JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2026
- Easy
- Numerical
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 \(\left(\frac{1}{ e }\right)\) of its maximum value is \(\alpha \pi \times \frac{1}{ e ^2} J / m ^3\). The value of \(\alpha\) is __________ . \(\left(\mu_0=4 \pi \times 10^{-7} Tm / A \right)\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 20
Step-by-step Solution
Detailed explanation
\(L =10 \times 10^{-3} H\)
N=104
\(I _0=\frac{10}{10}=1 A\) (max current)
\(I =\frac{1}{ e }\)
\(\begin{array}{l} E _{ d }=\frac{ B ^2}{2 \mu_0} \\ B=\mu_0 nI \\ L =\mu_0 n ^2 \pi R ^2 \ell \\ E _{ d }=\frac{\mu_0 n ^2 I ^2}{2}\end{array}\)
\(=\frac{4 \pi \times 10^{-7} \times 10^8 \times \frac{1}{ e ^2}}{2}\)
\(=\frac{20 \pi}{ e ^2}\)
N=104
\(I _0=\frac{10}{10}=1 A\) (max current)
\(I =\frac{1}{ e }\)
\(\begin{array}{l} E _{ d }=\frac{ B ^2}{2 \mu_0} \\ B=\mu_0 nI \\ L =\mu_0 n ^2 \pi R ^2 \ell \\ E _{ d }=\frac{\mu_0 n ^2 I ^2}{2}\end{array}\)
\(=\frac{4 \pi \times 10^{-7} \times 10^8 \times \frac{1}{ e ^2}}{2}\)
\(=\frac{20 \pi}{ e ^2}\)
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