A straight wire of mass 500 g and length 7 m carries a current of 1.4 A. It is suspended in mid air by a uniform horizontal magnetic field B perpendicular to the length of the wire. The magnitude of magnetic field is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.5 T
The wire hangs in mid-air when the upward magnetic force on the current-carrying wire, \( F = BIl \) (field perpendicular to the wire), balances its weight mg.
\( m = 500\,\mathrm{g} = 0.5\,\mathrm{kg},\quad g = 9.8\,\mathrm{m\,s^{-2}} \)
\( BIl = mg \)
\( \displaystyle B = \frac{mg}{Il} = \frac{0.5\times9.8}{1.4\times7} = \frac{4.9}{9.8} \)
\( B = 0.5\,\mathrm{T} \)
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 – …
