A straight wire of mass 100 g and length 1.5 m carries a current of 2 A. It is suspended in mid air by a uniform horizontal magnetic field as shown in the figure. What is the magnitude of the magnetic field? (acceleration due to gravity = \( 10\,\mathrm{m\,s^{-2}} \).)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.33 T
The wire hangs in mid air, so the magnetic force on it, \( F = BIl \) (the field being perpendicular to the wire), is directed upward and balances its weight W shown in the figure.
\( m = 100\,\mathrm{g} = 0.1\,\mathrm{kg} \)
\( BIl = mg \)
\( \displaystyle B = \frac{mg}{Il} = \frac{0.1\times10}{2\times1.5} = \frac{1}{3}\,\mathrm{T} \approx 0.33\,\mathrm{T} \)
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 – …
