Question

A proton is projected with a speed of 2 × 10⁶(m/s) at an angle of 60° to the X-axis. If a uniform mag. field of 0.104 tesla is applied along Y-axis, the path of proton is _______ 

(a) A circle of r = 0.2 m and time period π × 10^–7sec 

(b) A circle of r = 0.1m and time period 2π × 10^–7sec 

(c) A helix of r = 0.1m and time period 2π × 10^–7sec 

(d) A helix of r = 0.2 m and time period 4π × 10^–7sec

Answer 1

The correct option (c) A helix of r = 0.1m and time period 2π × 10^–7 sec

Explanation:

Path of proton will be helix. [if angle is other than 0°, 90° or 180° path followed in helix]

as  qBr = mVsinθ

∴  r = [(mVsinθ)/qB]

V = 2 × 10⁶ m/s

θ = 60°

 B = 0 – 10⁴ T

q = 1.6 × 10^–19

m = 1.6 × 10^–27 kg

∴  r = [{1.6 × 10^–27 × 2 × 10⁶ × (√3/2)}/(1.6 × 10^–19 × 0.104)]

= 16.66 × 10^–2 = 0.16 m

time period = T = [(2πm)/qB]

= [(2π × 1.6 × 10^–27)/(1.6 × 10^–19 × 0.104)]

[T = (2π/w) and w = (qB/m)]

= 2π × 10^–7sec