A beam of protons with a velocity 4.0×105 ms-1 enters a uniform magnetic field of 0.2 T at an angle of 300 to the magnetic field. If the radius of the helical path taken by the proton beam is p×10−2 m, then find p (Take mproton=1.6×10−27kg)
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Detailed Solution
The radius of helical path r=mv⊥Bq r=mvsinθBq=1.6×10−274×105sin3000.21.6×10−19=1×10−2 m