Question

An electron moves in a circle of radius $1.0 \,cm$ with a constant speed of $ 4.0\times {{10}^{6}}m/s $ , the electric current at a point on the circle will be $ (e=1.6\times {{10}^{-19}}C) $

• A. $ 1\times {{10}^{-11}}\Omega $
• B. $ 1.1\times {{10}^{-7}}\Omega $
• C. $ 5.1\times {{10}^{-7}}\Omega $
• D. $ 2.1\times {{10}^{-7}}\Omega $

Answer 1

Answer: (A)

Using the releation
$ i=\frac{q}{t}=\frac{electronic\text{ }charge}{time\text{ }period}=\frac{e}{t} $
Hence, $ t=\frac{2\pi r}{v} $
So, $ i=\frac{e}{2\pi rv} $
$ =\frac{ev}{2\pi r} $
$ =\frac{(1.6\times {{10}^{-19}})(4\times {{10}^{6}})}{2\times 3.14\times 1.0\times {{10}^{-2}}} $ $ =1\times {{10}^{-11}}\Omega $