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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  • #2
You could also mention that the response should be antisymmetric about the midway point because of the geometrical symmetry.
 
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  • #3
And perhaps sketch what the emf looks like as a function of time.
 
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  • #4
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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1. How does the direction of current change as a bar magnet moves through a copper ring?

As the bar magnet moves through the copper ring, the magnetic field lines cut across the copper ring, inducing an electromotive force (EMF) according to Faraday's law of electromagnetic induction. This induced EMF causes a current to flow in the copper ring in a direction that opposes the change in magnetic flux, following Lenz's law.

2. What factors determine the direction of the induced current in the copper ring?

The direction of the induced current in the copper ring is determined by the direction of the change in magnetic flux. Lenz's law states that the induced current will flow in a direction that creates a magnetic field opposing the change in magnetic flux caused by the moving bar magnet.

3. Does the direction of current depend on the speed of the bar magnet through the copper ring?

The direction of the induced current in the copper ring does not depend on the speed of the bar magnet. It is solely determined by the change in magnetic flux, which is caused by the movement of the magnet through the ring, regardless of the speed.

4. Why does the current flow in a closed loop in the copper ring?

The current flows in a closed loop in the copper ring because the induced EMF causes charges to move in the conductor, creating a continuous flow of current. This closed loop ensures that the induced current follows Lenz's law and opposes the change in magnetic flux.

5. How does the direction of current affect the magnetic field produced by the copper ring?

The direction of the induced current in the copper ring produces a magnetic field that opposes the change in magnetic flux caused by the moving bar magnet. This opposing magnetic field interacts with the magnetic field of the bar magnet, resulting in a damping effect on the motion of the magnet through the ring.

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