I know Faraday-Newmann-Lenz's law is...(easy form)

$$\mathcal E_{\text{inducted}}=-\frac{d\Phi_S(\mathbf{B})}{dt} \tag 1$$ The flow through a Gaussian surface is given by a surface integral through $S$, $$\Phi_S(\mathbf{B})\equiv\int_S \mathbf{B} \cdot d\mathbf{a}$$

Why the relation $$\boxed{\frac{dB(t)}{dt}=-kI(t)} \tag 2$$ derives from Faraday-Henry's law that given the induced electromotive force $\mathcal E_{\text{inducted}} $ in a spiral of surface $S$? Note that $k$ is a positive dimensional constant.

For the $(2)$ I have not a flux for $B$! What is the proof of $(2)$: I don't remember doing something like this when I was in my University. That was 23 years ago.

I'm adding the image just to show users the exercise (which doesn't care if someone executes it) where there is the green expression that I have called the (2).

enter image description here

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    $\begingroup$ What is $k$? And I doubt this law is true for any arbitrary surface, so I assume the surface $S$ is a constant? $\endgroup$ – Philip Jun 9 at 21:17
  • $\begingroup$ @Philip Very kind Philip I have edited my question for $k$. I don't know the terms of this law when it applies or not. But I don't remember studying it. $\endgroup$ – Sebastiano Jun 9 at 21:30
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    $\begingroup$ So where did you see it? $\endgroup$ – mike stone Jun 9 at 21:32
  • $\begingroup$ @mikestone Into an exercise for the high school to solve it. If you want I add the image into my question for the (2). $\endgroup$ – Sebastiano Jun 9 at 21:34

I've actually never seen this form of the law before, so if you have a reference, please let me know. I just worked out how to derive it from the "standard" form of Faraday's Law you've given at the start of your question, but I'm going to make a couple of assumptions, let me know if they aren't justified:

  1. The magnetic field is constant in space, it only varies in time,
  2. The area of the spiral or coil is fixed and unchanging,
  3. The constant $k$ depends on the parameters like the resistance of the wire $R$ and the area of the loop $S$.

(If anyone can figure out how to do it without these assumptions, I'd be very interested to know.)

Given these assumptions, the flux simply reduces to the component of the magnetic field along the surface, since $\vec{B}$ is constant over $S$ and so $\Phi_B = \oint \vec{B}\cdot\text{d}\vec{S} = \vec{B}\cdot\vec{S} = B S \cos{\theta}$, where $\theta$ is the angle between the magnetic field and the area $S$.

Using this in Faraday's Law, $$\mathcal{E}_\text{induced} = -\frac{\text{d}}{\text{d}t}\left( B S \cos{\theta}\right) = -\frac{\text{d}B}{\text{d}t}\times \left(S \cos{\theta}\right).$$

Now, this is the induced electromotive force (emf) in the loop, the actual current that flows through it will depend on its resistance. The current flowing through the loop will thus be (from Ohm's Law) $$I(t) = \frac{\mathcal{E}_\text{induced}}{R}.$$


\begin{equation*} \begin{aligned} I(t) = \frac{\mathcal{E}_\text{induced}}{R} = -\frac{\text{d}B}{\text{d}t}\times \left(\frac{S \cos{\theta}}{R}\right) \end{aligned} \end{equation*}

Rearranging, I get: $$\frac{\text{d}B(t)}{\text{d}t} = - \left(\frac{R}{S\cos{\theta}}\right) I(t).$$

I suppose the constant $k$ should then be $$k = \frac{R}{S\cos{\theta}},$$

which is indeed positive, as $0<\theta<\pi/2$.

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  • $\begingroup$ +1 surely. The exercise is an italian language. $\endgroup$ – Sebastiano Jun 9 at 21:40
  • $\begingroup$ I have adding the photo: I hope that the formula into the green rectangle is visible. $\endgroup$ – Sebastiano Jun 9 at 21:45
  • $\begingroup$ Yep. I barely understand Italian, but it looks like all my assumptions are given in the question ;) In fact, (unless I'm wrong) they say that the field is perpendicular to the surface, so $\theta=\pi/2$, so it's even simpler. Whoops, now this looks like a homework-style problem! $\endgroup$ – Philip Jun 9 at 21:47
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    $\begingroup$ Yes :-) naturally :-) done now; it's very rare that I don't put positive votes and green check marks :-) I am very happy when others are happy and help me :-). $\endgroup$ – Sebastiano Jun 9 at 22:02
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    $\begingroup$ I noticed :) I'm glad I could help! $\endgroup$ – Philip Jun 9 at 22:19

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