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 Bx = 3.0 T, By = 1.5 T, and Bz = 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

The magnitude of the acceleration of an electron moving in a magnetic field is given by the equation: 
a = (e/m)*(v x B), 
where e is the charge of the electron, m is its mass, v is its velocity, and B is the magnetic field vector. 
Taking the cross product of v and B, we get: 
v x B = (v_y*B_z - v_z*B_y, v_z*B_x - v_x*B_z, v_x*B_y - v_y*B_x)
Plugging in the given values, we get: 
v x B = (-1.5x106*2.0, -2.0x3.0x106, 6.0x106*1.5) = (-3.0, -6.0, 9.0) x 1012 m/s2 
Multiplying by the charge-to-mass ratio of the electron (1.76x1011 C/kg), we get: 
a = (1.76x1011)*(3.0, 6.0, -9.0) x 1012 m/s2 = (5.28, 10.56, -15.84) x 1018 m/s2 
Taking the magnitude of the acceleration, we get: 
|a| = sqrt(5.28^2 + 10.56^2 + 15.84^2) x 1018 m/s2 ≈ 2.6 x 1018 m/s2, 
which is closest to answer choice C.

An Electron Has a Velocity of 6