Can someone help please?

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 t

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

or:A velocity selector is built with a magnetic field of magnitude 5.2 T and an electric field of strength 4.6 \u00d7 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 = \u22121.6 \u00d7 10 ^ \u221219A.8.8 \u00d7 10^3 m/sB 2.4 \u00d7 10 ^ - 15 m/sC 7.9 \u00d7 10 ^ 13 m/sD 2.0 x 10 ^ 4 m/s

or: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 You need of any our Technical Support agent. Please call Physics Tech Support team at +1 888 597 3962 Our experts are available 24/7 by phone.Right Answer Its (C)

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