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

Answer:

Velocity will be $v=1.291\times 10^8m/sec$

Explanation:

We have given magnetic field B = 0.00005 T

Mass m = 238 U

We know that $1u=1.66\times 10^{-27}kg$

So 238 U $=238\times 1.66\times 10^{-27}=395.08\times 10^{-27}kg$

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 $qvB=\frac{mv^2}{r}$

$1.6\times 10^{-19}\times v\times 0.00005=\frac{395.08\times 10^{-27}v^2}{6379.44}$

$v=1.291\times 10^8m/sec$