The magnitude of magnetic field inside a solenoid of length \( 0.3\,\mathrm{m} \) having 800 turns carrying a current of \( 6\,\mathrm{A} \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 20\,\mathrm{mT} \)
Inside a long solenoid the field is uniform and is given by \( B = \mu_0 n I \), where \( n \) is the number of turns per unit length.
\( \displaystyle n = \frac{800}{0.3} = 2.67\times10^{3}\,\mathrm{m^{-1}} \)
\( \displaystyle B = 4\pi\times10^{-7} \times 2.67\times10^{3} \times 6 = 2.01\times10^{-2}\,\mathrm{T} \)
\( \displaystyle B \approx 20\,\mathrm{mT} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
The Solenoid – …
