How to determine the voltage across an LR circuit?

Question

Imagine a circuit with an inductor and a resistor rotating in a constant magnetic field. What is the relationship between the voltage induced across it's inductor ,resistor individually and it's generator voltage at a particular instance in time?and why?

Answer 1

I give an answer based on the link figure which OP provided in the comment, to explain the rotation of the phase diagram.

First a quick solution to the circuit with a sinusoidal applying voltage, $V_g = V_0 \cos(\omega t)$: \begin{align} V_R + V_L &= V_g.\\ i(t) R + L\frac{d i}{dt} &= V_0 \cos \left(\omega t\right). \tag{1}\\ \end{align}

Then assume a solution $i(t) = i_0 \cos \left(\omega t - \theta\right)$, the angle $\theta$ is called the phase delay. Plug this solution into Eq.(1), solving two unknown parameters $i_0$ and $\theta$. We have two equations, one for $\cos\left(\omega t\right)$, and the other for $\sin\left(\omega t\right)$ \begin{align} R i_0 \cos \left(\omega t - \theta\right) - L i_0 \omega \sin \left(\omega t - \theta\right) &= V_0 \cos \left(\omega t\right).\tag{2}\\ i_0 R\cos\theta + i_0 \omega L\sin\theta &= V_0; \tag{3}\\ i_0 R \sin\theta - i_0 \omega L\cos\theta &= 0. \tag{4} \end{align}

Parameter $i_0$ and $\theta$ can be solve using equations (3) and (4). And according to the linked figure, the equation (2) can be interpreted as $V_R$ in the $\hat x$ axis and $V_L$ in the $\hat y$ axis, their vector sum is always equal to the voltage of the generator, which is with a phase $\theta$ ahead of the current voltage. These three vectors are coherently rotate with angular frequency $\omega$ driven by the applying alternated voltage. This is called a phase diagram.