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Bar magnet entering a solenoid

by leah3000
Tags: solenoid
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leah3000
#1
Apr25-10, 09:43 PM
P: 43
I've attached a diagram of a bar magnet being plunged into a solenoid.

I'd like someone to explain the direction of the current please.

Why do both arrows indicating current move upwards?

Using the right hand grip rule i understand the north pole being in the position it's in. I get a little confused when using the letters. But I know that the north represents the current flows in an anti- clockwise direction and the south clockwise. So I can understand why the arrow moves upward where the north pole is but why does it move upward at the south? Shouldn't it move downward to complete the circuit?

Got this diagram in class.
Attached Thumbnails
bar magnet- solenoid.jpg  
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Stonebridge
#2
Apr26-10, 03:05 AM
P: 648
The diagram is incorrect. It looks like a mistake in the direction arrow on the right. It should be pointing downwards. Everything else is fine.
The diagram shows Lenz's Law; and shows the current in the coil producing a magnetic field that opposes the north pole of the magnet moving towards the left end of the coil.

If you were to move the magnet away from the left end of the coil, the current would move in the opposite direction in the coil.
leah3000
#3
Apr26-10, 03:12 PM
P: 43
Quote Quote by Stonebridge View Post
The diagram is incorrect. It looks like a mistake in the direction arrow on the right. It should be pointing downwards. Everything else is fine.
The diagram shows Lenz's Law; and shows the current in the coil producing a magnetic field that opposes the north pole of the magnet moving towards the left end of the coil.

If you were to move the magnet away from the left end of the coil, the current would move in the opposite direction in the coil.
Thank you for helping.

So moving the magnet away causes the current to flow in the opposite direction...I get it.

What if I had a 2nd coil in place of the magnet? The 1st coil would induce a current in the 2nd. By Lenz's law they would oppose each other. If the circuit of the first coil is just opened, Would the direction of the current in the 2nd coil still flow in opposition?

This is what I'm thinking: If the cct is just opened then the current in the cct would instantly fall to zero. Although no current is flowing, wouldn't there be a change in flux? I was thinking that this flux change would then induce an e.m.f in the 2nd coil and the induced e.m.f would be in the same direction as the current.

But then I'm not sure because the other part of Lenz's law states that the induced e.m.f opposes the flux change causing it if an induced current flows...and no current flows in the 1st. So what happens to the current in the 2nd coil?

Stonebridge
#4
Apr26-10, 03:47 PM
P: 648
Bar magnet entering a solenoid

If you replaced the magnet by another coil, then you could produce the same effect in the right hand coil in a number of ways. The important thing about Lenz's Law is that it says that the current in the right coil will create a magnetic field that will try to oppose the change in the left coil.
You could:
a) arrange that the left coil has a N pole at its right end nearest the other coil, keep the current steady, but actually move the left coil towards the other one. The right coil would not know the difference between the N pole produced by the magnet, and the one just produced by the left coil. The result would be the right coil getting an induced current that creates a N pole that opposes the N pole it sees arriving from the left.
b) Keep the left coil stationary but slowly increase the current in it so that the N pole grows at its right end nearest the other coil. The right coil "sees" the growing N pole and gets an induced current that produces a N pole at its left end to oppose the one "arriving" from the other coil.
The key is that what the right coil does is always designed to oppose any change in the magnetic field it sees. In both cases above, the change is a growing N pole at the left end of the right hand coil. In both cases, the right hand coil produces a N pole at its left end.
The direction the current flows in the right coil depends on the way the coils are wound. Clockwise produces a S pole. Anti(counter) clockwise produces a N.
leah3000
#5
Apr26-10, 04:40 PM
P: 43
ok so i understand the effect of the growing north pole (on the right end) of the left coil-->inducing a current in the right coil which in turn produces the north pole (on the left end) to oppose that of the left coil.- as indicated by Lenz's law.

what I still don't catch is shutting off the current altogether in the left coil (breaking the cct). How does this affect the current in the right coil as soon as the cct of the left coil is opened?

Also if the current in the left coil slowly decreases does this mean the current in the right coil would gradually reverse itself? meaning the effect of the North pole decreases eventually becoming a south? i.e when the current reaches 0 in the left coil.
Stonebridge
#6
Apr27-10, 04:49 AM
P: 648
Quote Quote by leah3000 View Post
ok so i understand the effect of the growing north pole (on the right end) of the left coil-->inducing a current in the right coil which in turn produces the north pole (on the left end) to oppose that of the left coil.- as indicated by Lenz's law.

what I still don't catch is shutting off the current altogether in the left coil (breaking the cct). How does this affect the current in the right coil as soon as the cct of the left coil is opened?

Also if the current in the left coil slowly decreases does this mean the current in the right coil would gradually reverse itself? meaning the effect of the North pole decreases eventually becoming a south? i.e when the current reaches 0 in the left coil.
(Same setup as before)
When you reduce (or switch off) the current in the left coil, the effect is one of a disappearing N pole. The N pole that is at the right end of the left coil gets weaker.
This is exactly the same (as far as the right coil is concerned) as moving the bar magnet with its N pole away from that end.
The current induced in the right coil is opposite this time.
The right coil reacts to an increasing N pole (magnetic field) by inducing a N pole at its left end. It reacts to a decreasing N pole by inducing a S pole at its left end.
I always used to picture this as the right coil trying to stop the left coil from moving. N repels N but S attracts N.
leah3000
#7
Apr27-10, 11:29 AM
P: 43
Quote Quote by Stonebridge View Post
(Same setup as before)
When you reduce (or switch off) the current in the left coil, the effect is one of a disappearing N pole. The N pole that is at the right end of the left coil gets weaker.
This is exactly the same (as far as the right coil is concerned) as moving the bar magnet with its N pole away from that end.
The current induced in the right coil is opposite this time.
The right coil reacts to an increasing N pole (magnetic field) by inducing a N pole at its left end. It reacts to a decreasing N pole by inducing a S pole at its left end.
I always used to picture this as the right coil trying to stop the left coil from moving. N repels N but S attracts N.
Thank you for helping i understand alot better now
mishi003
#8
Jan23-12, 08:08 AM
P: 3
on similar lines, i have a doubt.
i have a permanent magnet that moves due to current that is applied to the solenoid. based on the polarity due to the current ofcourse. Say it attracts. now since the bar magnet is moving with some acceleration towards the coil, there would be a change in the magnetic flux and this wud lead to a back emf or eddy current. How do i calculate this current.
emf = -N (dB/dT)A
A is the area of the magnet
N is the number of coils

my problem is tht i cannot calculate the dB/dT accurately. because i cannot calculate the force with which it gets attracted accurately.
Please let mek now if u have ne suggestion.


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