A particle of charge q and mass m is moving with a velocity - vî (v ≠ 0) towards a large screen placed in the Y - Z plane at a distance d. If there is a magnetic field B⃗ = B₀k̂ , the maximum value of v for which the particle will not hit the screen is :

Solution

Physics sir · 4 steps

The particle moves on a circle, and it gets at most one radius closer to the screen before turning back.

1. Circular path. The velocity -vî is perpendicular to B⃗ = B₀k̂, so the magnetic force qv⃗ × B⃗ points along ĵ. It always stays perpendicular to the velocity, so the particle moves on a circle in the X-Y plane.

   qvB₀ = mv²/r ⇒ r = mv/qB₀

2. How close it gets. At the start the velocity points straight at the screen, so the centre of the circle lies along the y-direction from the particle. The particle moves a distance r towards the screen and then curves back.
3. Condition to miss. The particle does not reach the screen if r ≤ d, that is mv/qB₀ ≤ d. This gives v ≤ qdB₀/m.
4. So. The maximum speed is v = qdB₀/m, which is option (d).

The idea to keep. If a charge starts moving straight at a boundary perpendicular to B, it gets at most one radius closer before turning back. Compare r = mv/qB with the distance, not the diameter.

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