A charged particle moves through a region of space that has both a uniform electric field and a uniform magnetic field. In order for the particle to move through this region at a constant velocity,
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The two main forces are the electric force (\( \vec{F}_E \)) due to the electric field (\( \vec{E} \)) and the magnetic force (\( \vec{F}_B \)) due to the magnetic field (\( \vec{B} \)). Show more…
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A charged particle can move with constant velocity through a region containing both an electric field and a magnetic field only if the electric field is perpendicular to the magnetic field. None of the given options: magnetic field is perpendicular to the velocity vector, electric field is parallel to the magnetic field, electric field is parallel to the velocity vector.
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Now consider the case in which a positively charged particle is moving in a uniform magnetic and electric field perpendicular to each other. The particle velocity v is perpendicular to both fields (as is shown in the figure below). Determine the magnitude of the velocity v for which the sum of the two forces, i.e. magnetic and electric, on the charged particle is zero. (c) If we increase the magnitude of the magnetic field by a factor 1000 (from B to 1000 B), determine the magnitude of the velocity, call it v2, for which the two forces are in equilibrium again. (d) A mass equal to the mass of a proton, but a net charge twice the proton charge (qproton=1.6 x 10^-19 C, and mproton=1.67 x 10^-27 Kg) is moving on a circular orbit of radius 100 cm in a uniform magnetic field of 1 kT. What is the speed of the charged particle? NOTE: The magnitude of the acceleration of the particle is a=v^2/r NOTE: Work out the proper units
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