Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In a circuit with a solenoid/inductor and a resistor and a battery .

Books say that $\Sigma \Delta V=0$ around a closed loop . That means work done by electrostatic field per unit charge is $0$ around a closed loop .

Now as we go pass through a solenoid $\Delta V= -L\frac{di}{dt}$ . Suppose I take charge $idt$ through it , the work done by me against the field will be $\frac{1}{2}Li^2$.

Then it is said this energy is stored in the magnetic field and not the electrostatic field . Then how are we using the loop law ? When $\Delta V$ across an inductor is actually because of work done by magnetic field but loop hole holds for work done by electrostatic field only .

If there is such a electrostatic field that does the same work as the magnetic field , then $\frac{1}{2}Li^2$ must also be stored in electrostatic field in the conductor .

share|improve this question
add comment

1 Answer 1

Books say that ΣΔV=0 around a closed loop

KVL holds only if the magnetic flux linking the circuit is unchanging.

In ideal circuit theory, it is assumed that circuit elements are ideal lumped elements and the self inductance of the circuit is zero.

In other words, we assume that the dimensions of the circuit and circuit elements are arbitrarily small and the electric and magnetic fields associated with a circuit element are confined to that element.

Without these assumptions, KVL is only approximate.

When ΔV across an inductor is actually because of work done by magnetic field but loop hole holds for work done by electrostatic field only.

The energy stored in the magnetic field of an inductor is unrelated to the voltage, at any instant, across the inductor. For a steady current, the voltage across the inductor is zero while the energy stored is proportional the square of the current.

The voltage across the inductor, at any instant, is proportional to the time rate of change of the current through the inductor. Since the power associated with a circuit element is given by the product of the voltage across and the current through that element, work is done only when the current through the inductor is changing.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.