Magnetic Effects Of Current Question 269
Question: An $ \alpha - $ particle travels in a circular path of radius 0.45 m in a magnetic field $ B=1.2,Wb/m^{2} $ with a speed of $ 2.6\times 10^{7},m/\sec $ . The period of revolution of the $ \alpha - $ particle is
Options:
A) $ 1.1\times {{10}^{-5}},\sec $
B) $ 1.1\times {{10}^{-6}}\sec $
C) $ 1.1\times {{10}^{-7}},\sec $
D) $ 1.1\times {{10}^{-8}},\sec $
Show Answer
Answer:
Correct Answer: C
Solution:
$ T=\frac{2\pi m}{qB}=\frac{2\pi r}{v}=\frac{2\times 3.14\times 0.45}{2.6\times 10^{7}}=1.08\times {{10}^{-7}}\sec $
BETA
