A charged particle with a velocity of 2 km/s along the + x axis enters a region

Question

A charged particle with a velocity of 2 km/s along the + x axis enters a region of uniform electric and magnetic fields. The magnetic field is upward along the + y axis with a magnitude of 1.5 T. Find the magnitude and direction of the electric field if the particle is to pass through without deflection

Explanation / Answer

e1 = unit vector along positive x-axis … ……… e2 = unit vector along positive y-axis … ……… e3 = unit vector along positive z-axis … The total force on the particle is … … F = Q E + Q ( v0 × B ) = m a … so that to pass through undeflected … m a = 0 … which means … Q E + Q ( v0 × B ) … or … E = - ( v0 × B ) … … v0 × B … --> … e2 × ( - e3 ) = - e2 × e3 = - e1 That is … the electric field vector must be along the negative x-axis … its magnitude must be This is a straigtforward application of the Lorentz force. F = 0 = q(E + v × B) E = -v × B =

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