Magnetic Effects Of Current Question 283
Question: A proton of mass $ 1.67\times {{10}^{-27}}kg $ and charge $ 1.6\times {{10}^{-19}},C $ is projected with a speed of $ 2\times 10^{6},m/s $ at an angle of $ 60{}^\circ $ to the $ X- $ axis. If a uniform magnetic field of 0.104 Tesla is applied along $ Y- $ axis, the path of proton is [IIT-JEE 1995]
Options:
A) A circle of radius = 0.2 m and time period $ \pi \times {{10}^{-7}}s $
B) A circle of radius = 0.1 m and time period $ 2\pi \times {{10}^{-7}}s $
C) A helix of radius = 0.1 m and time period $ 2\pi \times {{10}^{-7}}s $
D) A helix of radius = 0.2 m and time period $ 4\pi \times {{10}^{-7}}s $
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Answer:
Correct Answer: C
Solution:
Path of the proton will be a helix of radius $ r=\frac{mv\sin \theta }{qB} $ (where q = Angle between $ \overrightarrow{B},\text{and},\overrightarrow{v,} $ )
$ \Rightarrow \ r=\frac{1.67\times {{10}^{-27}}\times 2\times 10^{6}\times \sin 30^{o}}{1.6\times {{10}^{-19}}\times 0.104} $ $ =0.1m $ Time period $ T=\frac{2\pi m}{qB}=\frac{2\pi \times 1.67\times {{10}^{-27}}}{1.6\times {{10}^{-19}}\times 0.104} $ $ =2\pi \times {{10}^{-7}}sec $
BETA
