91Ó°ÊÓ

Inductors, like the coil in our exercise, store energy in their magnetic fields. When a battery is connected to an inductor, it causes a current to flow and the inductor begins to store energy. The energy stored in an inductor at any time can be calculated using the formula \[ U = \frac{1}{2} L I^2 \]where \( U \) is the energy stored, \( L \) is the inductance, and \( I \) is the current. Unlike capacitors which store energy in an electric field, inductors use a magnetic field to store energy. An increasing current signifies more energy being stored. In our problem, we need to calculate how quickly energy is being stored at a specific time. The formula for the rate of energy storage \( \frac{dU}{dt} \) involves both the current \( I \) and its rate of change \( \frac{dI}{dt} \). This is given by:\[\frac{dU}{dt} = LI \frac{dI}{dt}\]Substituting in all relevant values and calculated figures gives us the rate at which energy is stored in the magnetic field.

As current passes through a resistor, like the resistance in our coil, it turns electrical energy into heat, a process known as Joule heating. The thermal energy produced per unit time, also called power (\( P \)), is determined by the current and the resistance, using the formula:\[ P = I^2 R \]In this equation, \( I \) is the current through the resistor and \( R \) is the resistance. This is part of the energy transferred from the battery that is not stored and instead dissipated as heat.For our task, we calculated the current earlier, allowing us to compute the thermal power generated. Since resistance converts electrical energy to thermal energy based on the square of the current, even a modest current can produce substantial heat. Here, substituting the calculated current into this equation gives the thermal power being dissipated as heat in the resistor.

The battery in this scenario provides energy to the entire circuit. This energy is split between being stored in the inductor's magnetic field and dissipated as thermal energy in the resistor. The total power delivered by the battery can be expressed with:\[ P_{\text{battery}} = \mathscr{E} I \]where \( \mathscr{E} \) is the electromotive force (emf) of the battery and \( I \) is the current flowing through the circuit. The exercise solution involves calculating this using our earlier determined current.It's worth noting that the power provided by the battery not only contributes to storing energy in the magnetic field but also overcomes the resistive losses. Simultaneously considering these aspects helps understand how energy is split and utilized within the circuit.