Is there current induced in a solenoid in an open circuit?

[Jam Jay]

A bar magnet is being pulled away from one end of a solenoid, eg N pole of magnet nearest to solenoid. I've learned that the current induced in a solenoid is such that the magnetic field it creates opposes the change in flux through the loop. But what if the solenoid is in an open circuit? Will there still be current induced? Will there be induced emf and induced magnetic S pole at that end of the solenoid?

How do i differentiate between induced emf (voltage) and induced current? I'm normally clear about voltage and current but when it comes to this, I'm confused between the two.
Faraday's law: induced emf?
Lenz's law: induced current?

 

[berkeman]

A bar magnet is being pulled away from one end of a solenoid, eg N pole of magnet nearest to solenoid. I've learned that the current induced in a solenoid is such that the magnetic field it creates opposes the change in flux through the loop. But what if the solenoid is in an open circuit? Will there still be current induced? Will there be induced emf and induced magnetic S pole at that end of the solenoid?

How do i differentiate between induced emf (voltage) and induced current? I'm normally clear about voltage and current but when it comes to this, I'm confused between the two.
Faraday's law: induced emf?
Lenz's law: induced current?

Welcome to the PF.

Since it is an open circuit, the only current that can flow is the current that charges up the parasitic capacitance of the coil.

 

[Jam Jay]

Welcome to the PF.

Since it is an open circuit, the only current that can flow is the current that charges up the parasitic capacitance of the coil.

So, no to all induced emf, current and magnetic pole?
Thanks.

 

[berkeman]

So, no to all induced emf, current and magnetic pole?
Thanks.

Not sure I understand the question. There will be an EMF and a small current. The small current interacts with the very real parasitic capacitance of the coil.

 

[cabraham]

A bar magnet is being pulled away from one end of a solenoid, eg N pole of magnet nearest to solenoid. I've learned that the current induced in a solenoid is such that the magnetic field it creates opposes the change in flux through the loop. But what if the solenoid is in an open circuit? Will there still be current induced? Will there be induced emf and induced magnetic S pole at that end of the solenoid?

How do i differentiate between induced emf (voltage) and induced current? I'm normally clear about voltage and current but when it comes to this, I'm confused between the two.
Faraday's law: induced emf?
Lenz's law: induced current?

Faraday's law describes the induced emf/voltage. Ampere's law describes the induced mmf/current. Lenz' law describes the polarity, i.e. the induced flux/emf/mmf is always oriented opposite to the existing flux/emf/mmf. In other words if an emf/mmf is induced, any current in the circuit produces a magnetic flux with a polarity opposite to the inducing flux. This is consistent with conservation of energy.

In an open circuit there is still a displacement current induced, since capacitance is always greater than zero. In a short there is induced voltage due to inductance. Induction always involves both current and voltage. Ampere's law. Faradays law, Lenz' law, and Ohm's law all apply simultaneously. Does this help?

Claude

 

[bjacoby]

A bar magnet is being pulled away from one end of a solenoid, eg N pole of magnet nearest to solenoid. I've learned that the current induced in a solenoid is such that the magnetic field it creates opposes the change in flux through the loop. But what if the solenoid is in an open circuit? Will there still be current induced? Will there be induced emf and induced magnetic S pole at that end of the solenoid?

How do i differentiate between induced emf (voltage) and induced current? I'm normally clear about voltage and current but when it comes to this, I'm confused between the two.
Faraday's law: induced emf?
Lenz's law: induced current?

It's pretty simple really. You need to know that the basis of generator action is the Lorentz force equation F = qV x B. The vector math may be advanced for you but just imagine the lines of force from the end of the magnet cutting the solenoid wires as you pull it out. This then causes a force on the free charges in the wires (electrons). They try to go round and round the solenoid. If the solenoid is shorted or has a low resistance load they do go round and round which is a current. If the wire is open circuited the charges only move far enough until an excess of electrons is built up at one end (-) and too few at the other end (+) of the wire. This causes an emf to appear in the wire. But since nothing can flow there is no current and the charges just build until they exactly cancel the forces due to pulling the magnet out. If you were to carefully examine all the forces created here, you'd find that even though no current actually flows, the emf is in such a direction that if it were allowed to flow it would be as Lenz predicts.

