Energy Stored in a Magnetic Field — Question 1

Question 1

Integrating inductor power gives a quadratic energy law

An ideal constant inductance L=.250 H has current raised quasistatically from zero to I=3.00 A. Its geometry is fixed and there are no losses.

1. Derive stored energy by integrating passive electrical power.

2. Find total energy at 3.00 A and the additional energy required to increase current from 2.00 A to 3.00 A.

Question 1 - Solution

Integrating inductor power gives a quadratic energy law

(a) Method and calculation

$$p=vi=Li\dot i\Rightarrow U(I)=\int_0^I Li\,di.$$$$\boxed{U=\tfrac12LI^2}.$$

(b) Interpretation and further calculation

$$\boxed{U(3.00)=\tfrac12(.250)(3.00)^2=1.125\,J}.$$$$\boxed{\Delta U=\tfrac12(.250)(3.00^2-2.00^2)=.625\,J}.$$

Physical check

Equal current increments do not require equal energy increments because the instantaneous required power is proportional to current.