In a cloud chamber experiment, a proton enters a uniform 0.280 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 6.12 cm. How fast was the proton moving?

Given Information:

Magnetic field = B = 0.280 T

Radius = r = 6.12 cm = 0.0612 m

Required Information:  

Speed of proton = v = ?

Speed of proton = v = 1641.77x10³ m/s

Explanation:

Since the proton is moving in a circular path we can model it in terms of magnetic force and centripetal force.

The magnetic force is given by

F = qvB

The centripetal force is given by

F = mv²/r

equating the both equations yields,

mv²/r = qvB

mv = qBr

v = qBr/m

Where q = 1.6x10⁻¹⁹ C is the of the proton, m = 1.67x10⁻²⁷ kg is the mass of proton, B is the magnetic field and r is the radius o circular arc around which proton is moving.

v = 1.6x10⁻¹⁹*0.280*0.0612/1.67x10⁻²⁷

v = 1641772.45 m/s

v = 1641.77x10³ m/s

Therefore, the proton is moving at the speed of 1641.77x10³ m/s.

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