Question

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.5 meters in the first 2.84 μs after it is released. What is magnitude of the electric field? What is the direction of the magnetic field? Are we justified if ignoring gravity on the electron in this situation?

Answer 1

The electric and magnetic field are 6.34 N/C and
`2.11*10^(-8)\ T`.

Step-by-step 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

`s=ut+(1)/(2)at^2`

When, the electron at rest

`s = (1)/(2)at^2`

Where, s = distance

a = acceleration

t = time

Put the value into the formula

`4.5=(1)/(2)* a*(2.84*10^(-6))^2`

`a=(2*4.5)/((2.84*10^(-6))^2)`

`a=1.116*10^(12)\ m/s^2`

We need to calculate the electric field

Using formula of the electric field

`E=(F)/(q)`

`E=(ma)/(q)`

Put the value into the formula

`E=(9.1*10^(-31)*1.116*10^(12))/(1.6*10^(-19))`

`E=6.34\ N/C`

We need to calculate the magnetic field

Using formula of magnetic field

`B = (E)/(c)`

Put the value into the formula

`B=(6.34)/(3*10^(8))`

`B=2.11*10^(-8)\ T`

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
`2.11*10^(-8)\ T`.