A very long straight wire carries a current of 5 A. An electron moves with a velocity of `10^(6)ms^(-1)` remaining parallel to the wire at a distance of 10 cm from wire in a direction opposite to that of electric current. Find the force on this electron. (Here the mass of electron is taken as constant)
`(e=-1.6xx10^(-19)C,mu_(0)=4pixx10^(-7)SI)`

Explanation

1. I = 5 A,
V = `10^(6)` m/s
y = 0.1 m
e = `1.6xx10^(-19)C`
`mu_(0)=4pixx10^(-7)SI`
Magnetic field at a distance y from the conducting wire,
`B=(mu_(0)I)/(2piy)`
= `(4pixx10^(-7)xx5)/(2pixx0.1)`
`B=10^(-5)T`
2. This magnetic field is perpendicular to page. Electron moves in the direction opposite to current so v and B are perpendicular to each other.
`thereforetheta=pi/2" "thereforesintheta="sin"pi/2=1`
Now, `vecF=e(vecvxxvecB)=BeVsintheta=BeV`
= `10^(-5)xx1.6xx10^(-19)xx10^(6)`
= `16xx10^(-19)N`