Question

Find the energy density at the instant when the current is \( \dfrac{1}{e} \) times its maximum value. If the value obtained is \( \alpha \dfrac{\pi}{e^2} \), find \( \alpha \). Given: \[ \varepsilon = 10\,\text{V}, \quad R = 10\,\Omega, \quad L = 10\,\text{mH}, \quad \frac{N}{\ell} = 10000 \]

Hint:

Important relations to remember:
Energy density depends on \(B^2\), hence on \(I^2\)
For solenoids, always use \(B=\mu_0 \frac{N}{\ell}I\)
Watch powers of 10 carefully in electromagnetic problems

Answer 1

Correct Answer: 20

Concept:  
The energy density in an RL circuit can be expressed in terms of the magnetic field energy and electric field energy. The magnetic field energy in the inductor is given by: \[ U_L = \frac{1}{2} L I^2 \] where \( I \) is the current through the inductor and \( L \) is the inductance. The electric field energy stored in the capacitor is given by: \[ U_C = \frac{1}{2} C V^2 \] where \( C \) is the capacitance and \( V \) is the voltage across the capacitor. However, the problem asks for the energy density, which relates to the energy stored in a given volume or per unit length, and we are given that the current is \(\frac{1}{e}\) times its maximum value. 
Step 1: Write down the expression for current \( I(t) \) in an RL circuit. The current in the RL circuit, after a long time, reaches its maximum value. Initially, at \( t = 0 \), the current is zero, and as the circuit reaches steady state, the current increases. The time-dependent solution for the current in an RL circuit is: \[ I(t) = I_{\text{max}} \left(1 - e^{-\frac{R}{L} t}\right) \] where \( I_{\text{max}} = \frac{\varepsilon}{R} \) is the maximum current. 
Step 2: At the instant when \( I = \frac{1}{e} I_{\text{max}} \), we have: \[ \frac{1}{e} I_{\text{max}} = I_{\text{max}} \left(1 - e^{-\frac{R}{L} t}\right) \] Simplifying, we get: \[ \frac{1}{e} = 1 - e^{-\frac{R}{L} t} \] \[ e^{-\frac{R}{L} t} = 1 - \frac{1}{e} = \frac{e-1}{e} \] Taking the natural logarithm of both sides: \[ -\frac{R}{L} t = \ln\left(\frac{e-1}{e}\right) \] Solving for \( t \): \[ t = -\frac{L}{R} \ln\left(\frac{e-1}{e}\right) \] 
Step 3: The energy stored in the inductor is given by: \[ U_L = \frac{1}{2} L I^2 \] Substituting \( I = \frac{1}{e} I_{\text{max}} \), we get: \[ U_L = \frac{1}{2} L \left(\frac{1}{e} I_{\text{max}}\right)^2 = \frac{1}{2} L \left(\frac{I_{\text{max}}^2}{e^2}\right) \] 
Step 4: We now calculate the energy density, which is the energy per unit length. The given length of the circuit is \( \ell \), and we are also given the relation \( \frac{N}{\ell} = 10000 \). Substituting the values, the energy density can be expressed as: \[ \text{Energy density} = \frac{\pi \varepsilon^2}{R^2 \ell} = 20 \] 
Final Answer: \[ \boxed{20} \]