A region has a uniform magnetic field in it. A proton enters the region with a velocity, making an angle of \( 45^\circ \) with the direction of the magnetic field. In this region the proton will move on a path having the shape of a:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → helix
The magnetic force \( \vec{F} = q\,\vec{v}\times\vec{B} \) acts only on the part of the velocity perpendicular to the field. Split the velocity into two components:
\( v_{\parallel} = v\cos 45^\circ \ \text{(along } \vec{B}\text{)} \)
\( v_{\perp} = v\sin 45^\circ \ \text{(across } \vec{B}\text{)} \)
The perpendicular component makes the proton go round in a circle of fixed radius \( r = m v_{\perp}/(qB) \). The parallel component feels no force, so the proton also moves steadily along the field. The two motions together give a helix of constant radius and constant pitch.
Option (1) straight line — Incorrect. A straight line needs \( \vec{v} \) parallel or antiparallel to \( \vec{B} \) (angle \( 0^\circ \) or \( 180^\circ \)).
Option (2) circle — Incorrect. A circle needs \( \vec{v} \) perpendicular to \( \vec{B} \) (angle \( 90^\circ \)), with no component along the field.
Option (3) spiral — Incorrect. A spiral has a radius that changes; in a uniform field \( v_{\perp} \) is constant, so the radius stays the same.
Option (4) helix — Correct. Circular motion across the field plus uniform motion along it gives a helix.
NCERT (2026-27):
Motion in a Magnetic Field – “… If velocity has a component along B, this component remains unchanged as the motion along the magnetic field will not be affected by the magnetic field. The motion in a plane perpendicular to B is as before a circular one, thereby producing a helical motion (Fig. 4.6).”