Undergrad Angular momentum associated with a current carrying circular wire

[gurbir_s]

How should I calculate the angular momentum carried by a current carrying circular wire? Is it correct to consider the angular momentum of the electrons moving with drift velocity? Like
$L = n m_e v_{drift} r$ where $r$ is radius of the loop, and $n$ is total number of electrons moving in the wire?

 

[berkeman]

Can you say more about your setup? You can calculate the forces and torque on such a current-carrying coil in the presence of a B-field, but what do you mean by "angular momentum" in this context? If the coil is free to spin up under the influence of the torque, the angular momentum will increase with time...

Also, is there a commutator in this setup (like with a DC motor)?

 

[vanhees71]

There's angular momentum of both the electrons and the electromagnetic field!

 

[gurbir_s]

Can you say more about your setup? You can calculate the forces and torque on such a current-carrying coil in the presence of a B-field, but what do you mean by "angular momentum" in this context? If the coil is free to spin up under the influence of the torque, the angular momentum will increase with time...

Also, is there a commutator in this setup (like with a DC motor)?

It is simply a resistive wire in the shape of a circle connected to a DC source.
Actually, I read somewhere that such a loop placed in a magnetic field will simply get aligned in the direction of field. But if it carried an angular momentum, it should have precessed around the applied field.

 

[gurbir_s]

There's angular momentum of both the electrons and the electromagnetic field!

How to quantify it?

 

[berkeman]

It is simply a resistive wire in the shape of a circle connected to a DC source.
Actually, I read somewhere that such a loop placed in a magnetic field will simply get aligned in the direction of field. But if it carried an angular momentum, it should have precessed around the applied field.

If it carried some angular momentum before the B-field is turned on? I'm still not understanding the question. The loop will experience a torque to try to align it with the B-field, and depending on the damping of the setup, will oscillate for several cycles during that alignment (when there is no commutator).

 

[gurbir_s]

If it carried some angular momentum before the B-field is turned on? I'm still not understanding the question. The loop will experience a torque to try to align it with the B-field, and depending on the damping of the setup, will oscillate for several cycles during that alignment (when there is no commutator).

Yes. The angular momentum carried by the circular current carrying wire without applying any field.

Just like in the classical picture, the electrons revolving around nucleus, have an angular momentum and give rise to magnetic moment, by that analogy, since the wire loop has magnetic moment, I want to calculate the angular momentum associated with the current flowing through the wire.

 

[berkeman]

Ah, so the angular momentum of the electrons in the conduction band moving at the drift current velocity induced by the voltage applied to the coil terminals...

https://en.wikipedia.org/wiki/Drift_velocity

 

[gurbir_s]

Ah, so the angular momentum of the electrons in the conduction band moving at the drift current velocity induced by the voltage applied to the coil terminals...

https://en.wikipedia.org/wiki/Drift_velocity

Yes. Sorry, it took me so long to make the question clear.

However, I can't find any reference to the angular momentum in the above mentioned Wikipedia article.

 

[nasu]

Actually, the magnetic moment of the current loop can be seen as the magnetic momentum of the electrons moving around the loop with the drift velocity. If we consider a loop of radius R, made from wire with cross section S, we can start with the magnetic moment $$\mu=IA$$
and express the current, I, as $I=nev_d S$ and the area of the loop as $A=\pi R^2$. Considering that the volume of the wire is $V_w= 2\pi R S$, we'll have$$ \mu=(nev_d a) (\pi R^2) = (2\pi R S) n \frac{e}{2m} (m v_d R) = N \frac{e}{2m}(m v_d R)$$ where N is the total number of electrons, $N=n V_w$ and $\frac{e}{2m}$ is the gyromagnetic factor for the orbital motion of the electrons.
So, it looks like $\mu=\gamma N l$ or $\mu=\gamma L$ where l is the angular momentum for one electron and L is for all of them.