Direction of current as bar magnet moves uniformly through copper ring

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Homework Statement
Analyze the direction of electric current in a stationary copper ring as a bar magnet moves through it at constant velocity north-pole first.
Relevant Equations
##\Phi_B=\vec{B}\cdot\vec{A}##

##\displaystyle\epsilon=-\frac{d\Phi_B}{dt}##
magnet-ring.png

Suppose the ring is held stationary such that its area vector points upward. The magnet moves downward toward the ring at constant velocity north-pole first. (The starting position is shown in the figure.)

Since ##\vec{B}## is downward while ##\vec{A}## is upward, ##\Phi_B=\vec{B}\cdot\vec{A}## is negative and decreasing (becoming more negative). So, ##\displaystyle\frac{d\Phi_B}{dt}## is negative. Thus, ##\displaystyle\epsilon=-\frac{d\Phi_B}{dt}## is positive. By the right-hand rule, the induced current in the ring flows counterclockwise when viewed from above.

##\Phi_B## keeps decreasing until it reaches its minimum (##\displaystyle\frac{d\Phi_B}{dt}=0##) when the magnet is midway through the ring. Then, ##\Phi_B## starts increasing (becoming less negative). So, ##\displaystyle\frac{d\Phi_B}{dt}## is positive. ##\displaystyle\epsilon=-\frac{d\Phi_B}{dt}## becomes zero (at the minimum of ##\Phi_B##), then becomes negative, and eventually becomes zero when the magnet is far away from the ring. By the right-hand rule, the induced current stops momentarily, reverses direction (flows clockwise when viewed from above), and eventually dies out.

Does this look okay? Thanks in advance.
 
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You could also mention that the response should be antisymmetric about the midway point because of the geometrical symmetry.
 
And perhaps sketch what the emf looks like as a function of time.
 
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One quick way to get the current-direction is to use Lenz’s law.

As the bar magnet moves down (N-pole first) towards the ring, its motion will be opposed by the field produced by the induced current. So we know the induced field will have the 'N-pole' at the top, causing repulsion.

Applying the RH rule, the current's direction - to make the induced field’s N-pole at the top - will be anticlockwise (UK!) (viewed from above).

Similarly when the bar magnet is below the ring and moving down, the induced field must have the 'N-pole' at the bottom causing attraction.