The winding of the solenoid with 1000 turns per meter, core material with relative permeability 400 and carry a current of 2 A, the magnetisation (M) is approximately given by:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 8 \times 10^{5}\,\mathrm{A/m} \)
The current in the winding sets up the magnetic intensity \( H = nI \) in the core, and the core is magnetised in proportion to it: \( M = \chi H \), where the magnetic susceptibility is \( \chi = \mu_r - 1 \).
\( n = 1000\,\mathrm{m^{-1}} \)
\( I = 2\,\mathrm{A} \)
\( \displaystyle H = nI = 1000 \times 2 = 2000\,\mathrm{A\,m^{-1}} \)
\( \displaystyle \chi = \mu_r - 1 = 400 - 1 = 399 \)
\( \displaystyle M = \chi H = 399 \times 2000 = 7.98\times10^{5}\,\mathrm{A\,m^{-1}} \approx 8\times10^{5}\,\mathrm{A\,m^{-1}} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Magnetisation and Magnetic Intensity – …
