Question

Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________. 
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]

Hint:

Energy density in magnetic fields depends on the square of current and inversely on magnetic permeability.

Answer 1

Correct Answer: 500

Step 1: Find maximum current.
\[ I_{\max} = \frac{V}{R} = \frac{10}{10} = 1\ \text{A}. \] Step 2: Current at given instant.
\[ I = \frac{1}{e} I_{\max} = \frac{1}{e}. \] Step 3: Magnetic energy density formula.
Energy density in an inductor is given by
\[ u = \frac{B^2}{2\mu_0}. \] For a solenoid,
\[ B = \mu_0 n I, \quad n = \frac{N}{l}. \] Step 4: Substitute values.
\[ u = \frac{(\mu_0 n I)^2}{2\mu_0} = \frac{\mu_0 n^2 I^2}{2}. \] Using \( L = \mu_0 n^2 A l \), we get
\[ u = \frac{1}{2} \frac{L I^2}{A l}. \] Given values lead to
\[ u = 500 \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] Hence,
\[ \alpha = 500. \] Final Answer:
\[ \boxed{500} \]