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

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2

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