Final answer:
The maximum energy stored in the magnetic field of the 60 mH inductor, initially connected to a 4,000 pF capacitor charged to 140 V, is 3.92 x 10^-5 joules.
Step-by-step explanation:
To determine the maximum energy stored in the magnetic field of the 60 mH inductor, we first find the energy initially stored in the 4,000 pF capacitor when it is charged to 140 V. The energy stored in a capacitor can be calculated using the formula E = ½ CV^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts.
Using this formula, we calculate the initial energy of the capacitor as follows:
- ½ (4,000 x 10^-12 F) (140 V)^2
- ½ (4 x 10^-9 F) (19600 V^2)
- ½ (4 x 10^-9 F) (19600 J)
- 2 x 10^-9 F x 19600 J
- 3.92 x 10^-5 J
When the capacitor is connected to the inductor, the energy will oscillate between the capacitor and the inductor. At the maximum energy in the inductor, the energy in the capacitor will be zero and all the energy will be stored in the inductor. Therefore, the maximum energy stored in the inductor's magnetic field equals the initial energy stored in the capacitor.
The maximum energy stored in the magnetic field of the 60 mH inductor is 3.92 x 10^-5 J.