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 $\left(\frac{1}{\mathrm{e}}\right)$ of its maximum value is $\alpha \pi \times \frac{1}{\mathrm{e}^{2}} \mathrm{~J} / \mathrm{m}^{3}$. The value of $\alpha$ is ______. $(\mu_{0}=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A})$.

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Electromagnetic Induction

Inductance of a coil with

1 0^{4}

turns is 10 mH and it is connected to a dc source of 10 V with internal resistance of

10Ω

. The energy density in the inductor when the current reaches

(\frac{1}{e})

of its maximum value is

α π \times \frac{1}{e ^{2}} J / m^{3}

. The value of

α

is ______.

(μ_{0} = 4 π \times 1 0^{- 7} Tm / A)

Answer: 20

The steady (maximum) current is

I_{0} = \frac{V}{R} = \frac{10}{10} = 1 A

, so at the stated instant

I = \frac{I _{0}}{e} = \frac{1}{e} A

. Inside the long coil the field is

B = μ_{0} n I

, and the magnetic energy density is

u = \frac{B ^{2}}{2 μ _{0}} = \frac{μ _{0} n ^{2} I ^{2}}{2}

. With

n = 1 0^{4}

turns per metre,

u = \frac{( 4 π \times 1 0 ^{- 7} ) ( 1 0 ^{4} ) ^{2}}{2} \cdot \frac{1}{e ^{2}} = \frac{20 π}{e ^{2}} J/ m^{3}

. Comparing with

α π \times \frac{1}{e ^{2}}

, the value of

α

20

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Integer answer