so the amount of work done to drive the current through the inductor is given by dw=Eidt where E is the emf of the variable voltage source.We put E=+L (di/dt). can you please tell me why we put that value?? (i know the induced e.m.f on the inductor is - L (di/dt) thanks
while calculating the amount of work done we have to take the integration. For that we have to reduce the expression of dw in single variable So dw = Eidt = L*i*di/dt*dt = L*i*di. Now take integrations to get the energy stored in the inductor.
Consider the equation E = -L di/dt. The minus sign in the equation indicates that the induced EMF has an orientation that opposes the the change in current. The reason why it's positive for your given equation, is because positive work must be done to induce the current. We need only the magnitude of E then.
thanks but why to drive the current through the circuit so as to defeat the induced e.m.f -L di/dt we need Ldi/dt; dont we not need a greater value?? thanks again.
The induced emf is Ldi/dt; the negative sign just indicates it opposes the battery's emf. The electrons pushed through the inductor must overcome a voltage of Ldi/dt (that is, they must overcome an increase in potential of magnitude Ldi/dt), so they gain Lidi/dt of potential energy per unit time. This gain in potential is what the energy stored in the inductor really is.
I don't think that this is quite correct. Consider a circuit with an applied emf, resistor, and inductor, where the current varies with time. If the current is increasing (di/dt > 0 ), then the self-induced emf opposes the direction of the applied emf. If the current is decreasing (di/dt < 0), the self-induced emf is actually oriented in the same direction as the applied emf. In both cases, the self-induced emf has orientation that always opposes the "change in current", not the "applied emf."