A straight current carrying copper wire with 28 Ampere current is held normal to the magnetic field such that copper wire is balanced against gravity acting on it. Find the value of magnetic field if mass per unit length of wire is (46.6 gram/meter).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.6\times10^{-2}\,\mathrm{T} \)
The wire is balanced when the magnetic force on it, \( F = BIl \) (field normal to the wire), is equal and opposite to its weight \( mg \). Dividing by the length l, the condition becomes \( BI = (m/l)\,g \).
Convert the mass per unit length: \( 46.6\,\mathrm{g\,m^{-1}} = 0.0466\,\mathrm{kg\,m^{-1}} \). Take \( g = 9.8\,\mathrm{m\,s^{-2}} \).
\( \displaystyle B = \frac{(m/l)\,g}{I} = \frac{0.0466\times9.8}{28} \)
\( B \approx 1.6\times10^{-2}\,\mathrm{T} \)
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 – …
