The force experienced by a current carrying wire placed in a magnetic field will be maximum when it is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Perpendicular to the direction of magnetic field
The force on a current-carrying wire of length \( l \) in a magnetic field is \( \vec{F} = I\,\vec{l}\times\vec{B} \), so its magnitude depends on the angle \( \theta \) between the wire and the field.
\( \displaystyle F = B I l \sin\theta \)
Option (1) At an angle of \( 45^\circ \) to the direction of magnetic field — Incorrect. At \( 45^\circ \), \( F = BIl \sin 45^\circ \approx 0.71\,BIl \), less than the maximum.
Option (2) Parallel to the direction of magnetic field — Incorrect. At \( \theta = 0^\circ \), \( \sin\theta = 0 \), so the force is zero.
Option (3) Anti-parallel to the direction of magnetic field — Incorrect. At \( \theta = 180^\circ \), \( \sin\theta = 0 \), so the force is again zero.
Option (4) Perpendicular to the direction of magnetic field — Correct. At \( \theta = 90^\circ \), \( \sin\theta = 1 \), so the force has its maximum value \( BIl \).