Solveeit Logo
Login

Question

Physics Question on Current electricity

An electron moves in a circle of radius 1.0 cm1.0 \,cm with a constant speed of 4.0×106m/s4.0\times {{10}^{6}}m/s , the electric current at a point on the circle will be (e=1.6×10−19C)(e=1.6\times {{10}^{-19}}C)

A

1×10−11Ω1\times {{10}^{-11}}\Omega

B

1.1×10−7Ω1.1\times {{10}^{-7}}\Omega

C

5.1×10−7Ω5.1\times {{10}^{-7}}\Omega

D

2.1×10−7Ω2.1\times {{10}^{-7}}\Omega

Answer

1×10−11Ω1\times {{10}^{-11}}\Omega

Explanation

Solution

Using the releation i=qt=electronic chargetime period=eti=\frac{q}{t}=\frac{electronic\text{ }charge}{time\text{ }period}=\frac{e}{t} Hence, t=2πrvt=\frac{2\pi r}{v} So, i=e2πrvi=\frac{e}{2\pi rv} =ev2πr=\frac{ev}{2\pi r} =(1.6×10−19)(4×106)2×3.14×1.0×10−2=\frac{(1.6\times {{10}^{-19}})(4\times {{10}^{6}})}{2\times 3.14\times 1.0\times {{10}^{-2}}} =1×10−11Ω=1\times {{10}^{-11}}\Omega