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I've recently been reading the Griffiths Electrodynamics book, and came across this problem (7.17, 4th edition) and its solution:

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I don't have any objections for the solution to part a). I have, however, a hard time justifying the solution to part b). Shouldn't the magnetic field (and hence the magnetic flux) by the solenoid at a distance far away from the loop be 0? And therefore shouldn't the change in flux be $\Delta\Phi = \pi{a^2}\mu_0nI - 0 = \pi{a^2}\mu_0nI$ instead of $\Delta\Phi = \pi{a^2}\mu_0nI - (-\pi{a^2}\mu_0nI) = 2\pi{a^2}\mu_0nI?$

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Oh, it seems that it was an errata that was fixed 8+ years ago! Somehow never noticed.

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As the solenoid moves toward the left, we can imagine it to be a solenoid with $-n$ turns per unit length, therefore the final flux is $-\pi{a^2}\mu_0nI$.

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  • $\begingroup$ I don't see how moving the solenoid could turn the number of turns negative... $\endgroup$
    – davidaddisonsenjaya
    Commented Feb 21 at 7:34

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