A long straight wire suspended in the air carries a conventional current of 8.2

Question

A long straight wire suspended in the air carries a conventional current of 8.2 amperes in the ?x direction as shown in the figure (the wire runs along the x axis). At a particular instant an electron at location <0,-0.003,0>m has velocity< -1.5e5,1.8e5,0>m/s.

(a) What is the (vector) magnetic field due to the wire at the location of the electron?

(b) What is the (vector) magnetic force on the electron due to the current in the wire?

(c) If the moving particle were a proton instead of an electron, what would be the (vector) magnetic force on the proton?

So I know :

I just don't understand in Fmag where they got 2.34e5 m/s ? Could someone explain that to me. And then how they got their final answer for Fmag, I can read that it says cross product but I need to see these two things worked out.

Explanation / Answer

A) B = uo*i/(2*pi*r) = 2*10^-7*i/r = 2*10^-7*8.2/0.003 = 5.467*10^-4 T..along +Z -axis..

B) F = q*(V X B) = 1.6*10^-19*(98.406 i + 82.005 j) = (157.45810^-19 i + 131.208*10^-19 j)..

Fmag = 204.95*10^-19 N

C) F mag = 204.95*10^-19 N

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