Question

A charged particle moves in a uniform magnetic field. The velocity of the particle at some instant makes an acute angle with the magnetic field. The path of the particle will be
A. A straight line
B. A circle
C. A helix with uniform pitch
D. A helix with non-uniform pitch

Answer 1

Hint:If a charged particle is moving in a magnetic field then it experiences the magnetic force. As we know that the force causes acceleration and hence the change in velocity leads to the change in path of the motion.

Formula used:
\[\overrightarrow F = q\left( {\vec v \times \vec B} \right)\]
here \[\vec F\] is the magnetic force vector, \[\vec v\] is the velocity of the charged particle and \[\vec B\] is the magnetic field in the region.

Complete step by step solution:
It is given that the charged particle enters into the region of the magnetic field with the magnetic field making an acute angle. The angle between the velocity and the magnetic field is less than \[90^\circ \]. The components of the velocity relative to the magnetic field is,\[{v_x}\] and \[{v_y}\] which are horizontal and perpendicular. As the magnetic force is proportional to the vector product of the velocity and the magnetic field, so the horizontal component will not experience the magnetic force and it remains constant.

The perpendicular component of the force will cause the circular path of the motion and the horizontal velocity component will make the charged particle move forward making a circular revolution. So, the path of the motion of the charged particle will be circular motion in the vertical plane and linear in horizontal plane, i.e. the path will be helix.

The radius of the path will be,
\[r = \dfrac{{m{v_y}}}{{Bq}}\]
Then the period of revolution will be,
\[T = \dfrac{{2\pi r}}{v}\]
\[\Rightarrow T = \dfrac{{2\pi \left( {\dfrac{{m{v_y}}}{{Bq}}} \right)}}{{{v_y}}}\]
\[\therefore T = \dfrac{{2\pi m}}{{Bq}}\]
That is the period is constant. So the distance travelled along the horizontal will be constant.
The distance covered by the charged particle along the horizontal axis within the period of revolution is called the pitch of the helix. So, the path of the motion of the charged particle is a helix with uniform pitch.

Therefore, the correct option is C.

Note: We should be careful that the uniformity of the pitch of the helix is independent of the nature of the angle made by the velocity with the magnetic field. If there is a horizontal component of velocity, the pitch will be uniform.