Respuesta :
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
[tex]F=qv\ \times B\\B=\frac{\mu_0I}{2\pi r}[/tex]
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
[tex]F=qvB=(1.6*10^{-19}C)(9.75*10^{7}m/s)(1.6*10^{-4}T)=2.15*10^{-15}N[/tex]
And by replacing this factor in F=ma we have
[tex]a=\frac{F}{m}=\frac{2.5*10^{-15}N}{9.1*10^{-31}kg}=2.75*10^{15}\frac{m}{s^2}[/tex]
where we have used the mass of the electron.
hope this helps!!
Answer:
a = 2.8 × 10¹⁵m/s²
Explanation:
The expression of magnetic field is
[tex]F=qv\ \times B\\B=\frac{\mu_0I}{2\pi r}[/tex]
[tex]\mu[/tex] is the permitivity space
[tex]I[/tex] is the current in wire
r is the distance from the wire
substitute
[tex]\mu=4\pi*10^{-7}T.m/a[/tex]
I = 18.9A
r = 2.35cm
[tex]B=\frac{\mu_0I}{2\pi r}[/tex]
[tex]B = \frac{(4\pi \times10^-^7)(18.9)}{2\pi (2.35\times10^-^2)} \\\\B = 1.6435\times10^-^4T[/tex]
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
[tex]F=qvB[/tex]
[tex]=(1.6*10^{-19}C)(9.75*10^{7}m/s)(1.6435*10^{-4}T)\\\\=2.56*10^{-15}N[/tex]
And by replacing this factor in
F=ma we have
[tex]a=\frac{F}{m}[/tex]
[tex]=\frac{2.564*10^{-15}N}{9.11*10^{-31}kg}\\\\=2.8*10^{15}\frac{m}{s^2}[/tex]