Question

An $ \alpha - $ particle with a specific charge of $ 2.5\times {{10}^{7}}C\,k{{g}^{-1}} $ moves with a speed of $ 2\times {{10}^{5}} $ $ m{{s}^{-1}} $ in a perpendicular magnetic field of $ 0.05 \,T $ . Then the radius of the circular path described by it is:

• A. 8 cm
• B. 4 cm
• C. 16cm
• D. 2cm
• E. 32 cm

Answer 1

Answer: (C)

$ r=\frac{mv}{Bq} $
$ \Rightarrow $ $ r=\frac{v}{B\frac{q}{m}}=\frac{2\times {{10}^{5}}}{0.005\times 2.5\times {{10}^{7}}} $
$ =\frac{2\times {{10}^{7}}}{12.5\times {{10}^{7}}}=\frac{200}{12.5}cm=16\,cm $