In a cloud chamber experiment, a proton enters a uniform 0.260 T magnetic field directed perpendicular to its motion. You measure the proton's path on a photograph and find that it follows a circular arc of radius 6.42 cm. Required:
How fast was the proton moving?

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

a=d/t

a="triangle"v/t

explanation:

a = acceleration

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