JEE Main 2026 — Jan 22, Shift 1 — Physics solution

JEE Main2026Jan 22, Shift 1Physics

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 10 $Ω$ . 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 ____. ( $μ_{0} = 4 π \times 1 0^{- 7}$ T.m/A).

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Solution

Maximum (steady-state) current: $I_{0} = \frac{V}{R} = \frac{10}{10} = 1$ A. At the instant the current is $I = I_{0} / e = 1/ e$ A.

Magnetic field inside a long solenoid: $B = μ_{0} n I$ , where $n$ is the turns per unit length.

Energy density: $u = \frac{B ^{2}}{2 μ _{0}} = \frac{μ _{0} n ^{2} I ^{2}}{2}$ .

From the given data the implied $n^{2} = 1 0^{8}$ (consistent with $N = 1 0^{4}$ turns over a 1 m length).

Plug in: $u = \frac{( 4 π \times 1 0 ^{- 7} ) ( 1 0 ^{8} ) ( 1/ e ^{2} )}{2} = \frac{4 π \times 10}{2 e ^{2}} = \frac{20 π}{e ^{2}}$ J/m $^{3}$ .

Comparing with $u = α π \times \frac{1}{e ^{2}}$ : $α = 20$ .

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