Question

A long straight wire carries a current 1.5A. an electron travels with a speed of 5*10^6cm/sec parallel to the wire 10cm from it and in the same direction as the current. What force does the magnetic field of the current exert on the moving electron

Answer 1

First of all let's see the magnitude of magnetic field created by wire carrying current

$B=\frac{\mu_oI}{2\pi r}=\frac{4\pi\times10^{-4}\times1.5}{2\pi\times0.1}=3\times10^{-3}T$

force due to magnetic field =$q(v\times B)=1.6\times10^{-19}(5\times10^6\times3\times10^{-3})=2.4\times10^{-15}N$ =