Question
A proton moving with a velocity 3 10^ 5 ~m / s enters a magnetic field of 0.3 tesla at an angle of 30^ with the field. The radius of curvature of its path will be ( e / m for proton =10^ 8 C / kg )
BITSAT
Physics
Magnetic Effects of Current
2021
BITSAT 2021
BITSAT Physics Question (2021) — Solution
Options
- A. 2 ~cm
- B. 0.5 ~cm
- C. 0.02 ~cm
- D. 1.25 ~cm
Answer
B. 0.5 ~cm
Step-by-step solution
r= m v B e = 3 10^ 5 30^ 0.3 10^ 8 3 10^ 5 1 2 3 10^ 7 =0.5 10^ -2 m=0.5 ~cm
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