Faraday's Law: Current loop and proton A single circular loop of wire with radius R carries a large clockwise current I(loop)=I0, which constrains a proton of mass M and charge e to travel in a small circle of radius r at constant speed around the center of the loop, in the plane of the loop. The orbit radius r is much smaller than the loop radius R: r << R
What direction does the proton travel, clockwise or counterclockwise? What is the speed of the proton in terms of the known quantities I0, R, r, M, and e (and what approximations, if any, must be made)?
Then, if the current the begins to decrease slowly at a certain time t=t0, so that the current was I(loop)=I0-k(t-t0), does the proton speed up or slow down?
 A: I'm a little new here, so I can't comment yet. However, I do think you're supposed to indicate what you've already tried, so please try to give this problem you're best shot before looking to my answer below and next time, give some indication that you've put some effort into the problem.
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(1) Find the magnetic field produced from the loop of wire using a right hand rule: 
a) Curl the fingers of your right hand into a fist in the direction that current flows through the loop (clockwise), and you should find that your thumb (the direction of the magnetic field) is pointing down (z-direction; "into" the page).
b) "Grab" a piece of the loop so that your thumb points in the direction of the current. Whatever piece you grab, you should find that your fingers curl down towards the center of the loop; again you find that the magnetic field inside the loop is down (z-direction; "into" the page).
(2) The magnetic force is F = q$\vec{v} \times \vec{B}$. Guess that the proton is travelling in a certain direction (say clockwise) and is at a certain position of its orbit (near the "top" of the plane [y-direction]).
Use a right hand rule: stick your hand in the direction of $\vec{v}$ (to the right for the above position and direction chosen) and curl your fingers in the direction of $\vec{B}$ which we determined was down in (1). To do so, you should find your thumb is pointing "up" (positive y-direction; in the plane). That's not right; we know the centripetal acceleration (what keeps the proton orbiting in a circle) is pointed towards the center of the circle.
That means our initial guess for the direction of the proton's orbit was wrong. If you next try counter-clockwise, you'll find that the force points towards the center of the orbit as it should.
I'm going to let you try to find the equation for the speed; if you have trouble, leave a comment about what you've tried and I'll try to help you out.

For the second part of your question, know that the magnetic field from the loop of wire is proportional to the strength of the current (stronger current means stronger magnetic field and vice versa).
Lenz's law says that nature resists changes in the magnetic flux through a surface.
The proton can be thought of as smeared out over its orbit forming another a loop, which also produces a magnetic field at the center of its orbit. (Use another right hand rule: will it be up or down inside the orbit of the proton?)
If the magnetic field from the proton is up inside its orbit (opposite that of the loop of wire) then the proton must make its magnetic field weaker (otherwise, there would be net up flux from the proton) by slowing down and vice versa for the other case.
I hope you won't be too upset I didn't directly answer your question, but I want to make sure you give this problem some effort and actually understand the solution. 
Reference for above: Serway and Jewett's Physics for Scientists and Engineers.
