A long wire having linear mass density of 0.25 g/cm, carrying a current of 5 A, is to be suspended in air in the magnetic field created around it as shown. The magnitude of the required magnetic field is
(Neglect earth's magnetic field, take \( g \) = 10 m/s2)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 5 × 10−2 T
The wire stays suspended when the magnetic force balances its weight. The field is into the page (the crosses) and the current flows to the right, so \( \vec{F} = I\vec{l}\times\vec{B} \) points upwards, as required. Work per unit length, using the linear mass density \( \lambda \) in kg/m.
\( \displaystyle \lambda = 0.25\,\mathrm{g/cm} = \frac{0.25\times10^{-3}\,\mathrm{kg}}{10^{-2}\,\mathrm{m}} = 2.5\times10^{-2}\,\mathrm{kg/m} \)
\( \displaystyle BIl = mg = \lambda lg \quad\Rightarrow\quad B = \frac{\lambda g}{I} \)
\( \displaystyle B = \frac{(2.5\times10^{-2}\,\mathrm{kg/m}) \times (10\,\mathrm{m/s^{2}})}{5\,\mathrm{A}} = 5\times10^{-2}\,\mathrm{T} \)
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 – …
