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A proton wit a kinetic energy of 9. 3 x 10^-15 j moves perpendicular to a magnetic field of 0. 87 t. What is the radius of its circular path?

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onyebuchinnaji

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.

Magnetic force of the proton

The magnitude of the magnetic force on the proton is determined using the following formula;

F = qvBsin(θ)

Centripetal force on the proton

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;

  • v is the speed of the proton

Speed of the proton

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]

Radius of the circular path

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]

Learn more about magnetic force here: https://brainly.com/question/13277365

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