A proton of charge +1.60×10-19 C and mass 1.67×10-27 kg is introduced into a reg

Question

A proton of charge +1.60×10-19 C and mass 1.67×10-27 kg is introduced into a region of B = 0.146 T with an initial velocity of 1.03×106 m/s perpendicular to B. What is the radius of the proton's path? Suppose the initial velocity of the proton makes an angle of 50 o with the direction of the magnetic field B. What is the pitch of the helix that describes the path of the proton? The pitch is the distance the proton travels along the axis of the helix in the time required for the proton to make one complete loop around the axis.

Explanation / Answer

A.

Centripetal force = Magnetic force of the proton
Fc = Fm
(mv^2)/r = qvB
(mv)/qB = r

r = 1.67*10^-27*1.03*10^6/(1.6*10^-19*0.146)

r = 0.0736 m

r = 7.36 cm

B. Pitch = Vx*T

Vx = V*cos A

T = 2*pi*m/(qB*sin A)

Pitch = 2*pi*m*v*cot A/(qB)

Pitch = 2*3.14*1.67*10^-27*1.03*10^6*cot 50 deg/(1.6*10^-19*0.146)

Pitch = 0.388 m

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