A proton with a kinetic energy of 2.0 eV moves into a region of uniform magnetic field of magnitude π/2 × 10⁻³ T. The angle between the direction of magnetic field and velocity of proton is 60°. The pitch of the helical path taken by the proton is __________ cm. (Take, mass of proton =1.6 × 10⁻²⁷ kg and Charge on proton =1.6 × 10⁻¹⁹ C ).

Solution

Physics sir · 3 steps

Pitch is the distance covered along B in one turn, so use the component of velocity parallel to the field.

1. Find the speed. K = 2.0 eV = 2 × 1.6 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ J. Then v = √(2K/m) = √((6.4 × 10⁻¹⁹)/(1.6 × 10⁻²⁷)) = √(4 × 10⁸) = 2 × 10⁴ m s⁻¹.
2. Time for one turn. The perpendicular component makes the circle, and the time period does not depend on speed: T = (2π m)/qB = (2π × 1.6 × 10⁻²⁷)/(1.6 × 10⁻¹⁹ × π/2 × 10⁻³) = 4 × 10⁻⁵ s.

   p = (v cos 60°) T = (2 × 10⁴ × 1/2) × 4 × 10⁻⁵ m

3. So. p = 10⁴ × 4 × 10⁻⁵ = 0.4 m, which is 40 cm.

The common slip: 69. Using v sin 60° instead of v cos 60° for the motion along the field gives about 69 cm, not 40.

The idea to keep. In a helix, v sin θ sets the radius and v cos θ sets the pitch. The period 2π m/qB is the same for every speed.

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