Question

A proton of magnetic field 0.2 T enters a magnetic field perpendicular with a velocity = 6.0 × 105 m/sec. Calculate the acceleration of the proton and radius of the path.
Solution:
Given,
Magnetic field B = 0.2T
Speed of proton v = 6.0 × 105 m/s
Charge q = e = 1.6 × 10-19 C
θ = 90° ‘
∴ F = qvBsinθ
F = 1.6 × 10-19 × 6.0 × 105 × 0.2 × sin90°
F = 1.92 × 10-14 N
∵ F = ma

Answer 1

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