A beam of protons with a velocity of 4×105 ms−1  enters a uniform magnetic field of 0.3T.  The velocity makes an angle of  60° with the magnetic field. Find the radius of the helical path taken by the proton beam

A beam of protons with a velocity of 4×105ms1  enters a uniform magnetic field of 0.3T.  The velocity makes an angle of  60° with the magnetic field. Find the radius of the helical path taken by the proton beam

  1. A

    1.2 cm

  2. B

    1.6cm

  3. C

    1.7cm

  4. D

    1.8cm

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    Solution:

    The components of the proton’s velocity along and perpendicular to the magnetic field are v=4×105ms1cos600=2×105ms1.
     


    And   v=4×105ms1sin60=23×105ms1
    As the force qv×B is perpendicular to the magnetic field, the component v will remain constant. In the plane perpendicular to the field, the proton will describe a circle whose radius is obtained from the equation qvB=mv2r
    Or   r=mvqB=1.67×1027kg×23×105ms11.6×1019C×(0.3T)=0.012m=1.2cm
     

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