A straight wire of mass 300 g and length 1.5 m carries a current of 2 A. It is suspended in mid air by a uniform horizontal magnetic field. The magnitude of the magnetic field is (Neglect the earth's magnetic field and \( g = 10\,\mathrm{m/s^2} \))
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1 T
For the wire to hang in mid air, the magnetic force on it must balance its weight. The field is horizontal and perpendicular to the wire, so the force has its largest value \( BIL \) and acts vertically upward.
\( \displaystyle B I L = m g \;\Rightarrow\; B = \frac{m g}{I L} \)
\( m = 300\,\mathrm{g} = 0.3\,\mathrm{kg} \)
\( I = 2\,\mathrm{A} \)
\( L = 1.5\,\mathrm{m} \)
\( \displaystyle B = \frac{0.3 \times 10}{2 \times 1.5} = \frac{3}{3} = 1\,\mathrm{T} \)
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 – …
