Respuesta :
Answer:
The electric and magnetic field are 6.34 N/C and [tex]2.11\times10^{-8}\ T[/tex].
Explanation:
Given that,
Distance = 4.5 m
Time = 2.84 μs
We need to calculate the acceleration
Using equation of motion
The distance covered by the electron is
[tex]s=ut+\dfrac{1}{2}at^2[/tex]
When, the electron at rest
[tex]s = \dfrac{1}{2}at^2[/tex]
Where, s = distance
a = acceleration
t = time
Put the value into the formula
[tex]4.5=\dfrac{1}{2}\times a\times(2.84\times10^{-6})^2 [/tex]
[tex]a=\dfrac{2\times4.5}{(2.84\times10^{-6})^2}[/tex]
[tex]a=1.116\times10^{12}\ m/s^2[/tex]
We need to calculate the electric field
Using formula of the electric field
[tex]E=\dfrac{F}{q}[/tex]
[tex]E=\dfrac{ma}{q}[/tex]
Put the value into the formula
[tex]E=\dfrac{9.1\times10^{-31}\times1.116\times10^{12}}{1.6\times10^{-19}}[/tex]
[tex]E=6.34\ N/C[/tex]
We need to calculate the magnetic field
Using formula of magnetic field
[tex]B = \dfrac{E}{c}[/tex]
Put the value into the formula
[tex]B=\dfrac{6.34}{3\times10^{8}}[/tex]
[tex]B=2.11\times10^{-8}\ T[/tex]
According to the right hand rule,
The direction of magnetic field is outward because the direction of force is upward.
Hence, The electric and magnetic field are 6.34 N/C and [tex]2.11\times10^{-8}\ T[/tex].
