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Dmitry_Shevchenko [17]
3 years ago
5

Un protón se acelera por una diferencia de potencial de 2.6k V y se dirige a una región donde existe un campo magnético uniforme

de 4.5T. ¿Cuál es el valor?
Physics
1 answer:
Bond [772]3 years ago
7 0

Answer:

5.08*10^{-13}N

Explanation:

With this information we can calculate the velocity of the proton by taking into account the kinetic energy of the proton:

qV=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(1.6*10^{-19}C)(2.6*10^3V)}{1.67*10^{-27}kg}}=705835.3m/s

The magnitude of the magnetic force will be:

F=qvB=(1.6*10^{-19}C)(705835.3m/s)(4.5T)=5.08*10^{-13}N

hope this helps!

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(a)

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Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2

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Q=75\times 10^{-6}\ \mu C, r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2

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