A straight wire of mass \( 0.25\,\mathrm{kg} \) and length \( 1.25\,\mathrm{m} \) carries a current. It is suspended in mid air by a uniform horizontal magnetic field of \( 0.8\,\mathrm{T} \). The magnitude of the current flowing in the straight wire is:
[Given, acceleration due to gravity \( = 10\,\mathrm{m/s^{2}} \)]
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2.5\,\mathrm{A} \)
The wire hangs in mid air when the upward magnetic force on the current balances its weight. The field is horizontal and the wire is horizontal and perpendicular to it, so the force \( F = BIl \) (with \( \sin 90^\circ = 1 \)) is vertical.
\( BIl = mg \)
\( \displaystyle I = \frac{mg}{Bl} = \frac{0.25\,\mathrm{kg}\times10\,\mathrm{m/s^{2}}}{0.8\,\mathrm{T}\times1.25\,\mathrm{m}} \)
\( \displaystyle I = \frac{2.5\,\mathrm{N}}{1.0\,\mathrm{T\,m}} = 2.5\,\mathrm{A} \)
The direction of the current must be such that \( I\vec{l}\times\vec{B} \) points upward.