# How do you calculate the back emf?

I am trying to compute the back emf in a solenoid of radius $a$, with length $l$ and $N_1$ turns and a driving current $$I(t)=I_1\cos(\omega t).$$

I thought it would be a good idea to begin by computing the induced emf in the solenoid, so given that the magnetic field inside the solenoid is $\vec{B}=\mu_0nI_1\cos(\omega t)$, the flux is $$\Phi_B=\vec{B}\cdotp\vec{A}=\mu_0 n \pi a^2 I_1\cos(\omega t).$$

Then the induced emf is the time derivative of the magnetic flux, so $$\epsilon=\mu_0\frac{N_1}{l}\pi a^2I_1\omega\sin(\omega t).$$

The plan was then to somehow use this to compute the back emf, since the back emf makes up for the change in magnetic field. However, I am stuck. On the solutions it says that the back emf is $$\epsilon=\mu_0\frac{N_1^2}{l}\pi a^2I_1\omega\sin(\omega t),$$ which is my answer for the induced emf multiplied by the number of turns of wire around the solenoid.

How has this answer been obtained?

When calculating the flux through a coil of wire (such as your solenoid), you need to multiply the flux from one turn (your $\Phi_B$ above) by the number of turns. This assumes that each turn of the coil encloses the same flux (i.e., field lines don't pass through the "sides" of the coil), which is a pretty good approximation for a long coil.
Once you multiply $\Phi_B$ by $N_1$ to account for this, your result agrees with the book's result.