What magnitude of electric current can store 2.5 J of energy in a 3 H induction coil?

Answer Details

A current-carrying inductor stores energy in the magnetic field of its coil. The energy stored is

\[E = \tfrac{1}{2}LI^2,\]

where \(L\) is the inductance in henries and \(I\) the steady current. This is the magnetic counterpart of the energy \(\tfrac{1}{2}CV^2\) stored in a capacitor's electric field, and like it the energy depends on the square of the current.

Rearrange for the current before substituting:

\[I = \sqrt{\frac{2E}{L}} = \sqrt{\frac{2\times 2.5}{3}} = \sqrt{\frac{5}{3}} = \sqrt{1.667} = 1.29\ \text{A}.\]

So a steady current of about \(1.29\ \text{A}\) stores \(2.5\ \text{J}\) in a \(3\ \text{H}\) coil.

The trap is forgetting the square root and dividing instead, for example \(2E/L = 1.67\) or \(E/L\) style combinations, or forgetting the factor \(\tfrac{1}{2}\), which would give \(\sqrt{2.5/3}=0.91\ \text{A}\). Because the relationship is quadratic, doubling the current stores four times the energy, and that squared dependence is exactly what the examiner is checking. Write the formula down, make the unknown the subject, then substitute.