A charge particle moving in magnetic field B, has the components of velocity along B as well as perpendicular to B. The path of the charge particle will be

Solution

Physics sir · 4 steps

Split the velocity into two parts: one part goes round in a circle, the other just slides along B.

1. Split the velocity. Write the velocity as v_∥ along B plus v_⊥ perpendicular to B. The magnetic force q v × B treats these two parts very differently.

   F = q(v_∥ + v_⊥) × B = q v_⊥ × B

2. Along B. v_∥ × B = 0, so there is no force along the field. The particle keeps drifting along B at constant speed v_∥.
3. Across B. The force from v_⊥ is always at right angles to v_⊥, so it bends that motion into a circle of radius r = (m v_⊥)/qB, in the plane perpendicular to B.
4. So. A circle in the perpendicular plane combined with steady motion along B gives a helix whose axis lies along B, which is option (a).

The idea to keep. In a uniform magnetic field, v_∥ never changes and v_⊥ makes a circle. Both present gives a helix along B, only v_⊥ gives a circle, only v_∥ gives a straight line.

Answer from the NTA final answer key. Solution by PrepWiser.