A horizontal wire of length 10 cm and mass 0.3 g carries a current of 5 A. The magnitude of the magnetic field which can keep the wire in suspension is
(Take: g = 10 m/s2)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 6 × 10-3 T
The wire stays suspended when the magnetic force balances its weight. The least field does this when it is horizontal and perpendicular to the wire, so that \( F = BIl \) acts vertically upwards.
\( m = 0.3\,\mathrm{g} = 3\times10^{-4}\,\mathrm{kg} \)
\( l = 10\,\mathrm{cm} = 0.1\,\mathrm{m} \)
\( \displaystyle BIl = mg \Rightarrow B = \frac{mg}{Il} \)
\( \displaystyle B = \frac{3\times10^{-4}\,\mathrm{kg} \times 10\,\mathrm{m\,s^{-2}}}{5\,\mathrm{A} \times 0.1\,\mathrm{m}} \)
\( \displaystyle B = \frac{3\times10^{-3}}{0.5} = 6\times10^{-3}\,\mathrm{T} \)
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):
Magnetic Force on a Current-Carrying Conductor – …
