Answered

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A charged particle (charge 1.6x10-19 C and mass 1.67x10-27 kg) is initially moving with a velocity of 2x105 m/s and then moves into a region having a magnetic field and an electric field (6x104 V/m). The direction of initial velocity is perpendicular to the electric field and magnetic field. If the charged particle keep moving in the original direction without being deflected, what is the magnitude of the magnetic field

Sagot :

boffeemadrid

Answer:

[tex]0.3\ \text{T}[/tex]

Explanation:

q = Charge = [tex]1.6\times 10^{-19}\ \text{C}[/tex]

m = Mass of particle = [tex]1.67\times 10^{-27}\ \text{kg}[/tex]

v = Velocity = [tex]2\times 10^5\ \text{m/s}[/tex]

E = Electric field = [tex]6\times 10^4\ \text{V/m}[/tex]

B = Magnetic field

Magnetic field is given by

[tex]B=\dfrac{E}{v}\\\Rightarrow B=\dfrac{6\times 10^4}{2\times 10^5}\\\Rightarrow B=0.3\ \text{T}[/tex]

The magnitude of magnetic field is [tex]0.3\ \text{T}[/tex]

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