A conducting wire of length \( 0.1\,\mathrm{m} \) and mass \( 3\times10^{-3}\,\mathrm{kg} \), carrying a current of \( 5\,\mathrm{A} \), is suspended freely in a horizontal direction in a uniform magnetic field. The magnitude of magnetic field perpendicular to the length of conductor is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 0.06\,\mathrm{T} \)
The wire floats when the upward magnetic force on it balances its weight. With the field perpendicular to the wire, the force is \( BIl \). Take \( g = 10\,\mathrm{m\,s^{-2}} \).
\( \displaystyle BIl = mg \;\Rightarrow\; B = \frac{mg}{Il} \)
\( \displaystyle B = \frac{3\times10^{-3} \times 10}{5 \times 0.1} = \frac{0.03}{0.5} = 0.06\,\mathrm{T} \)
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 – …
