Copper with a slit braking question (1 Viewer)

[MinaSaw]

Hey guys,
So this is a concept that has really been confusing and it is one of a pretty common HSC question so it is as follows:
When a horse shoe made of copper is dropped onto a strong magnet, it experiences electromagnetic braking as there is a changing magnetic field, induced emf, it is complete so induced current and hence induced magnet field which opposes the bar magnet and so slows it down.
If we have a copper horseshoe with a slit, there is a changing magnetic field, and I would assume there is an induced emf, but there will be no induced current and hence no induced magnetic field and hence electromagnetic braking.
Is this correct?
Thanks in advance

 

[1729]

Hey guys,
So this is a concept that has really been confusing and it is one of a pretty common HSC question so it is as follows:
When a horse shoe made of copper is dropped onto a strong magnet, it experiences electromagnetic braking as there is a changing magnetic field, induced emf, it is complete so induced current and hence induced magnet field which opposes the bar magnet and so slows it down.
If we have a copper horseshoe with a slit, there is a changing magnetic field, and I would assume there is an induced emf, but there will be no induced current and hence no induced magnetic field and hence electromagnetic braking.
Is this correct?
Thanks in advance

Using a copper cylinder instead of a horseshoe,

If there is no vertical slit,

If there is a vertical slit, eddy currents are still induced in the thickness of the conductor but are weaker. For the HSC though I think you have to assume that they aren't induced, and that they can ony be induced in the flat surface of a conductor.

 

[MinaSaw]

Thanks for your explanation!
My question is, will electromagnetic braking still occur or will it free fall due to gravity?

 

[pikachu975]

Thanks for your explanation!
My question is, will electromagnetic braking still occur or will it free fall due to gravity?

Free fall since in HSC slits = no eddy currents or negligible eddy currents.

Also just a note I don't think this is electromagnetic braking, this is just an application Faraday + Lenz etc.

 

[MinaSaw]

Alright cheers!