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.

Answer 1

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