 

[cabraham]

It's pretty simple really. You need to know that the basis of generator action is the Lorentz force equation F = qV x B. The vector math may be advanced for you but just imagine the lines of force from the end of the magnet cutting the solenoid wires as you pull it out. This then causes a force on the free charges in the wires (electrons). They try to go round and round the solenoid. If the solenoid is shorted or has a low resistance load they do go round and round which is a current. If the wire is open circuited the charges only move far enough until an excess of electrons is built up at one end (-) and too few at the other end (+) of the wire. This causes an emf to appear in the wire. But since nothing can flow there is no current and the charges just build until they exactly cancel the forces due to pulling the magnet out. If you were to carefully examine all the forces created here, you'd find that even though no current actually flows, the emf is in such a direction that if it were allowed to flow it would be as Lenz predicts.

But current is flowing! The charges which accumulate at the ends of the wire are moving within the confines of the wire. This is displacement current.

The ends of the wire have an area. The space between the wires is an insulator. Two conducting areas separated by an insulator constitutes a capacitor. Displacement current flows per i = C*dv/dt. If the emf/voltage is changing there has to be a current. Induction can only produce current & voltage together in unison. Induction cannot produce just one without the other. It's a fact known since the 19th century. Cheers.

Claude

 

[Dadface]

But current is flowing! The charges which accumulate at the ends of the wire are moving within the confines of the wire. This is displacement current.

The ends of the wire have an area. The space between the wires is an insulator. Two conducting areas separated by an insulator constitutes a capacitor. Displacement current flows per i = C*dv/dt. If the emf/voltage is changing there has to be a current. Induction can only produce current & voltage together in unison. Induction cannot produce just one without the other. It's a fact known since the 19th century. Cheers.

Claude

On open circuit a current will flow only if the emf changes

 

[cabraham]

On open circuit a current will flow only if the emf changes

The emf is changing. Induction does not happen at dc. Induction only happens in the ac domain. If the time-changing flux is a sinusoid, so is the induced emf & mmf. Even open circuited there has to be a current. A time changing voltage cannot be unless charges are moving. Static charges result in static voltage and no current. But time varying charges constitute both a time varying voltage and a current.

In the ac domain, one cannot exist w/o the other, short circuited, open, or anything in between.

Claude

 

[Jam Jay]

Thanks all for the explanations! Pardon me as i alternate between understanding and confusion as i read the replies. i was facing this question below. Appreciate all help. Thanks!

Q: A bar magnet is pulled away from a solenoid as shown in the diagram below.
Which of the following statements is true?
A. An e.m.f. is induced in the solenoid.
B. A magnetic north pole is induced at end X.
C. A current is induced in the solenoid.
D. The galvanometer pointer is deflected to the left.

 

[Dadface]

The emf is changing. Induction does not happen at dc. Induction only happens in the ac domain. If the time-changing flux is a sinusoid, so is the induced emf & mmf. Even open circuited there has to be a current. A time changing voltage cannot be unless charges are moving. Static charges result in static voltage and no current. But time varying charges constitute both a time varying voltage and a current.

In the ac domain, one cannot exist w/o the other, short circuited, open, or anything in between.

Claude

Thanks all for the explanations! Pardon me as i alternate between understanding and confusion as i read the replies. i was facing this question below. Appreciate all help. Thanks!

Q: A bar magnet is pulled away from a solenoid as shown in the diagram below.
Which of the following statements is true?
A. An e.m.f. is induced in the solenoid.
B. A magnetic north pole is induced at end X.
C. A current is induced in the solenoid.
D. The galvanometer pointer is deflected to the left.

Hello Jam Jay having seen the question and realized the level it is aimed at I am not surprised that you are confused.An emf is induced whether the circuit is open or closed so the answer is A.Answers B,Cand D can only apply if a current flows but since the coil is on open circuit there is no current.If you want to be really fussy and go beyond the level of the question you could argue that there is an extremely tiny current due to parasitic capacitance etc which others here have referred to.You could also complain that the question itself is faulty and discriminates against the more deep thinking student who realizes that there could be a tiny current.Should it be an exam question,the examiners normally brush off such complaints by stating that students were instructed to pick the most appropriate answer.

