A closely wound circular coil of radius 10 cm carries a current of 6.4 A. If the number of turns in coil are 10 then magnetic moment of this coil is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2.0096\,\mathrm{A\,m^{2}} \)
The magnetic moment of a closely wound coil of N turns is N times that of a single loop, \( m = NIA \). The options show that \( \pi = 3.14 \) is intended.
\( r = 10\,\mathrm{cm} = 0.1\,\mathrm{m} \)
\( A = \pi r^{2} = 3.14\times(0.1)^{2}\,\mathrm{m^{2}} = 0.0314\,\mathrm{m^{2}} \)
\( m = NIA = 10\times6.4\,\mathrm{A}\times0.0314\,\mathrm{m^{2}} = 2.0096\,\mathrm{A\,m^{2}} \)
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 – …
