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:
In an \(RL\) circuit connected to a DC source:
Maximum (steady-state) current: \[ I_0 = \frac{\varepsilon}{R} \]
Current growth with time: \[ I(t) = I_0\left(1 - e^{-t/\tau}\right), \quad \tau = \frac{L}{R} \]
Magnetic energy density in an inductor: \[ u = \frac{B^2}{2\mu_0} \]
Magnetic field inside a solenoid: \[ B = \mu_0 \frac{N}{\ell} I \]
Step 1: Find maximum current. \[ I_0 = \frac{\varepsilon}{R} = \frac{10}{10} = 1\,\text{A} \] Given condition: \[ I = \frac{I_0}{e} = \frac{1}{e}\,\text{A} \]
Step 2: Find magnetic field at this instant. \[ B = \mu_0 \frac{N}{\ell} I = (4\pi \times 10^{-7}) \times 10000 \times \frac{1}{e} \] \[ B = \frac{4\pi \times 10^{-3}}{e} \]
Step 3: Write expression for energy density. \[ u = \frac{B^2}{2\mu_0} = \frac{1}{2\mu_0}\left(\frac{4\pi \times 10^{-3}}{e}\right)^2 \] \[ u = \frac{16\pi^2 \times 10^{-6}}{2\mu_0 e^2} \] Substitute \( \mu_0 = 4\pi \times 10^{-7} \): \[ u = \frac{16\pi^2 \times 10^{-6}}{2(4\pi \times 10^{-7})e^2} \] \[ u = \frac{16\pi^2}{8\pi}\frac{10}{e^2} \] \[ u = \frac{20\pi}{e^2} \]
Step 4: Compare with given form. Given: \[ u = \alpha \frac{\pi}{e^2} \] Thus, \[ \alpha = 20 \] \[ \boxed{\alpha = 20} \]