Question

# A proton of mass 1.67×10−27 kg and charge 1.68×10−19 C is projected with a speed of 2×106ms−1 at an angle of 60o to the X−axis. If a uniform magnetic field of 0.10 T is applied along Y−axis, the path of proton is :

A
a circle of radius 0.2 m and time period π×107s
B
a circle of radius 0.1 m and time period 2π×107s
C
a helix of radius 0.1 m and time period 2π×107s
D
a helix of radius 0.2 m and time period 4π×107s
Solution
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#### vx=vcosθvy=vsinθSince vy≠0, this component will be responsible for motion along the direction B( i.e y−axis).The path will be helical and pitch will be constant.Radius r of the circle perpendicular to the magnetic field Br=mvcosθqB=1.6×10−27×(2×106×cos600)1.6×10−19×0.1=0.1m

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