Given :
Potential ,
[tex]P=644 \times 10^6\ V\\ \\P=6.44\times 10^8\ V[/tex]
Uniform magnetic field , B = 0.02 T .
Mass of proton , [tex]M_p=1.67\times 10^{-27}\ kg[/tex] .
Charge on proton , [tex]e=1.6\times 10^{-19}\ C[/tex] .
Also , velocity of proton is perpendicular to magnetic field .
To Find :
The radius of the proton's resulting orbit .
Solution :
Now , when velocity is perpendicular to magnetic field radius of orbit is given by :
[tex]r=\sqrt{\dfrac{2VM_p}{eB^2}}[/tex]
Putting all given values above :
[tex]r=\sqrt{\dfrac{2\times 6.44\times 10^8\times 1.67\times 10^{-27}}{1.6\times 10^{-19}\times 0.02^2}}\\\\r=183.33 \ m[/tex]
Therefore , radius of orbit is 183.33 m .
Hence , this is the required solution .
