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Moving Charges And Magnetism

Question
CBSEENPH12037624

A beam of protons with a velocity 4 x 105 m/s enters a uniform magnetic field of 0.3 T at an angle 60° to the magnetic field. Find the radius of the helical path taken by the proton beam. Also find the pitch of the helix mp = 1.67 x 10–27 kg.

Solution
Given, the velocity of protons, v = 4 x 105 m/s 
Uniform magnetuc field, B = 0.3 T 
Angle made by particle with magnetic field = 60o 
The magnetic force will act like centripetal force.
Therefore, 

Using the formula,
                           r = mvqB

           r = 1.67 × 10-27 × 4 × 105 sin 60°1.6 × 10-19× 0.3   = 1.2 × 10-2    = 1.2 cm  

Pitch is the distance moved along the magnetic field in one rotation.

Therefore,

 
T =2πrv sin θ = 2.175 × 10-7 S

               P = v cos θ. T 

                 = 4 × 105 × 12×2.175 × 10-7 = 4.35 cm