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

• mymodded
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 $$\frac{\mu_{0}^{2} n^{3} l A_{2} i_{0} \omega sin(\omega t)}{R}$$

Last edited:
mymodded said:
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.]

Last edited:
mymodded

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