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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): $$\oint \vec B \cdot d\vec \ell=\mu_0I_{\ell}=\mu_0NI_0(t)\tag{1}$$ 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 $$\Phi =BS=\mu_0 \frac{N}{L}I_0(t)S=4\pi \times 10^{-7}\times 4\times 10^3 \times \pi \times 10^{-4}I_0(t)=16\pi^2\times 10^{-8}I_0(t)$$ $$\approx 1.5 \times 10^{-6}I_0(t)$$ 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 $$\oint \vec E \cdot d \vec \ell= \frac{d\Phi}{dt}=-\mu_0 \frac{N}{L}S\frac{d I_0(t)}{dt}\approx -1.5 \times 10^{-6}\frac{d I_0(t)}{dt}\tag{2}$$ So the total voltage is $$V=0.1\oint \vec E \cdot d \vec \ell\approx -1.5 \times 10^{-7}\frac{d I_0(t)}{dt}$$

This is the wrong answer and the right answer is basically $$\bbox[5px,border:2px solid red]{V=400\oint \vec E \cdot d \vec \ell\approx 6 \times 10^{-4}\frac{d I_0(t)}{dt}}$$


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 are given by

$\bbox[yellow, 5px] {\text{number of turns}\times \text{rate of change of flux} \propto N \times \frac{d\Phi}{dt}\propto \mu_0 \frac{N}{L}S\frac{d I_0(t)}{dt}\times L}$

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

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

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 $$\mathcal{E} = -N\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}\ne -N^2\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}$$ Clearly, I'm missing the point, so if someone could please explain it to me that would be great.

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  • $\begingroup$ Magnetic flux through a surface $\mathcal{S}$ is given by $\Phi_{B} = \int_{\mathcal{S}} \mathbf{B} \cdot d\mathbf{A}$. Do you know why the number of turns $N$ appears in the formula? $\endgroup$ – K_inverse Oct 19 '18 at 5:39
  • $\begingroup$ @K_inverse No, I don't know why. Could you please tell me? $\endgroup$ – BLAZE Oct 19 '18 at 6:03
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I checked Faraday's law of induction so I know that $$\mathcal{E} = -N\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}\ne -N^2\frac{\mathrm{d}\Phi_B}{\mathrm{d}t}$$

In fact $$\mathcal{E} = -N\frac{\mathrm{d}\Phi_B(N)}{\mathrm{d}t}$$ where $\Phi_B(N)\propto N$

The magnetic field due to a solenoid is proportional to the number of turns, $N$
$\Rightarrow$ the magnetic flux through one turn of the solenoidis proportional to the number of turns, $N$
$\Rightarrow$ the magnetic flux through $N$ turns is proportional to the number of turns squared, $N^2$.

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In practical one need to calculate the thickness and weidth, however, notice that many fields have additive properity, or say superposition principle, same with $B, G$ and $E$.

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  • $\begingroup$ Are you sure this is answering the question? $\endgroup$ – BLAZE Oct 19 '18 at 6:04
  • $\begingroup$ @BLAZE I thougth I did, you have the answer in your post, Notice there already $N$ in your $V$. You just lost track of the multiplicaiton of $N$s, if you want, write it in symbolic represtation. $\endgroup$ – user9976437 Oct 19 '18 at 6:09

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