Question

An emf of \(20.0 \mathrm{~V}\) is applied to a coil with an inductance of \(40.0 \mathrm{mH}\) and a resistance of \(0.500 \Omega\). a) Determine the energy stored in the magnetic field when the current reaches \(\frac{1}{4}\) of its maximum value. b) How long does it take for the current to reach this value?

Short answer

The energy stored in the magnetic field when the current reaches 1/4 of its maximum value is 40 x 10^-3 J (40 mJ). b) How long does it take for the current to reach 1/4 of its maximum value? It takes approximately 0.0230 seconds for the current to reach 1/4 of its maximum value.

Step-by-step solution

Find the time constant

The time constant of an inductor-resistor circuit is calculated as the ratio of the inductance to the resistance: $$ \tau = \frac{L}{R} $$ Using the given values for inductance (40 mH) and resistance (0.5 ohms), we can plug in the values and calculate the time constant: $$ \tau = \frac{40 \times 10^{-3} \mathrm{H}}{0.5 \mathrm{\Omega}} = 0.08 \mathrm{s} $$

Determine the maximum current

The maximum current in this circuit occurs when the circuit reaches a steady state. Ohm's law relates the current, resistance, and voltage in the circuit: $$ I_{max} = \frac{V}{R} $$ Using the given values for the applied emf (20 V) and resistance (0.5 ohms), we can plug in the values and calculate the maximum current: $$ I_{max} = \frac{20 \mathrm{V}}{0.5 \mathrm{\Omega}} = 40 \mathrm{A} $$

Calculate the energy stored in the magnetic field

For the energy stored in the magnetic field when the current reaches 1/4 of its maximum value, we can use the following equation: $$ E = \frac{1}{2} LI^2 $$ Since we want the energy at 1/4 of the maximum current, we have \(I = \frac{1}{4}I_{max}\). Plugging in the given value for inductance (40 mH) and the calculated value for the maximum current, we get: $$ E = \frac{1}{2} (40 \times 10^{-3} \mathrm{H}) \left(\frac{1}{4}(40 \mathrm{A})\right)^2 = 40 \times 10^{-3} \mathrm{J} $$ So the energy stored in the magnetic field when the current reaches 1/4 of its maximum value is \(40 \times 10^{-3}\) J.

Calculate the time it takes for the current to reach 1/4 of its maximum value

The equation for the current in an inductor-resistor circuit as a function of time is given by: $$ I(t) = I_{max}(1 - e^{-t/\tau}) $$ We want to find the time (\(t\)) it takes for the current to reach 1/4 of its maximum value. Therefore, we can set up the following equation: $$ \frac{1}{4}I_{max} = I_{max}(1 - e^{-t/\tau}) $$ Dividing both sides by \(I_{max}\) and solving for \(t\), we get: $$ \frac{1}{4} = 1 - e^{-t/\tau} \\ e^{-t/\tau} = \frac{3}{4} \\ t = -\tau \ln{\frac{3}{4}} $$ Using our calculated value for the time constant and \(\ln{\frac{3}{4}} \approx -0.2877\), we find the time: $$ t = -0.08 \mathrm{s} \times -0.2877 \approx 0.0230 \mathrm{s} $$ So it takes about 0.0230 seconds for the current to reach 1/4 of its maximum value.