqrt{\left(\frac L2\right)^2+R^2}}
$$
Note : There is no change in material (permeability)
Here $\mu = \mu_r \mu_0 = k \mu_0$ and $n$ is no of turns per unit length.
In an MRI scanner the fields are produced around objects which are not in contact ! i.e. The fields have strengths ranging from 1 to 7 T in air, it is not very difficult to produce these fields in material like iron as they boost magnetic fields by factors of 1000. But to make such fields in air 1000 times energy is required and therefore these scanners utilise high power, coolants and nowadays superconducting wires too.
Addendum :
For an example (as requested in comments), suppose a solenoid of length $1 m$ and radius of cross section $0.1m$, having $1000$ no of turns per unit length($1 m$) carrying current $1 A$ and filled with air which makes $\mu = \mu_0$. Then, 
$$B = \frac{1000\mu_0}{2} \left[\frac{x}{\sqrt{x^2+0.1^2}} + \frac{(1-x)}{\sqrt{(1-x)^2+0.1^2}}\right].$$
$$B_{\rm end} = \frac{1000\mu_0}{2\sqrt{1^2+0.1^2}}$$
$$B_{\rm end} = 0.0025008012 T$$
$$B_{\rm middle} = \frac{1000\mu_0}{2\sqrt{\left(\frac{1}{2}\right)^2+0.1^2}}$$
$$B_{\rm middle} = 0.0049289361 T$$
