# A question about induced current in LC circuits

• B
• mymodded
mymodded
TL;DR Summary
when a capacitor starts discharging, why is the induced magnetic field in the inductor in the same direction of the current? isn't it supposed to resist changes in current?
Let's say we have an LC circuit and we begin with the capacitor fully charged, it then starts discharging, and currents begins to flow in the circuit, and a magnetic field is generated in the inductor because of the current flowing through it. My question is, why is the magnetic field in this direction (figure b)? this magnetic field implies that the induced current is in the same direction as the current coming from the discharging of the capacitor, but the inductor is supposed to resist changes in current, so shouldn't its magnetic field be in the opposite direction? same with figure f.

Sorry, that figure just doesn't make sense to me either. Maybe if you showed us the source of that we could sort it out.

I guess you mean the potential (voltage) across the inductor, not really the field. The inductor voltage is the same as the capacitor voltage, it's not "induced" it's applied externally by the capacitor. Then the current in the loop is determined by the inductor equation ##i(t) = \frac{1}{L} \int_0^t v(\tau) \, d\tau + i(0)## or ## v(t) = L \frac{di(t)}{dt} ##.

I would look for another source to explain this circuit. Khan Academy is pretty good, but there are hundreds of others on the web.

The bar graphs do not show polarity.
There is no reference ground marked.

The red arrows, B, consistently point in the same direction as the current. But the polarity of B, will depend on whether the inductor is wound with a left-hand or a right-hand helix, and if the flux shown is internal or external to the solenoid.

DaveE said:
Sorry, that figure just doesn't make sense to me either. Maybe if you showed us the source of that we could sort it out.

I guess you mean the potential (voltage) across the inductor, not really the field. The inductor voltage is the same as the capacitor voltage, it's not "induced" it's applied externally by the capacitor. Then the current in the loop is determined by the inductor equation ##i(t) = \frac{1}{L} \int_0^t v(\tau) \, d\tau + i(0)## or ## v(t) = L \frac{di(t)}{dt} ##.

I would look for another source to explain this circuit. Khan Academy is pretty good, but there are hundreds of others on the web
sorry I think I didn't make my question very clear, I was talking about the polarity of the inductor, that in figure b it should have been positive in the lower side and negative in the upper side, and in my textbook's circuit, that arrow should indicate something about the polarity (even though it shows the field lines). I looked at many sources, and almost nobody talked about the polarity of the inductor. Libretexts pretty much explained it the same way my textbook did. And I noticed something, my textbook and libretexts' current graphs start negative, while the one at khan academy starts positive

mymodded said:
I didn't make my question very clear, I was talking about the polarity of the inductor
Umm... still not clear to me. Inductors rarely have polarity, they work bidirectionally. The polarity is in the equation. If you apply a voltage across the inductor, the current will start to change, increasing the flow from the positive voltage side to the negative voltage side. If you switch the voltage polarity the direction of the current change with switch with it. For example:

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