Final answer:
The energy stored in the inductor after 2.0 ms can be determined by calculating the current at that time and then using it to calculate the energy with the formula E = (1/2)LI^2. The current in an RL circuit follows an exponential growth, and the rate of growth is determined by the time constant τ = L/R. Once we have the current, we can plug it into the energy formula to find the answer.
Step-by-step explanation:
When the switch is closed, the current starts to flow through the circuit, causing the inductor to store energy in its magnetic field. The energy stored in an inductor is given by the formula: E = (1/2)LI^2. Here, L is the inductance of the inductor and I is the current flowing through it. In this case, the inductance is 6.0 mH and the current is zero, so the energy stored in the inductor is zero initially.
After the switch is closed, the current starts to increase over time. When the current reaches its maximum value, the energy stored in the inductor is at its maximum as well. However, since the question asks for the energy after a specific time period (2.0 ms), it is important to consider the rate at which the current increases. The current in an RL circuit follows an exponential growth, given by the equation: I = I_0 * (1 - e^(-t/τ)). Here, I_0 is the initial current, t is the time, and τ is the time constant of the circuit, which is given by τ = L/R, where R is the resistance. In this case, since the current is initially zero, we can simplify the equation to: I = I_0 * (1 - e^(-t/τ)) = I_0 * (1 - e^(-t/(L/R))).
To find the energy stored in the inductor after 2.0 ms, we need to find the current at that time and then calculate the energy using the formula E = (1/2)LI^2. Given that the inductance is 6.0 mH, the resistance is 3.0 Ω, and the battery voltage is 12 V, we can use the formula above to calculate the current at 2.0 ms, and then use that current to calculate the energy stored in the inductor.