The figure shows a region of length ‘l’ with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity 4 × 10⁵ ms –1 making an angle 60ᵒ with the field. If the proton completes 10 revolution by the time it cross the region shown, ‘l’ is close to
(mass of proton = 1.67 × 10 –27 kg, charge of the proton = 1.6 × 10 –19 C)

Solution

Physics sir · 4 steps

Split the velocity: the part along B carries the proton forward, the part across B makes it circle.

1. Split the velocity. The component across the field, v sin 60°, makes the proton go round in circles. The component along the field, v cos 60° = 4 × 10⁵ × 1/2 = 2 × 10⁵ m s⁻¹, is untouched by B and carries it along the field. Together they give a helix.
2. Time for one turn. The period does not depend on speed: T = (2π m)/qB = (2π × 1.67 × 10⁻²⁷)/(1.6 × 10⁻¹⁹ × 0.3) = 2.19 × 10⁻⁷ s.

   p = (v cos 60°) T = 2 × 10⁵ × 2.19 × 10⁻⁷ ≈ 0.0437 m

3. Ten pitches. Each revolution moves the proton forward by one pitch p, so 10 revolutions cover l = 10p ≈ 0.437 m.
4. So. l is close to 0.44 m, option (d).

The idea to keep. In a helix, the period comes from T = 2π m/qB, but the distance covered along B per turn uses only v cos θ, the velocity component parallel to the field.

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