In all both my physics textbooks, the number of loops $N$ is left out of the equation for magnetic flux $\phi_B = \int_s \vec B \cdotp\hat n dA$ and only when calculating the Emf induced by an inductor is the number of turns taken into account.

That leads to the question : does the number of loops in a coil effect the magnetic flux?

Consider the following setup. You have a coil of 5 turns/loops placed perpendicularly to a magnetic field. There is no current going through the coil. With the magnetic field, area of the coil, and position of the coil kept constant, the number of loops/turns in the coil is increased to 50. Is an Emf generated across the coil?

In other words, if you were to change the number of loops in a coil, would the magnetic flux change (thus inducing an Emf)?

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    $\begingroup$ You are not explaining the scenario very well. Yes, for a coil with a current, the number of turns influences the flux by changing the magnetic field. $\endgroup$
    – ProfRob
    Commented Nov 22, 2015 at 0:40
  • $\begingroup$ I've added an example for the particular senario am now wondering about. Is that more clear? Thank you for your comment. Atleast half my question is now answered. $\endgroup$ Commented Nov 22, 2015 at 0:55

1 Answer 1


Yes, if you go around N times then the emf will be N times greater than if you went around once.

Why? Suppose one loop bounds a surface with area A, then N loops will bound a surface with area perpendicular to the field NA. Imagine you're looking directly through the solenoid and the surface it bounds is made out coloured glass, how many layers of glass are you looking through? (Answer: N)

Why don't your books include it? Actually they do! It's the the "S" term, i.e. the surface that your integrating over.

What about the E field? The E field doesn't change because although your emf increases by N, the tangential distance of the curve you integrate over also increases by N. The electric field doesn't care what curve you integrate over, or if you call it one big wire or many loops. The electric field is going to do what it wants.

  • $\begingroup$ Its in the "s" term. The answer was hiding in plain site. Thank you. $\endgroup$ Commented Nov 22, 2015 at 3:07

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