A beam of protons enters a uniform magnetic field of 0.3 Tesla with a velocity of 4 x 10⁵ m/sec at an angle of 60° to the field. The radius of the helical path taken by the beam is
  • a)
    47.1 mm
  • b)
    32 mm
  • c)
    18 mm
  • d)
    46 mm
Correct answer is option 'A'. Can you explain this answer?

IIT JAM Question

RITU GUPTA
May 20, 2020
v1 is responsible for horizontal motion of proton
v2 is responsible for circular motion of proton
Mass of proton =  1.67 * 10-27 Kg
Charge on proton= 1.6 * 10-19 C
∴ mv22 / r = qv2 B
r = mv2 /qB
= 1.67 * 10-27 *  4 x 10⁵ / 1.6 * 10-19 * 0.3
= 0.013 m
Pitch of helix = v1 x T
Where T = 2πr / v2
                ­ = 2πr / v sin θ
⇒ Pitch of helix = v cos θ x 2πr / v sin θ
                     = 2 πr cotθ
                     = 2x 3.14 x 0.013 x cot 60°
                     = 0.0471 m
                     = 47.1 mm

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