Question

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.

Answer 1

Answer:

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

$F=qv\ \times B\\B=\frac{\mu_0I}{2\pi r}$

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

$F=qvB=(1.6*10^{-19}C)(9.75*10^{7}m/s)(1.6*10^{-4}T)=2.15*10^{-15}N$

And by replacing this factor in F=ma we have

$a=\frac{F}{m}=\frac{2.5*10^{-15}N}{9.1*10^{-31}kg}=2.75*10^{15}\frac{m}{s^2}$

where we have used the mass of the electron.

hope this helps!!

Answer 2

Answer:

a = 2.8 × 10¹⁵m/s²

Explanation:

The expression of magnetic field is

$F=qv\ \times B\\B=\frac{\mu_0I}{2\pi r}$

$\mu$ is the permitivity space

$I$ is the current in wire

r is the distance from the wire

substitute

$\mu=4\pi*10^{-7}T.m/a$

I = 18.9A

r = 2.35cm

$B=\frac{\mu_0I}{2\pi r}$

$B = \frac{(4\pi \times10^-^7)(18.9)}{2\pi (2.35\times10^-^2)} \\\\B = 1.6435\times10^-^4T$

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

$F=qvB$

$=(1.6*10^{-19}C)(9.75*10^{7}m/s)(1.6435*10^{-4}T)\\\\=2.56*10^{-15}N$

And by replacing this factor in

F=ma we have

$a=\frac{F}{m}$

$=\frac{2.564*10^{-15}N}{9.11*10^{-31}kg}\\\\=2.8*10^{15}\frac{m}{s^2}$