In Conquering the Physics GRE 2nd edition, Exam 3 problem 20, the statement is as follows:

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A metal bar is pulled at constant velocity $v\mathbf{\hat{x}}$ along two metal rails a distance $d$ apart connected by a resistor or resistance R, as shown in the diagram. There is a magnetic field, pointing into the page, of magnitude $B = Cx$, where $x=0$ is the initial position of the bar. At time $T$, how much energy has been dissipated in the resistor thus far, as a function of $T$?

The solution is to first find the flux as a function of time, then the EMF, then the power dissipated by the resistor, then the energy. My method is to compute the flux as:

$$\Phi = \iint \mathbf{B}\cdot d\mathbf{S} = \iint (Cx\mathbf{\hat{z}})\cdot(dx\,dy\,\mathbf{\hat{z}}) = \int_0^d dy \int_0^{vt} dx\,Cx = \frac{1}{2}Cd(vt)^2 = \frac{1}{2}Cv^2t^2d$$

However, the solution says "Since the magnetic field is perpendicular to the loop, the flux through the loop is $\Phi = BA = C(vt)(vtd) = C v^2t^2 d$." The errata states that one of the answers was misprinted but that the solution is correct. Am I crazy or is the solution wrong? $\Phi = BA$ should only be true if $B = const.$ correct?


closed as off-topic by John Rennie, Kyle Kanos, ZeroTheHero, Cosmas Zachos, stafusa Dec 27 '18 at 23:06

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  • 2
    $\begingroup$ On the very last page of the linked errata, their correction is precisely what you say here. $\endgroup$ – J. Murray Dec 22 '18 at 22:28
  • $\begingroup$ I didn’t read your question, but I found an error in one of their solutions and just sent them an email. They’re very quick to respond $\endgroup$ – InertialObserver Dec 22 '18 at 22:43
  • $\begingroup$ Oh I did not realize that the errata was separated into versions, I only saw the version 2 correction. Thanks for pointing that out @J.Murray, glad I'm not crazy $\endgroup$ – Kai Dec 22 '18 at 22:52
  • $\begingroup$ It seems to depend on what is meant. Perhaps they mean there is a time changing (but uniform) B field? Because the way you have interpreted it B is not divergence-less (and the problem is also sensitive to the location of the origin, which is not well-specified). $\endgroup$ – Green Apples Dec 22 '18 at 22:58