Question

Suppose the Earth's magnetic field at the equator has magnitude 0.00005 T and a northerly direction at all points. How fast must a singly ionized uranium atom (m=238u, q=e) move so as to circle the Earth 1.44 km above the equator? Give your answer in meters/second.

Answer 1

Velocity will be
`v=1.291* 10^8m/sec`

Step-by-step explanation:

We have given magnetic field B = 0.00005 T

Mass m = 238 U

We know that
`1u=1.66* 10^(-27)kg`

So 238 U
`=238* 1.66* 10^(-27)=395.08* 10^(-27)kg`

Radius
`=R+1.44=6378+1.44=6379.44KM`

We know that magnetic force is given by

`F=qvB` which is equal to the centripetal force

So
`qvB=(mv^2)/(r)`

`1.6* 10^(-19)* v* 0.00005=(395.08* 10^(-27)v^2)/(6379.44)`

`v=1.291* 10^8m/sec`