Respuesta :
Answer:
The magnitude of electron acceleration is [tex]2.34 \times 10^{15}[/tex] [tex]\frac{m}{s^{2} }[/tex]
Explanation:
Given:
Distance from the wire to the field point [tex]r = 2.83 \times 10^{-2}[/tex] m
Speed of electron [tex]v = 35.5 \%c[/tex]
Current [tex]I = 17.7[/tex] A
For finding the acceleration,
First find the magnetic field due to wire,
[tex]B = \frac{\mu _{o}I }{2\pi r }[/tex]
Where [tex]\mu_{o} = 4\pi \times 10^{-7}[/tex]
[tex]B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }[/tex]
[tex]B = 12.50 \times 10^{-5}[/tex] T
The magnetic force exerted on the electron passing through straight wire,
[tex]F = qvB[/tex]
[tex]F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}[/tex]
[tex]F = 21.3 \times 10^{-16}[/tex] N
From the newton's second law
[tex]F = ma[/tex]
Where [tex]m =[/tex] mass of electron [tex]= 9.1 \times 10^{-31}[/tex] kg
So acceleration is given by,
[tex]a = \frac{F}{m}[/tex]
[tex]a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }[/tex]
[tex]a = 2.34 \times 10^{15}[/tex] [tex]\frac{m}{s^{2} }[/tex]
Therefore, the magnitude of electron acceleration is [tex]2.34 \times 10^{15}[/tex] [tex]\frac{m}{s^{2} }[/tex]
