A straight wire of mass 200 g and length 1.5 m carrying a current of 2 A is placed in a uniform magnetic field at right angles to its direction, is suspended in mid air. The magnitude of magnetic field is: (\( g = 9.8\,\mathrm{m/s^2} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0.65 T
The wire stays suspended when the magnetic force on it balances its weight. With the field at right angles to the wire, the force is \( F = BIL \).
\( BIL = mg \)
\( m = 200\,\mathrm{g} = 0.2\,\mathrm{kg} \)
\( \displaystyle B = \frac{mg}{IL} = \frac{0.2 \times 9.8}{2 \times 1.5} = \frac{1.96}{3} \approx 0.65\,\mathrm{T} \)
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Magnetic Force on a Current-Carrying Conductor – …
