Why is the Potential at the Right Higher in an Inductive Circuit?

[quietrain]

ok, let's say we have a battery connected to a resistor and an inductance in series

-|battery|+______ 00000(inductor)______resistance(R)

if the rate of change of current is decreasing , then the induced EMF in the inductor would be towards the right? because they would want to strengthened and thus oppose the change.

so since the induced emf is towards the right, why is the potential at the right higher? shouldn't the left be higher? so that the induced emf is from higher(left) to lower(right) ?

thanks

 

[Naty1]

I not sure I understand the description you give. But I think you can answer your own question this way:

At t = 0^-, the voltage across the inductor and the resistor is zero...right?? when a switch is closed and the battery connected, at t = infinity what is sign the voltage across the resistor...plus to battery plus right??...

so now you can tell how the voltage must have been changing across the inductor...increasing on the right...in other words, part of the voltage drop during the period between zero and infinity is across the inductor and part across the resistor...totalling the battery voltage...so voltage on the inductor is plus on the right.

 

[quietrain]

At t = 0^-, the voltage across the inductor and the resistor is zero...right?? when a switch is closed and the battery connected, at t = infinity what is sign the voltage across the resistor...plus to battery plus right??...

er, i am not sure what you mean by resistor plus to battery plus?

i think it should be -ve end of battery connected to the right side of the resistor. the left of the resistor is connected to the inductor

so basically, i rephrase my question,
-if current is decreasing,
-then an induced emf will be set up in the inductor to oppose the change in the drop of current
-so, it would act in a direction towards the right to strengthen the decreasing current
-with higher potential at left side of inductor and lower potential at right side of inductor?
-so shouldn't the left side of the inductor be at a higher potential? and right side lower, so that the EMF will be pointing towards the right?

 

[quietrain]

i quote the following abstract from masteringphysics.com

"In sum: when an inductor is in a circuit and the current is changing, the changing magnetic field in the inductor produces an electric field. This field opposes the change in current, but at the same time deposits charge, producing yet another electric field. The net effect of these electric fields is that the current changes, but not abruptly. The "direction of the EMF" refers to the direction of the first, induced, electric field."

ok , so let's say current is increasing, di/dt >0 ,

-battery+ _____ (left)inductor(right)

ok, so since current is towards right, -->, and increasing, an induced emf will be set up on the inductor to oppose this change, by the magnetic field due to the changing current, so it will be directed to the left <--- , so shouldn't the (right) side of the inductor be at a higer potential compared to the (left) side of the inductor? since the direction of EMF refers to the direction of the first , induced, electric field?

 

[sardar]

for the explanation

I can explain this phenomenon using the principles of electromagnetism. When a circuit is connected to a battery, it creates a flow of electrons which we call current. This current creates a magnetic field around the inductor, which is a coil of wire. When the current starts to decrease, the magnetic field also starts to decrease. This change in the magnetic field induces an opposing EMF (electromotive force) in the inductor, which tries to maintain the current flow and therefore opposes the decrease in current.

Now, according to Faraday's law, the induced EMF is directly proportional to the rate of change of the magnetic field. So, as the magnetic field decreases, the induced EMF also decreases. This means that the potential difference (voltage) across the inductor is lower on the right side where the current is decreasing, and higher on the left side where the current is still flowing from the battery. This is because the induced EMF is trying to maintain the current flow in the same direction, from higher potential to lower potential, as you correctly pointed out.

Therefore, the potential at the right side is higher because it is closer to the battery and has a higher potential difference compared to the left side. This potential difference is what drives the current through the circuit. In summary, the potential difference across an inductor is determined by the rate of change of current and the distance from the battery, and it is always higher on the side closer to the battery. I hope this explanation helps clarify the concept of voltage across an inductance.