An electron passes through a point 3.47 cm from a long straight wire as it moves at 30.5 % of the speed of light perpendicularly toward the wire. At that moment a switch is flipped, causing a current of 12.7 A to flow in the wire. Find the magnitude of the electron's acceleration a at that moment.

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 =

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