Problem 9 Solution
1. Understanding Kinetic Inductance:
When a current changes in a conductor, not only does the magnetic field change (self-inductance), but the charge carriers (electrons) must also be accelerated. The inertia of these electrons opposes the change in current, acting like an additional inductance.
The force equation for an electron of mass $m$ and charge $e$ in an electric field $E_{ind}$ is: $$ F = -e E_{net} = m \frac{dv_d}{dt} $$ Where $v_d$ is the drift velocity.
When a current changes in a conductor, not only does the magnetic field change (self-inductance), but the charge carriers (electrons) must also be accelerated. The inertia of these electrons opposes the change in current, acting like an additional inductance.
The force equation for an electron of mass $m$ and charge $e$ in an electric field $E_{ind}$ is: $$ F = -e E_{net} = m \frac{dv_d}{dt} $$ Where $v_d$ is the drift velocity.
2. Relating to Current:
The current is $I = n A e v_d$, so $v_d = \frac{I}{n A e}$. Differentiating with respect to time: $$ \frac{dv_d}{dt} = \frac{1}{n A e} \frac{dI}{dt} $$ Substituting this into the force equation: $$ -e E_{net} = m \left( \frac{1}{n A e} \frac{dI}{dt} \right) \implies E_{net} = – \frac{m}{n A e^2} \frac{dI}{dt} $$
The current is $I = n A e v_d$, so $v_d = \frac{I}{n A e}$. Differentiating with respect to time: $$ \frac{dv_d}{dt} = \frac{1}{n A e} \frac{dI}{dt} $$ Substituting this into the force equation: $$ -e E_{net} = m \left( \frac{1}{n A e} \frac{dI}{dt} \right) \implies E_{net} = – \frac{m}{n A e^2} \frac{dI}{dt} $$
3. Loop Equation:
Integrating around the ring of radius $r$ (circumference $2\pi r$): $$ \oint E_{net} \cdot dl = – \frac{m}{n A e^2} \frac{dI}{dt} (2\pi r) $$ The total EMF in the loop is due to the changing magnetic flux (Faraday’s Law) and the self-inductance back-EMF: $$ \mathcal{E}_{loop} = -\frac{d\Phi_B}{dt} – L\frac{dI}{dt} $$ Equating the field integral to the EMFs: $$ -\frac{d\Phi_B}{dt} – L\frac{dI}{dt} = \frac{2\pi r m}{n A e^2} \frac{dI}{dt} $$ Rearranging terms to group $dI/dt$: $$ -\frac{d\Phi_B}{dt} = \left( L + \frac{2\pi r m}{n A e^2} \right) \frac{dI}{dt} $$
Integrating around the ring of radius $r$ (circumference $2\pi r$): $$ \oint E_{net} \cdot dl = – \frac{m}{n A e^2} \frac{dI}{dt} (2\pi r) $$ The total EMF in the loop is due to the changing magnetic flux (Faraday’s Law) and the self-inductance back-EMF: $$ \mathcal{E}_{loop} = -\frac{d\Phi_B}{dt} – L\frac{dI}{dt} $$ Equating the field integral to the EMFs: $$ -\frac{d\Phi_B}{dt} – L\frac{dI}{dt} = \frac{2\pi r m}{n A e^2} \frac{dI}{dt} $$ Rearranging terms to group $dI/dt$: $$ -\frac{d\Phi_B}{dt} = \left( L + \frac{2\pi r m}{n A e^2} \right) \frac{dI}{dt} $$
4. Integration:
Integrating both sides with respect to time from the initial state (Angle 0, $I=0$) to the final state (Angle $90^\circ$, Current $I$).
The change in magnetic flux $\Delta \Phi_B$ as the ring rotates $90^\circ$ (from parallel to field to perpendicular): $$ \Delta \Phi_B = B \cdot (\pi r^2) – 0 = B \pi r^2 $$ (Note: The effective flux change inducing current is the cut of field lines). $$ \int d\Phi_B = \left( L + \frac{2\pi r m}{n A e^2} \right) \int dI $$ $$ B \pi r^2 = \left( L + \frac{2\pi r m}{n A e^2} \right) I $$
Integrating both sides with respect to time from the initial state (Angle 0, $I=0$) to the final state (Angle $90^\circ$, Current $I$).
The change in magnetic flux $\Delta \Phi_B$ as the ring rotates $90^\circ$ (from parallel to field to perpendicular): $$ \Delta \Phi_B = B \cdot (\pi r^2) – 0 = B \pi r^2 $$ (Note: The effective flux change inducing current is the cut of field lines). $$ \int d\Phi_B = \left( L + \frac{2\pi r m}{n A e^2} \right) \int dI $$ $$ B \pi r^2 = \left( L + \frac{2\pi r m}{n A e^2} \right) I $$
Final Answer:
$$ I = \frac{\pi r^2 B}{L + \frac{2\pi r m}{n A e^2}} $$