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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) |
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Answer» Solution :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` |
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