A current-carrying wire is placed parallel to the magnetic field line…

The correct answer is Option (4) → the wire does not experience any force

A straight wire of length \( l \) carries a current \( I \) in a uniform magnetic field. It feels the force \( \vec{F}=I\,\vec{l}\times\vec{B} \) of magnitude \( IlB\sin\theta \). Here \( \theta \) is the angle between the wire and the field.

Option (1) the wire will experience a force in the direction of the magnetic field — Incorrect. A cross product with \( \vec{B} \) is always perpendicular to \( \vec{B} \), so the magnetic force on a wire can never act along the field.

Option (2) the wire will experience a force opposite to the direction of the magnetic field — Incorrect. A force opposite to \( \vec{B} \) also lies along the field line. The cross product never gives such a force.

Option (3) the wire will experience a force perpendicular to the magnetic field — Incorrect. The force \( I\,\vec{l}\times\vec{B} \) is perpendicular to the field only when the wire makes an angle with it. For a wire along the field, \( \sin\theta=0 \), so there is no force at all.

Option (4) the wire does not experience any force — Correct. With the wire parallel to the field lines, \( \theta=0^\circ \) and \( \sin\theta=0 \), so \( IlB\sin\theta=0 \). The charge carriers drift along the field, and a magnetic field exerts no force on charges moving parallel to it.

NCERT (2026-27):
Magnetic Field, Lorentz Force – “(ii) The magnetic force q [ v × B ] includes a vector product of velocity and magnetic field. The vector product makes the force due to magnetic field vanish (become zero) if velocity and magnetic field are parallel or anti-parallel. The force acts in a (sideways) direction perpendicular to both the velocity and the magnetic field. …”

Magnetic Force on a Current-Carrying Conductor – “… Thus,

    \( \mathbf{F}=[(nq\,\mathbf{v}_d)\,lA]\times\mathbf{B}=[\,\mathbf{j}Al\,]\times\mathbf{B}= \)\( I\,\mathbf{l}\times\mathbf{B} \)\( \qquad (4.4) \)

where l is a vector of magnitude l, the length of the rod, and with a direction identical to the current I.”