Answer:
the magnitude of the electron's acceleration a at that moment =
Explanation:
The expression for the magnetic field is :
where:
= permeability of free space =
I = current in the wire = 12.7 A
r = distance from the wire to the field point = 3.47 cm =
The magnetic force exerted in the electrond passing through the straight conducting wire is:
F = Bvgsinθ
where :
B = magnetic field =
v = speed of light = (30.5 % )
q = charge of electrons =
θ = angle between speed of electron & field = 90°
F = () × (30.5 % ) × () × ( sin 90°)
F =
From Newtons second Law ;
F = ma
where m = mass =
Thus, the magnitude of the electron's acceleration a at that moment =