Faraday's law: sin and cos? I am looking at this paper (Hanna S. A., Varhue W. J. and Titcomb S., IEEE Trans. on Instrumentaion and Measurement, Vol. 58, No. 1, 2009). They claim that the voltage generated in a loop of $N$ turns and diameter $d$ when placed in a time varying magnetic field is:
$V\propto B_0j\omega e^{j\omega t}$
where $j$ is the imaginary unit. Is this correct? Specifically, I thought that:
$V\propto B_0\omega \cos{\omega t}$.
I don't know how did they have a magnetic field with $\cos$ and $\sin$ terms, i.e. $e^{j\omega t}$.
 A: Usually, when something varies harmonically in time (i.e., as $\sin \omega t$ or $\cos \omega t$), we write it with a complex exponential, because it's easier to work with. Suppose we have a (real) signal $V(t) = V_0 \cos (\omega t + \delta)$. We can write as $V(t) = V_0 e^{i(\omega t + \delta)} = V_0 e^{i\delta} e^{i\omega t} = \hat{V_0} e^{i \omega t}$, where $\hat{V_0} = V_0 e^{i\delta}$ is a complex constant; it has to be complex to allow the possibility of a phase. 
When we write things like this, it is understood that the actual physical quantity is the real part of the complex expression. This isn't a problem since most of the operations we do to signals (like addition, multiplication by constants, differentiation, integration, Fourier transform, etc.) are linear, and so it makes no difference whether you take the real part first and then apply the operation or you do it the other way around. The only exception to this is when multiplying complex quantities together, since the real part of a product isn't the product of the real parts.
As for your question in the comments, the reason the magnitude is constant is that the time dependent part of $V$ is $e^{i\omega t}$, which has magnitude equal to one, so $|V(t)| =  |V_0|$, which is simply the amplitude of the signal.
