Question

a proton is projected with a speed of 2*10^6m/sec at angle 60degree to the x-axis if a uniform magnetic field of 0.104T is applied along y-axis. calculate radius of helix and time period of proton.

Answer 1

Answer:2π × 10–7sec

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

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