The magnitude of a magnetic force on a current carrying conductor is given by:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( i\,lB\sin\theta \)
A current-carrying conductor of length \( l \) in a magnetic field \( \vec{B} \) feels the force \( \vec{F} = i\,\vec{l}\times\vec{B} \), the sum of the magnetic forces on all its moving charges. Its magnitude is
\( \displaystyle F = i\,l\,B\sin\theta \)
where \( \theta \) is the angle between the conductor and the field.
Option (1) \( q\displaystyle\frac{dV}{dx} \) — Incorrect. \( q\,dV/dx \) is the size of the electric force \( qE \) on a charge, since \( E = -dV/dx \); it is not a magnetic force.
Option (2) \( q\,vB\sin\theta \) — Incorrect. This is the magnetic force on a single moving charge, not on a current-carrying conductor.
Option (3) \( i\,lB\sin\theta \) — Correct. \( i\,lB\sin\theta \) is the magnitude of \( i\,\vec{l}\times\vec{B} \), the force on a conductor carrying current \( i \).
Option (4) \( q(E + VB\sin\theta) \) — Incorrect. This is the Lorentz force on a moving charge, which includes the electric force as well; it is not the force on a conductor.
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.”