Question

An electron travels through a particular point in an experimental apparatus with a velocity of 0.949 \times 10^6×10 ​6 ​​ m/s and at an angle of +61.5^\circ ​∘ ​​ (clockwise) relative to the positive x-axis in the x-y plane. If the magnetic field at this point has a magnitude of 0.01 T and is directed in along the negative y-axis, what is the magnitude of the acceleration of the electron at this point?

Answer 1

The magnitude of the acceleration of the electron at this point is
`3.94*10^(14)\ m/s^2`.

Step-by-step explanation:

Given that,

Velocity
`v= 0.949*10^(6)\ m/s`

Angle = 61.5°

Magnetic field = 0.01 T

We need to calculate the magnetic force

Using formula of magnetic force

`F=Bev\cos\theta`

Where, B = magnetic field

v = velocity

e = charge of electron

Put the value into the formula

`F=0.01*1.6*10^(-19)*0.949*10^(6)\cos61.5`

`F=3.59*10^(-16)\ N`

We need to calculate the acceleration

Using newton's second law

`F= ma`

`a = (F)/(m)`

Put the value into the formula

`a=(3.59*10^(-16))/(9.1*10^(-31))`

`a=3.94*10^(14)\ m/s^2`

Hence, The magnitude of the acceleration of the electron at this point is
`3.94*10^(14)\ m/s^2`.