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A velocity selector is built with a magnetic field of magnitude 5.2 T and an electric field of strength 4.6 × 10 ^4 N/C. The directions of the two fields are perpendicular to each other. At what speed will electrons pass through the selector without deflection? Let the charge of an electron q = −1.6 × 10 ^ −19
A.8.8 × 10^3 m/s
B 2.4 × 10 ^ - 15 m/s
C 7.9 × 10 ^ 13 m/s
D 2.0 x 10 ^ 4 m/s
Solutions - A velocity selector has a magnetic field that has a magnitude equal to 0.28 T and is perpendicular to an electric field that has a magnitude equal to 0.46 MV/m. What must the speed of a particle be for that particle to pass through the velocity selector undeflected? What kinetic energy must electrons have in order to pass through the velocity selector undeflected Because the magnetic force \boldsymbol{F} supplies the centripetal force \boldsymbol{F_c}, we have
\boldsymbol{qvB =} \boldsymbol{\frac{mv^2}{r}}.
Solving for \boldsymbol{r} yields
\boldsymbol{r =} \boldsymbol{\frac{mv}{qB}}.
Here, \boldsymbol{r} is the radius of curvature of the path of a charged particle with mass \boldsymbol{m} and charge \boldsymbol{q}, moving at a speed \boldsymbol{v} perpendicular to a magnetic field of strength \boldsymbol{B}. If the velocity is not perpendicular to the magnetic field, then \boldsymbol{v} is the component of the velocity perpendicular to the field. The component of the velocity parallel to the field is unaffected since the magnetic force is zero for motion parallel to the field. This produces a spiral motion rather than a circular one
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Right Answer Its (C)
