The radius of the given circular path moved by the proton which is perpendicular to a magnetic field of 0.87 T determined as 4 cm.
The magnitude of the magnetic force on the proton is determined using the following formula;
F = qvBsin(θ)
The centripetal force of the circular path is given as;
F = mv²/r
Solve the two equations together,
[tex]\frac{mv^2}{r} = qvBsin(90)\\\\r = \frac{mv}{qB}[/tex]
Where;
The speed of the proton is determined from its kinetic energy;
[tex]K.E = \frac{1}{2}mv^2\\\\ v= \sqrt{\frac{2K.E}{m} } \\\\v = \sqrt{\frac{2 \times 9.3 \times 10^{-15}}{1.67 \times 10^{-27}} } \\\\v = 3.33\times 10^6 \ m/s[/tex]
The radius of the circular path is calculated as follows;
[tex]r = \frac{(1.67 \times 10^{-27}) \times (3.33 \times 10^6)}{(1.6\times 10^{-19}) \times (0.87)} \\\\r = 0.04\ m\\\\r = 4 \ cm[/tex]
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