Energy Stored in a Magnetic Field — Question 3

Question 3

Stored energy decays twice as fast in its exponent as the RL current

An inductor L with initial current I0 discharges through resistance R. There is no source and the geometry is fixed.

1. Find stored energy as a function of time.

2. Compare the energy half-life with the current half-life and find heat released at the energy half-life.

Question 3 - Solution

Stored energy decays twice as fast in its exponent as the RL current

(a) Method and calculation

$$i=I_0e^{-t/\tau},\quad\tau=L/R\Rightarrow U=\tfrac12LI_0^2e^{-2t/\tau}.$$$$\boxed{U(t)=U_0e^{-2t/\tau}}.$$

(b) Interpretation and further calculation

$$\boxed{t_{1/2,U}=\tfrac12\tau\ln2=\tfrac12t_{1/2,i}}.$$$$\boxed{H(t_{1/2,U})=U_0/2=LI_0^2/4}.$$

Physical check

Energy depends on current squared; applying the current decay fraction directly to energy gives the wrong time scale.