Direction of current as bar magnet moves uniformly through copper ring

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In summary, when a bar magnet moves uniformly through a copper ring, it induces an electric current in the ring due to electromagnetic induction. The direction of the induced current depends on the relative motion between the magnet and the ring, following Lenz's Law, which states that the induced current will flow in a direction that opposes the change in magnetic flux. As the magnet approaches, the current flows in one direction, and when it moves away, the current reverses, illustrating the principles of electromagnetism and the interaction between magnetic fields and conductive materials.
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Meow12
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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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FAQ: Direction of current as bar magnet moves uniformly through copper ring

What happens to the direction of current when a bar magnet moves uniformly through a copper ring?

When a bar magnet moves uniformly through a copper ring, the direction of the induced current depends on the direction of the magnet's motion and its magnetic poles. According to Lenz's Law, the induced current will flow in a direction that opposes the change in magnetic flux. If the north pole of the magnet approaches the ring, the induced current will create a magnetic field opposing the north pole, and vice versa for the south pole.

How does Lenz's Law apply to the movement of a bar magnet through a copper ring?

Lenz's Law states that the direction of the induced current in a conductor will be such that it opposes the change in magnetic flux that caused it. When a bar magnet moves through a copper ring, the changing magnetic field induces a current in the ring. This induced current generates its own magnetic field, which opposes the motion of the magnet, thereby adhering to Lenz's Law.

Does the speed of the bar magnet affect the induced current in the copper ring?

Yes, the speed of the bar magnet affects the induced current in the copper ring. A faster-moving magnet will cause a more rapid change in magnetic flux, which results in a stronger induced current. Conversely, a slower-moving magnet will induce a weaker current due to the slower change in magnetic flux.

What role does Faraday's Law of Electromagnetic Induction play in this scenario?

Faraday's Law of Electromagnetic Induction states that the induced electromotive force (EMF) in a closed circuit is proportional to the rate of change of magnetic flux through the circuit. In the case of a bar magnet moving through a copper ring, Faraday's Law explains that the changing magnetic field as the magnet moves induces an EMF, which in turn drives the current around the ring.

How does the orientation of the bar magnet affect the induced current in the copper ring?

The orientation of the bar magnet affects the direction and magnitude of the induced current in the copper ring. If the magnet is oriented such that its poles move perpendicularly through the ring, the change in magnetic flux is maximized, resulting in a stronger current. The direction of the induced current will depend on which pole (north or south) is entering or leaving the ring, following Lenz's Law to oppose the change in flux.

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