Is this a mistake in my textbook's answer about induced voltage question?

[mymodded]

Homework Statement

A long solenoid with cross-sectional area A_1 surrounds another long solenoid with cross-sectional area A_2 < A_1 and resistance R. Both solenoids have the same length and the same number of turns. A current given by $i=i_{0}cos(\omega t)$ is flowing through the outer solenoid. Find an expression for the magnetic field in the inner solenoid due to the induced current.

Relevant Equations

$\Delta V_{ind} = -\frac{d\Phi}{dt}$
$B_{solenoid} = \mu_{0} n i$

My textbook solved it by first finding the induced voltage in the inner solenoid but they found it by saying $-\Delta V_{ind} = A_{2} \frac{d\Phi}{dt}$, but they did not include the number of turns in the solenoid, but I think they should have done that. their final answer is $\Large \frac{\mu_{0}^{2} n^{2} A_{2} i_{0} \omega sin(\omega t)}{R}$ but I think the right answer should be $$\frac{\mu_{0}^{2} n^{3} l A_{2} i_{0} \omega sin(\omega t)}{R}$$

 

[TSny]

their final answer is $\Large \frac{\mu_{0}^{2} n^{2} A_{2} i_{0} \omega sin(\omega t)}{R}$ but I think the right answer should be $$\frac{\mu_{0}^{2} n^{3} l A_{2} i_{0} \omega sin(\omega t)}{R}$$

The answer should be proportional to $n^2$, not $n^3$. Show the details of your calculation so we can help you identify any mistakes.

The answer that was provided to you has some typographical errors, but the $n^2$ is correct.

[EDIT: Nevermind, I was thinking of finding the current in the inner solenoid. You are correct for the induced magnetic field in the inner solenoid.]