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A 6.47 micro-coloumb particle moves through a region of space where an electric field of magnitude 1300 N/C points in the positive x direction, and a magnetic field of magnitude 1.33 T points in the positive z direction. If the net force acting on the particle is 6.27E^-3 N in the positive x direction>

Required:
Calculate the magnitude of the particle's velocity. Assume the particle's velocity is in the xy plane.

Sagot :

Answer:

v = 248.8 m/s

Explanation:

Given that,

Charge, q = [tex]6.47\ \mu C[/tex]

Electric field, E = 1300 N/C

Magnetic field, B =1.33 T

The force acting on the particle, [tex]F=6.27\times 10^{-3}\ N[/tex]

We need to find the magnitude of the particle's velocity. The net force on the particle is given by :

[tex]F=qE+qvB\\\\6.27\times 10^{-3}=6.47\times 10^{-6}\times 1300+6.47\times 10^{-6}\times 1.33v\\\\6.27\times 10^{-3}-6.47\times 10^{-6}\times 1300=6.47\times 10^{-6}\times 1.33v\\\\-0.002141=6.47\times 10^{-6}\times 1.33v\\\\v=\dfrac{0.002141}{6.47\times 10^{-6}\times 1.33}\\\\v=248.8\ m/s[/tex]

So, the magnitude of the particle's velocity is 248.8 m/s.

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