Claude the current induced here,whether it be on open or closed circuit,is dc not ac.Also, there is no current in this or any other similar system when the emf reaches a steady value.In principle a steady emf can be achieved with this system by pulling the magnet at an increasing speed such that the rate of change of flux remains constant.

 

[Bob_for_short]

To observe (to measure) the displacement current you can put an ampere-meter in the middle of the coil.

 

[cabraham]

Hello Jam Jay having seen the question and realized the level it is aimed at I am not surprised that you are confused.An emf is induced whether the circuit is open or closed so the answer is A.Answers B,Cand D can only apply if a current flows but since the coil is on open circuit there is no current.If you want to be really fussy and go beyond the level of the question you could argue that there is an extremely tiny current due to parasitic capacitance etc which others here have referred to.You could also complain that the question itself is faulty and discriminates against the more deep thinking student who realizes that there could be a tiny current.Should it be an exam question,the examiners normally brush off such complaints by stating that students were instructed to pick the most appropriate answer.

Claude the current induced here,whether it be on open or closed circuit,is dc not ac.Also, there is no current in this or any other similar system when the emf reaches a steady value.In principle a steady emf can be achieved with this system by pulling the magnet at an increasing speed such that the rate of change of flux remains constant.

The current may hold at a steady value for a brief time as the magnet passes through, but thus is hardly what we'd call "dc". True dc involves no time factor. If the emf reaches a "steady value" that means that the flux is a ramp function wrt time unlimited. The flux would literally have to linearly ramp towards infinity to get the steady emf (dc) that you describe. How can this happen by passing a magnet through a solenoid?

The flux will change with time resulting in an induced current and voltage. Even open the capacitance is present, so a tiny current is present. The situation you describe cannot be attained with a person passing a bar magnet through a coil. To get a dc emf requires a time ramp of flux unbounded. Pretty hard to do manually. This has been known for over 135 years.

Claude

 

[Dadface]

The current may hold at a steady value for a brief time as the magnet passes through, but thus is hardly what we'd call "dc". True dc involves no time factor. If the emf reaches a "steady value" that means that the flux is a ramp function wrt time unlimited. The flux would literally have to linearly ramp towards infinity to get the steady emf (dc) that you describe. How can this happen by passing a magnet through a solenoid?

The flux will change with time resulting in an induced current and voltage. Even open the capacitance is present, so a tiny current is present. The situation you describe cannot be attained with a person passing a bar magnet through a coil. To get a dc emf requires a time ramp of flux unbounded. Pretty hard to do manually. This has been known for over 135 years.

Claude

I think we are at cross purposes here Claude.It is interesting to discuss what happens when the magnet passes through the coil but I wasn't referring to this ,I was referring to the question, this having the the magnet being pulled away from the coil.I agree that it is hard to get a steady emf with the system referred to in the question, that's why I stated that it can be done in principle.

 

[xunxine]

The question is supposedly aimed at GCE O Levels. I guess it's suffice to give answer as A at this level? I understand that at higher level, the question itself can be questioned. Much of the later explanations are beyond me. By the way, what's mmf? Maybe it's better to change the question altogether.
Thanks much!

Jam Jay

 

[cabraham]

The *mmf* is "magnetomotive force". The mmf and the emf cannot exist independently. They are either both zero, or both non-zero. In the ac domain there is no such thing a a true short or true open. There is always inductance and capacitance, hence there will be a current even when "open", and a voltage even when "shorted".

What amazes me is how everyone from hobbyist to EE pro, understands and is aware of "emf", but only a few, mostly EE practitioners with power experience, are aware of "mmf". In any university EE curriculum, the 2 are taught in unison in any energy conversion class. The mmf concept is equally as important as the emf concept, no more or less.

Whenever a magnetic flux changes, both mmf and emf are produced in unison. Neither is the "cause" of the other. Both are inclusive, and neither is more fundamental. Also. the term "force" is used colloquially here. Neither mmf nor emf is actually a force.

Claude