A charged particle moves through a velocity selector at a constant speed in a st

Question

A charged particle moves through a velocity selector at a constant speed in a straight line. The electric field of the velocity selector is 3.65x10^3 N/C, while the magnetic field is 0.360 T. When the electric field is turned off, the charged particle travels on a circular path whose radius is 4.00 cm. Find the charge-to-mass ratio of the particle.

Explanation / Answer

We know that                  F = B q v     and        F = E q From these relations we have                 E q = B q v           ==> E = Bv          .........1 We also know that radius of the path r = m v / qB                                                   ==> v = q B r / m .....2 Substituting 2 in 1 we get                   E = B ( q B r / m)        ==> q /m = E / B2 r                        = 3.65 x 103 / 0.36 * 0.36 * 0.04                        = 7.04 x 105 C / kg

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