Inductance of a coil with $10^4$ turns is 10 mH and it is connected to a dc source of 10 V with inte

Question

Inductance of a coil with $10^4$ turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of $10 \Omega$. The energy density in the inductor when the current reaches $\left(\frac{1}{e}ight)$ of its maximum value is $\alpha \pi 	imes \frac{1}{e^2} \mathrm{~J} / \mathrm{m}^3$. The value of $\alpha$ is $\_\_\_\_$ . $ \left(\mu_0=4 \pi 	imes 10^{-7} \mathrm{Tm} / \mathrm{A}ight) . $

Professor PC Thomas Classes · Answer

The magnetic field inside the inductor is $B = \mu_0 n I$. Assuming the number of turns per unit length is $n = 10^4 	ext{ m}^{-1}$, the energy density is $u = \frac{B^2}{2\mu_0} = \frac{1}{2}\mu_0 n^2 I^2$. The maximum current in the circuit is $I_0 = \frac{V}{R} = \frac{10}{10} = 1	ext{ A}$. The given current is $I = \frac{I_0}{e} = \frac{1}{e}	ext{ A}$. Substituting the values: $u = \frac{1}{2} (4\pi 	imes 10^{-7}) (10^4)^2 \left(\frac{1}{e}ight)^2 = \frac{1}{2} (4\pi 	imes 10^{-7}) (10^8) \frac{1}{e^2} = 20\pi 	imes \frac{1}{e^2} 	ext{ J/m}^3$. Thus, $\alpha = 20$.