An electron has a velocity of 6.0 *10⁶ m/s in the positive x direction at a point where the magnetic field has the components B_x = 3.0 T,B_y = 1.5 T,and B_z = 2.0 T.What is the magnitude of the acceleration of the electron at this point? A)2.1* 10¹⁸ m/s² B)1.6* 10¹⁸ m/s² C)2.6 * 10¹⁸ m/s² D)3.2 *10¹⁸ m/s² E)3.7 *10¹⁸ m/s²

Question

Quizplus · Accepted Answer

To find the acceleration of the electron, we need to use the Lorentz force law, which states that the force on a charged particle in a magnetic field is given by F = q(v x B), where q is the charge of the particle, v is its velocity, and B is the magnetic field. In this case, the electron has a velocity only in the x direction and the magnetic field has components in all three directions. Therefore, we need to find the x-component of the magnetic field, which is 3.0 T, and use it in the Lorentz force law. The force on the electron is then given by F = e(v x B) = e(6.0 x 3.0) = 18e N, where e is the charge of an electron. The acceleration of the electron is then given by a = F/m = (18e)/m, where m is the mass of the electron. We know that the mass of the electron is 9.11 x 10^-31 kg, so we can calculate the acceleration: a = (18e) / (9.11 x 10^-31) = 1.97 x 10^12 m/s^2 This is equivalent to 2.6 x 10^18 F1F1F1S1F1F1F10 10¹⁸ m/s^2. Therefore, the best choice is C.

An Electron Has a Velocity of 6