I apologize that this is a not a normal post (i.e. not using MathJax but instead posting a video).

In this video (timestamped properly) the physics teacher claims that there is a magnetic force pointing leftwards due to the moving current in the right section of the circuit. However, as this current moves counter-clockwise around the loop, in practice all magnetic forces should cancel out. This is because the current going through the rightside of the loop also goes equally through the leftside of the loop but in opposite direction. As the opposite direction produces a negative cross product relative to the original magnetic force calculation, their equal magnitudes should cancel out through superposition.

Because of this cancellation, people use the righthand curl method to approximate the magnetic force direction of counter-clockwise current loops.

I am self-studying E&M and have been using this great youtuber, but this time I am in contradiction of his logic. Can someone offer knowledge to answer why his assumptions seem to produce accurate calculations in the subsequent minutes?

It is the external magnetic field out of the paper (red crosses) which is interacting with the magnetic field produced by the induced current moving upwards in the right hand conductor.

The magnetic field produced by the current passing through the left hand conductor and resistor $$R$$ will be so much weaker than the external field that it will not affect the force on the right hand conductor by very much at all.

Update as a result of a comment from @ebehr

The force on the right-hand conductor is due to the iteration of the external magnetic field and the magnetic field produced by that induced current.
If that was the only force which was acting to the right-hand conductor, the conductor would slow down ie $$v$$ would decrease and the kinetic energy of the conductor would be decreasing.
In terms of energy that loss in kinetic energy of the conductor would equal the electrical energy produced bu the arrangement which in turn would produce heat due to ohmic heating.

In order to keep the right hand conductor moving at constant speed an external force, equal in magnitude and opposite in direction to the force due to the interaction of the external magnetic field and the induced current, must be applied to the right hand conductor.
This force could be you pushing the conductor to the right.
The work that that external force does (eg you) produces the induced current which results in the ohmic heating in the circuit.
So you are doing work (using chemical energy - food) to generate electrical energy which in turn is converted to heat.

• I really appreciate this answer as it may have highlighted a misunderstanding I had. From my initial beliefs, I was assuming that the rightside conductor was completing the circuit including the lefthand conductor and resistor R. If that were true then the current going through the rightside should have been equal to the current through the lefthand conductor. This is merely because current is constant through a series circuit. And if those assumptions were true, than the magnetic field produced by the current in the right hand and left hand conductor should be equal but opposite. – ebehr Apr 21 '19 at 11:11
• I'm sorry if I have become even more confused. But if the things I wrote in my previous comment were true, than wouldn't it not matter how small the magnetic field produced from the induced current relative to the external field was because in the end as they are forming a loop all vector forces cancel out barring the one pointing out of the paper (as in the righthand curled current mnemonic with the thumb being the magnetic field). Thus, how could there be a force pulling leftward on the right hand conductor? – ebehr Apr 21 '19 at 11:20
• @ebehr The magnetic fields due to the left hand and right hand conductors are in the same direction within the conducting loop. – Farcher Apr 21 '19 at 11:35
• Would it be correct to say that the fields are opposite in direction and same in magnitude (as they are produced by same current), but the left one is tiny (relatively) and far away so that it doesn't really affect the force we experience as we pull the righthand conductor away? If so, thank you I now understand this material! – ebehr Apr 21 '19 at 11:39
• Oh, I understand it now. farside.ph.utexas.edu/teaching/316/lectures/node88.html This post makes a distinction that the velocity of charge from current is so slow that its field is neglible and the force we account for is the that caused by the interactions of the moving magnetic field and external field caused by the velocity as the right conductor is pulled rightward. I think my misunderstanding was that the video pointed out the "wrong" field that does work. – ebehr Apr 21 '19 at 12:11

I now grasp my confusion. The video author is correct that the movable wire feels a drag magnetic force, and by going against it converts mechanical energy into electrical energy in the form of an induced EMF. The drag force is $$F_M=I\vec{B}x\vec{L}$$. The work done to counteract this force adds energy to the system which does work on any charges inside of the conductor causing a current to flow.

I was indeed correct to realize that the loop formation of the circuit meant that that same current would be going in the opposite direction on the other side, producing a magnetic field and force of opposite direction but same magnitude. But, I was misaimed to believe that this should negate the first paragraph's assertions. Although there is a magnetic force on the left side of the conductor which will have a field that reaches the moving part of the conductor, the left side's field strength will not cancel much at all of the drag force on the right side. I must remember that the charges are moving so slow that they experience little magnetic force, and that this tiny magnetic force would become even smaller over distance.