Direction of current as bar magnet moves uniformly through copper ring

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As a bar magnet moves downward through a stationary copper ring, the magnetic flux through the ring decreases, resulting in a positive induced electromotive force (emf) and a counterclockwise current when viewed from above. The flux reaches a minimum as the magnet is midway through the ring, at which point the induced current momentarily stops, reverses direction, and eventually dies out as the magnet moves away. The induced current opposes the motion of the magnet, creating a repulsive force when the magnet approaches and an attractive force when it is below the ring. The response of the induced current is antisymmetric about the midway point due to geometrical symmetry. Overall, Lenz's law effectively describes the current's behavior in relation to the magnet's motion.
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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.
 
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