Question

A 4.80 g bullet moves with a speed of 170 m/s perpendicular to the Earth's magnetic field of 5.00×10−5T.

Part A
If the bullet possesses a net charge of 1.06×10−8 C , by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled 1.00 km ?

Answer 1

`3.24* 10^(-7)\ \text{m}`

Step-by-step explanation:

m = Mass of bullet = 4.8 g

v = Velocity of bullet = 170 m/s

B = Magnetic field of Earth =
`5* 10^(-5)\ \text{T}`

q = Charge of bullet =
`1.06* 10^(-8)\ \text{C}`

a = Acceleration

Time the bullet will be in the air for is
`t=(1000)/(170)=5.88\ \text{s}`

Force is given by

`F=ma`

Magnetic force is given by

`F=qvB`

So

`ma=qvB\\\Rightarrow a=(qvB)/(m)\\\Rightarrow a=(1.06* 10^(-8)* 170* 5* 10^(-5))/(4.8* 10^(-3))\ \text{m/s}^2`

From the linear equations of motion we have

`s=ut+(1)/(2)at^2\\\Rightarrow s=0+(1)/(2)* (1.06* 10^(-8)* 170* 5* 10^(-5))/(4.8* 10^(-3))* 5.88^2\\\Rightarrow s=3.24* 10^(-7)\ \text{m}`

The defelection of the bullet is
`3.24* 10^(-7)\ \text{m}`