Which of the following is the correct expression for force due to a magnetic field \( (\vec{B}) \) on a straight conductor of length \( (l) \) carrying current I?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \vec{F}=I(\vec{l}\times\vec{B}) \)
A straight conductor carrying current I in a uniform field \( \vec{B} \) experiences the force obtained by adding the Lorentz force on its moving charges. The length is treated as a vector \( \vec{l} \) whose direction is that of the current, and the force is
\( \vec{F} = I\,(\vec{l}\times\vec{B}) \)
Its magnitude is \( IlB\sin\theta \) and its direction is given by the right-hand rule, perpendicular to both \( \vec{l} \) and \( \vec{B} \).
Why the other options are incorrect
Option (1) \( \vec{F}=I(\vec{B}\times\vec{l}) \) — Incorrect. Interchanging the factors of a cross product reverses its direction, \( \vec{B}\times\vec{l} = -\,\vec{l}\times\vec{B} \), so this gives the force in the opposite direction.
Option (2) \( \vec{F}=\vec{B}(\vec{I}\times\vec{l}) \) — Incorrect. The current I is a scalar, so \( \vec{I}\times\vec{l} \) has no meaning, and the force would not be perpendicular to \( \vec{B} \).
Option (4) \( \vec{F}=l(\vec{I}\times\vec{B}) \) — Incorrect. Again the current is treated as a vector; it is the length that carries the direction in the correct expression.
NCERT (2026-27):
Summary – “… 2. A straight conductor of length l and carrying a steady current I experiences a force F in a uniform external magnetic field B,
\( \displaystyle \mathbf{F} = I\,\mathbf{l} \times \mathbf{B} \)
where |l| = l and the direction of l is given by the direction of the current.”