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A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field. What is the radius of the proton's resulting orbit? (mproton = 1.67 × 10-27 kg, e = 1.60 × 10-19 C) A proton starting from rest travels through a potential of 1.0 kV and then moves into a uniform 0.040-T magnetic field. What is the radius of the proton's resulting orbit? (proton = 1.67 × 10-27 kg, = 1.60 × 10-19 C) 0.14 m 0.11 m 0.080 m 0.19 m 0.17 m

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whitneytr12

Answer:

r = 0.11 m

Explanation:

Let's suppose that the magnetic field is perpendicular to the vector speed. In that case, the centripetal force would be the same as the Lorentz force.

[tex] F=ma_{c}=qvB [/tex]

[tex] m\frac{v^{2}}{r}=qvB [/tex].

Solving this equation for r, we will have:

[tex] r=\frac{mv}{qB} [/tex]

Now, to find  r we need to get the speed of the particle. Let's use the law of conservation of energy. The kinetic energy is equal to the electric potential energy:

[tex] U_{E}=K [/tex], but [tex] U_{E}=qV [/tex]

So we would have:

[tex] qV=1/2mv^{2} [/tex], then [tex] v=\sqrt{\frac{2qv}{m}} [/tex]

Finally [tex] r = m\frac{\sqrt{\frac{2qv}{m}}}{qB}= 0.11 m [/tex]     

Have a nice day!  

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