Magnetic Effects Of Current Question 308
Question: A proton of energy 200 MeV enters the magnetic field of 5 T. If direction of field is from south to north and motion is upward, the force acting on it will be [RPET 1997]
Options:
A) Zero
B) $ 1.6\times {{10}^{-10}}N $
C) $ 3.2\times {{10}^{-8}}N $
D) $ 1.6\times {{10}^{-6}}N $
Show Answer
Answer:
Correct Answer: B
Solution:
$ F=qvB $ also Kinetic energy $ K=\frac{1}{2}mv^{2} $
$ \Rightarrow v=\sqrt{\frac{2K}{m}} $ \ $ F=q\sqrt{\frac{2K}{m}}B $ $ =1.6\times {{10}^{-19}}\sqrt{\frac{2\times 200\times 10^{6}\times 1.6\times {{10}^{-19}}}{1.67\times {{10}^{-27}}}}\times 5 $ $ =1.6\times {{10}^{-10}}N $
BETA
