In order for this question to be clear I must first give some context:

Consider a 10cm long solenoid (solenoid A) of radius approximately 1cm with 400 turns.

Let the current in $A$, $I_A$, change with time slowly, $I_A = I_0(t)$ so that you may ignore the displacement current.

i) What is the instantaneous magnetic flux in the solenoid?

The instantaneous magnetic flux in the solenoid is given by $Phi =BS$.

The magnetic field $B$ can be found from Ampere’s law (ignoring displacement current):

$$oint vec B cdot dvec ell=mu_0I_{ell}=mu_0NI_0(t)tag{1}$$

where $I_{ell}=NI_0(t)$ is the total current due to $N$ turns each with current $I_0(t)$

The area is $S=pi times 10^{-4}$m${^2}$

Using $(1)$, $B,L=mu_0 N,I_0(t)$; the magnetic field $B=mu_0 frac{N}{L}I_0(t)$ where $L$ is the *length of the solenoid*. The flux $Phi$ is thus

$$Phi =BS=mu_0 frac{N}{L}I_0(t)S=4pi times 10^{-7}times 4times 10^3 times pi times 10^{-4}I_0(t)=16pi^2times 10^{-8}I_0(t)$$

$$approx 1.5 times 10^{-6}I_0(t)$$

which is the correct answer.

(ii) A voltmeter is connected across the solenoid. The meter will measure

$oint vec E cdot d vec ell$ integrated

along the wire. Derive an expression for the voltage measured (ignoring

the resistance).

By Faraday’s law the induced voltage per unit length is $$oint vec E cdot d vec ell= frac{dPhi}{dt}=-mu_0 frac{N}{L}Sfrac{d I_0(t)}{dt}approx -1.5 times 10^{-6}frac{d I_0(t)}{dt}tag{2}$$ So the total voltage is

$$V=0.1oint vec E cdot d vec ellapprox -1.5 times 10^{-7}frac{d I_0(t)}{dt}$$

This is the wrong answer and the right answer is basically

$$bbox[5px,border:2px solid red]{V=400oint vec E cdot d vec ellapprox 6 times 10^{-4}frac{d I_0(t)}{dt}}$$

On dimensional grounds, it was my understanding that the LHS of Faraday’s law, $(2)$ gives the **unit** circulation of the electric field, or, $unicode{x222F}(nabla times vec E)cdot d vec S$ (by Stokes theorem).

But to find the voltage $V$ **across** the solenoid we must integrate over its length, so that the units given by

$bbox[yellow, 5px]

{text{number of turns}times text{rate of change of flux} propto N times frac{dPhi}{dt}propto mu_0 frac{N}{L}Sfrac{d I_0(t)}{dt}times L}$

But in the box marked red the author is simply multiplying by the number of turns $N$, which means dimensionally the RHS of Faraday’s law is

$bbox[#F85, 5px]{text{number of turns}^2times text{rate of change of flux} propto N^2 times frac{dPhi}{dt}propto mu_0 frac{N^2}{L}Sfrac{d I_0(t)}{dt}}$

So no longer do we have unit circulation but more concerning is that the units go as the square of the number of turns. Is this really correct?

I checked Faraday’s law of induction so I know that $$mathcal{E} = -Nfrac{mathrm{d}Phi_B}{mathrm{d}t}ne -N^2frac{mathrm{d}Phi_B}{mathrm{d}t}$$

Clearly, I’m missing the point, so if someone could please explain it to me that would be great.