Question

A proton is projected with a velocity $ {{10}^{7}}m{{s}^{-1}}, $ at right angles to a uniform magnetic field of induction 100 mT. The time (in seconds) taken by the proton to traverse $ 90{}^\circ $ arc is: (Mass of proton $ =1.65\times {{10}^{-27}}kg $ and charge of proton $ =1.6\times {{10}^{-19}}C $ )

• A. $ 0.81\times {{10}^{-7}} $
• B. $ 1.62\times {{10}^{-7}} $
• C. $ 2.43\times {{10}^{-7}} $
• D. $ 3.24\times {{10}^{-7}} $

Answer 1

Answer: (B)

When proton enters in a uniform magnetic field at right angle then it moves on a circular path. In this case velocity of proton. $ v=\frac{Bqr}{m} $ where r = radius of circular path Distance covered by proton to traverse 90° arc $ =\frac{1}{4} $ circumference $ d=\frac{1}{4}\times 2\pi r=\pi r/2 $ Time taken by proton to cover distance d $ t=\frac{\pi r/2}{v}=\frac{\pi r}{2v} $ $ t=\frac{\pi }{2}\frac{r}{Bqr/m} $ $ t=\frac{\pi m}{2Bq} $ Putting $ m=1.65\times {{10}^{-27}}kg, $ $ B=100mT=100\times {{10}^{-3}}T,q=1.6\times {{10}^{-19}}C $ $ t=\frac{3.14\times 1.65\times {{10}^{-27}}}{2\times 100\times {{10}^{-3}}\times 1.6\times {{10}^{-19}}} $ $ t=1.62\times {{10}^{-7}}\sec $