This is a question on magnetic flux and induced voltage. I believe that this problem's answer key is incorrect, and I want to determine whether my opinion is sound. I also want to explore one sub-question of the problem.

  1. How can a bar being pulled through a magnetic field have any flux through it? For flux to exist, doesn't a magnetic field need to run through some area? Shouldn't the area also be completely enclosed by a wire?
  2. If the electric field is $V/l$, shouldn't $E$ = $Blv/2l$, where l is the length of one bar in the square figure? If this is true, shouldn't $E = vB/2$?
  3. The solution solves for electric field magnitude by reasoning that $|F_{B}|$ = $F_{E}$. This seems reasonable. In conjunction with the reasoning in my second question, that $ V/l = E$, can some conceptual explanation be understood? I have the impression that something unites the same answer being arrived at through two ways.

Scan of exercise

  • $\begingroup$ Since the magnetic field is constant, even if you had a closed loop of wire moving at some velocity, the change in flux would be zero. This is a simpler Lorentz Force law problem, $F=qv \times B$. When the bar reaches a steady state, magnetic and electric forces must balance. $\endgroup$ – MonkeysUncle Mar 5 '15 at 1:47

To have flux, you do need an area. The area does not need to be enclosed by a wire, however if you have a thin wire, and mobile charges are confined to flow through the wire, then you can relate the change in flux to the emf due to the lorentz force (magnetic and electric) per unit charge in the direction of the instantaneous direction of the wire. So flux exists for any area, but it might not be useful or related to anything.

In your case you have a conducting bar, so I don't see any natural area (unless you give your bar some thickness). And even if you give your bar thickness, then you no longer can equate the emf to a changing flux, that result only holds for thin wires.

But you can move a conductor through a magnetic field, and it will experience a magnetic force. The protons and bound electrons feel a force, the net force on those will be in the opposite direction of the force the mobile electrons feel, and the net effect of those net forces will be to strain the solid lattice due to the stress, so your bar might deform ever so sightly. The mobile charges however will start to accumulate on end (and leave a deficit on the other end). Eventually the charge imbalance will counteract the magnetic force and equilibrium will be achieved. That's part b,c, and d which all look fine (though note that the direction of the magnetic force listed as pointing down is really the direction of the magnetic force per unit charge, but that's fine, it is supposed to counterbalance the electric field which points in the direction of force per unit charge). Part a is not OK, but for that you can instead find the magnetic force as $q\vec v\times \vec B$ and then get a force per unit charge as $v\times \vec B$, also this is only a first approximation, the charges don't move exactly at $\vec v$ moving the the right, when there is a current they also have some motion up an down. For a thin wire there are forces keeping the charge inside the wire and they can (potentially) counter balance any force other in a direction other than the up and down force.

Your question 2 makes no sense whatsoever to me, I see no factor of 2.

As for three, I don't see two methods. I only see the Lorentz force, there is a Lorentz force. The magnetic force on the protons goes in one direction and stresses the lattice, but the lattice fights back. The magnetic force on the mobile charges causes a charge separation, this continues to happen until eventually the charge separation is large enough that the electric force counterbalances the magnetic force. Only after enough time has passed for the charge imbalance to get large enough do the electric and magnetic forces counter balance.

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  • $\begingroup$ Note that the vertical motion of the charges does not produce a vertical force, so your comment about "first approximation" seems superfluous. The force is $\vec{v} \times \vec{B}$, and the vertical force is $v \times B$ without provisos. $\endgroup$ – Floris Mar 5 '15 at 4:19
  • $\begingroup$ @Floris In the steady state when there is no current along the bar, then the magnetic force is up and down. However when you first start moving the bar a Hall voltage has to develop side to side as the current flows during the time the voltage from top to bottom forms. An EMF is defined as the force per unit charge along the instantaneous of the direction of the wire. But the magnetic force is the force, and can and does point in directions other than along the wire. $\endgroup$ – Timaeus Mar 13 '16 at 16:15

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