Final answer:
Using the formula for energy stored in an inductor (Eind = 1/2LI²), the energy stored in a 22 H inductor with a current of 12 A is 1584 J. This energy could melt approximately 4.74 grams of ice at 0°C, using the latent heat of fusion for ice which is 334 J/g.
Step-by-step explanation:
The question is asking how much ice can be melted using the energy stored in the magnetic field of an inductor. To find the energy stored, we can use the equation Eind = 1/2LI², where Eind is the energy in joules, L is the inductance in henrys (H), and I is the current in amperes (A). In this case, the inductance L is 22 H and the current I is 12 A, so the energy stored will be calculated as follows:
Eind = 1/2 × 22 H × (12 A)²
Eind = 1/2 × 22 × 144
Eind = 1584 J
Once we have the total energy, we can determine the mass of ice that could be melted by setting it equal to the heat required to melt the ice, which is given by Q = mLf where m is the mass in kilograms, and Lf is the latent heat of fusion in joules per gram. Using the given Lf of 334 J/g for ice, the equation for mass m becomes:
m = Eind / Lf
m = 1584 J / 334 J/g
m ≈ 4.74 g
Therefore, the energy stored in the magnetic field of the inductor could melt approximately 4.74 grams of ice.