An air-cored solenoid is 40 cm long. It has 1000 turns per metre and a cross-sectional area of \(10\ \mathrm{cm^{2}}\). It carries a current of 2 A. The magnetic energy stored in the solenoid is (take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1.01\times10^{-3}\ \mathrm{J}\)
The energy stored in a coil is \( \frac{1}{2}LI^{2} \). So the self-inductance of the solenoid is found first.
\( n=1000\ \mathrm{m^{-1}} \)
\( A=10\ \mathrm{cm^{2}}=1\times10^{-3}\ \mathrm{m^{2}} \)
\( l=40\ \mathrm{cm}=0.4\ \mathrm{m} \)
\( I=2\ \mathrm{A} \)
\( \begin{aligned} L &= \mu_0n^{2}Al \\ &= 4\pi\times10^{-7}\times(1000)^{2}\times10^{-3}\times0.4 \\ &= 5.03\times10^{-4}\ \mathrm{H} \end{aligned} \)
\( \begin{aligned} U &= \frac{1}{2}LI^{2} \\ &= \frac{1}{2}\times5.03\times10^{-4}\times2^{2} \\ &= 1.01\times10^{-3}\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
Self-inductance – “… Thus, the self-inductance is,”
\( \displaystyle L=\frac{N\Phi_B}{I}=\mu_0n^{2}Al \)\( \qquad (6.15) \)
Self-inductance – “Thus, the energy required to build up the current I is,”
\( \displaystyle W=\frac{1}{2}LI^{2} \)\( \qquad (6.17) \)