Answer:
the proton speed of the proton was 1.6 × 10⁶ m/s
Explanation:
Given the data in the question;
Radius r = 6.42 cm = 0.0642 m
magnetic field B = 0.260 T
we know that; charge of proton q = 1.602 × 10⁻¹⁹ C
And mass of proton m = 1.672 × 10⁻²⁷ kg
we know that; Magnetic Force F = qvBsinθ
where q is the charge of proton, v is velocity, B is the magnetic field and θ is angle ( 90° )
Also the Centripetal force experienced by the particle is;
F = mv² / r
where r is radius, m is mass of proton and v is velocity
hence;
qvBsinθ = mv² / r
we solve for v
rqvBsinθ = mv²
divide both sides by mv
rqvBsinθ / mv = mv² / mv
rqBsinθ / m = v
so we substitute
v = [ 0.0642 m × (1.602 × 10⁻¹⁹ C) × 0.260 T × sin(90°) ] / 1.67 × 10⁻²⁷ kg
v = 2.6740584 × 10⁻²¹ / 1.672 × 10⁻²⁷
v = 1.6 × 10⁶ m/s
Therefore, the proton speed of the proton was 1.6 × 10⁶ m/s
