Quantum Spin (4) - Classical Dynamics in Magnetic Field 

if all of the cosines should be sines...Then U = -μ . B. [sin (theta2)-sin (theta1] ..it implies U = -μ x B...i.e Cross product comes into picture ..Then U will be Vector quantity...Where is mistake?...

@@ravuruvasudevareddy3347 You have to be careful about which angle is being talked about. θ in the video has been defined as the angle between B and the square loop of current. However, the vector μ actually points perpendicular to the square, so you have to be careful.

@@noahexplainsphysics got it... Thanks for clearing my doubt..

in 44:19

@@user-jw6ff1qo2o I think you're right, and the figure is not correct. It should have another line for the horizontal, and the angle should be from that horizontal line. That angle between the wire and the horizontal is the same as the angle between the magnetic moment (which is normal to the area of the wire and also not depicted in the figure) and the magnetic field.

Remember that this is a square circuit of wire "edge on." If you imagine the current circulating around the loop, then the current will be travelling in different directions on a pair of opposite of sides. Try drawing a square loop of wire with current going around it, then imagine looking at it from "the side."

@@noahexplainsphysics Thanks a lot

Ah yes... the eternal question. You can find discussions of this in certain textbooks and online as well. I recall that Griffith's EM textbook has a section on it. Imagine having a square loop of current in a magnetic field. Say the current has a magnetic moment vector μ. Imagine trying to physically grab the wire and rotate it so μ changes, changing U = -μ . B. I will leave it to you to verify the following analysis with your own diagrams (using the right hand rule and cross products and so forth). As you are trying to rotate the square, the charges moving around the square will have some VELOCITY simply due to the fact that the wire's position is changing with time. Now, the charges moving around the square always have some velocity (just because they are flowing along the wire) but I mean that they'll now have a new component of velocity pointing in the direction that the wire is physically being pushed. If you then use the Lorentz force law to to find the direction of the force from the external magnetic field on these circulating charged particles, you'll find that the magnetic field will actually try to SLOW DOWN or SPEED UP the particles (depending on the direction you are rotating the square)! Therefore, in this situation, the magnetic field will try to change value "I" of the current itself. However, say that there is some battery keeping the current constant along the wire. This means that when the magnetic field tries to slow down the current, the battery must work extra hard to compensate. Therefore, while the magnetic field directly does no work on the charges, it causes the battery to work extra hard to keep the current constant, and it is the battery that does the work. Likewise, if the magnetic field tries the speed up the current, then the battery has to work less hard, and the same thing happens. So that's all fine and good for dipoles which are loops of current. But what about pointlike electrons, which are "intrinsic dipoles"? Well, for an electron there is no secret "battery" keeping its spin constant. Therefore, we shrug our shoulders and proclaim that the magnetic field cannot do any work EXCEPT on intrinsic dipoles, on which it can do work.

@@noahexplainsphysics So, it's not completely right to say that -DeltaU is the work done BY the magnetic field, right? Because I think I understand the argument you propose but what always confused me was the fact that people insist on calling it work done by a magnetic field when it is very much not (in my understanding)! One of my big flaws is the infinite desire for rigor, otherwise I don't feel like I really understanding. + Is there some advanced explanation on why the magnetic field should have such an exception? Thank you

@@luca7253 I would say that you need to consider two cases: 1. If your magnetic dipole is a loop of current, then yes, the magnetic field DOES NOT do work. The change in energy U = - μ . B is fundamentally caused by the battery which keeps the current "I" constant. 2. If your magnetic dipole is an "intrinsic dipole," then the magnetic field DOES do work. This is because the full force law on an intrinsic dipole is F = ∇(μ . B) + q(E + v x B) Notice the dipole term out front. While the q v x B term cannot do work, the ∇(μ . B) term can do work. I discuss this force law in parts 7 and 8 in the series. Imagine you have two ferromagnets which start out separated by some distance, but then attract each other and stick together. Clearly, work has been done! This is fundamentally because of the dipole term, because ferromagnets are just a bunch of intrinsic dipoles which are aligned. (However, at a certain point this all becomes semantics because there is no well-defined concept of "work" in quantum mechanics.)