A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid air by a uniform horizontal magnetic field B perpendicular to length of wire. The magnitude of B is (\( g = 10\,\mathrm{m\,s^{-2}} \))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.67 T
A current-carrying wire in a magnetic field feels the magnetic force \( F = B I L \sin\theta \); here \( \theta = 90^\circ \). The wire stays suspended when this upward force balances its weight.
\( \displaystyle B I L = m g \)
\( m = 200\,\mathrm{g} = 0.2\,\mathrm{kg} \)
\( I = 2\,\mathrm{A}, \quad L = 1.5\,\mathrm{m} \)
\( \displaystyle B = \frac{m g}{I L} = \frac{0.2 \times 10}{2 \times 1.5} = \frac{2}{3}\,\mathrm{T} \approx 0.67\,\mathrm{T} \)
Note: NTA picked this question from an NCERT (2026-27) example and added the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Magnetic Force on a Current-Carrying Conductor – …
