Answer:
Therefore the velocity of the proton is [tex]3.79 \times 10^{-3}[/tex] m/s towards east.
Explanation:
The mass of proton is m= [tex]1.67\times 10^{-27}[/tex] kg.
q= [tex]1.6\times 10^{-19}[/tex] C is the magnitude of the charge of a proton.
The earth's magnetic field is B=[tex]2.7\times 10^{-5}[/tex] T toward north.
Let v be the velocity of the proton towards east.
The angle between the velocity of proton towards east and magnetic field towards north is [tex]90^\circ[/tex].
We know that,
[tex]F=|q|Bvsin \theta[/tex]
[tex]\Rightarrow mg=|q|Bvsin \theta[/tex]
[tex]\Rightarrow v=\frac{mg}{|q|Bsin\theta}[/tex]
Now putting the all values
[tex]\therefore v=\frac{1.67\times 10^{-27}\times 9.8}{1.6\times 10^{-19}\times 2.7\times 10^{-5} sin 90^\circ}[/tex]
[tex]=3.79 \times 10^{-3}[/tex] m/s
Therefore the velocity of the proton is [tex]3.79 \times 10^{-3}[/tex] m/s towards east.