An electron passes through a point 2.35 cm 2.35 cm from a long straight wire as it moves at 32.5 % 32.5% of the speed of light perpendicularly toward the wire. At that moment a switch is flipped, causing a current of 18.9 A 18.9 A to flow in the wire. Find the magnitude of the electron's acceleration a a at that moment.

a = 2.75*10^{15}m/s^2

Explanation:

The acceleration of the electron is generated by the Lorenz's force, due to the magnetic field produced by the wire. Hence, we have

I=18.9 A

r=2.35cm=0.0235m

mu=4pi*10^{-7}

v=(0.325)(3*10^{8}m/s)=9.75*10^{7}m/s

q=1.6*10^{-19} C

By replacing these values in the expression for B we have

B=1.6*10^{-4} T

The direction of the magnetic field is perpendicular to the direction of the motion of the electron. Thus

And by replacing this factor in F=ma we have

where we have used the mass of the electron.

hope this helps!!

a = 2.8 × 10¹⁵m/s²

Explanation:

The expression of magnetic field is

is the permitivity space

is the current in wire

r is the distance from the wire

substitute

I = 18.9A

r = 2.35cm

The direction of the magnetic field is perpendicular to the direction of the motion of the electron. Thus

The magnetic force exerted on electrons passing through a straight conducting wire is

And by replacing this factor in

F=ma we have

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.