When a magnet is moved into a coil of $50$ turns at $0.12$ m/s, an E.M.F. of $3.60$ V is generated into the coil. What will be the E.M.F. generated when the magnet is removed from the coil at $0.48$ m/s?

(This question has been quoted verbatim.)

My answer : $-14.4$ V
Book's answer :$0.90$ V

As far as I understand, the induced E.M.F. should be proportional to the rate of change of magnetic flux :

E.M.F. $= - \frac{d \left( N \Phi \right)}{d t}$

So, as the speed increased by a factor of $4$, that is, the rate of change of flux has also increased by a factor of $4$, shouldn't the induced E.M.F. also be $4$ times as much, but in the opposite direction since the bar magnet is now being removed instead of moving into the coil?

Apparently, the correct answer according to the book is $0.90$ V, which is $4$ times lesser than the original induced E.M.F., which seems impossible. Am I missing out something crucial? Any help would be highly appreciated.

Thank you.

  • $\begingroup$ and the book is....? $\endgroup$
    – JEB
    Mar 19, 2022 at 19:41
  • $\begingroup$ how about if neither answer is correct? Your answer was $-14.4V$ not$+$. $\endgroup$
    – hyportnex
    Mar 19, 2022 at 21:38
  • $\begingroup$ The book is "Pearson Edexcel International A Level Physics Student Book 2" by Miles Hudson, and the question can be found on page 70. $\endgroup$ Mar 20, 2022 at 7:31
  • $\begingroup$ Assuming the magnet leaves the coil from the opposite end to where it entered, then yes, +14.4V would make sense. Thank you. $\endgroup$ Mar 20, 2022 at 7:34

1 Answer 1


With the understanding that we are talking about maximum value of a voltage pulse, I would tend to agree with your answer.


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