Q - 19

Asked by majethiyarishat9566.12sdatl | 24th Oct, 2020, 02:20: PM

Expert Answer:

if proton enters in magnetic field at an angle 60o with respect to magnetic field direction  and velocity magnitude of proton is v ,
 
then resolved componet of velocity ( v sin60) perpendicular to magnetic field direction makes the proton in circular path whereas
 
the parallel component ( v cos60) makes the proton to move along the magnetic field direction resulting a helical path for proton.
 
Radius R of circular path is obtained from ,  ( m v2 sin260) / R  = ( q B v sin60 )    or  R = ( m v sin60) / ( q B )
 
where m = mass of proton, q is charge of proton and B is magnetic field induction
 
Hence Radius of circular path , R = ( 1.673 × 10-27 × 4 × 105 × 0.866 ) / ( 1.602 × 10-19 × 0.3 ) 
 
R = 12.06 × 10-3 m = 12.06 mm
 
Period of revolution T = ( 2π R ) / ( v sin60 )
 
In one period of revolution, proton moves a distance d along the magnetic field direction as ,
 
d = vcos60 × T  = ( 2π R ) / ( tan60 ) = ( 2π × 12.06 × 10-3 ) / tan60 = 43.75 × 10-3 m = 43.75 mm
 
Radius R of helical path = 12.06 mm
 
Pitch d of helical path = 43.75 mm

Answered by Thiyagarajan K | 24th Oct, 2020, 05:45: PM

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