A 200 turns closely wound circular coil of radius 10 cm carries a current of 3.5 A. The magnetic moment of the coil is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 22\,\mathrm{A\,m^2} \)
The magnetic moment of a current loop is the current times the area, and for a closely wound coil the moments of all \( N \) turns add.
\( \displaystyle m = N I A = N I \pi r^{2} \)
\( N = 200 \)
\( I = 3.5\,\mathrm{A} \)
\( r = 10\,\mathrm{cm} = 0.1\,\mathrm{m} \)
\( \displaystyle m = 200 \times 3.5 \times \pi \times (0.1)^{2} = 7\pi\,\mathrm{A\,m^2} \approx 22\,\mathrm{A\,m^2} \)
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):
Torque on Current Loop, Magnetic Dipole – …
