Electron velocity and unit vector notation

[4eleven]

Homework Statement

An electron has an initial velocity with an x component of zero, a y component of 17.2 km/s, a z component 20.7 km/s, and a constant acceleration of 1.95 x 1012 m/s2 in the positive x direction in a region in which uniform electric and magnetic fields are present. If the magnetic field has a magnitude of 378 µT and is in the positive x direction, find the (a)x, (b)y, and (c)z components of the electric field. Give your answers in V/m

Homework Equations

F = q(E + v x B) = ma
E = (ma/q) + B x V

The Attempt at a Solution

E = [(9.11E-31 kg*1.95E12 m/s) / -1.60E-19 C] + (378 uT)i x [(17.2 km/s)j + (20.7 km/s)k]

this may seem silly, but I can't remember how to solve for each x, y, and z component now...

 

[Irid]

Search the web for cross product, for example here:

http://en.wikipedia.org/wiki/Cross_product#Matrix_notation"

 

[baba89]

I would suggest reviewing the basics of vector notation and the dot and cross product operations. The equation for electric field, E = (ma/q) + B x V, can be rewritten in vector notation as E = (m/q)(a + v x B). This means that to find the components of the electric field, we need to first calculate the acceleration due to the electric field, a, and the cross product of the velocity and magnetic field, v x B.

To start, we can calculate the acceleration due to the electric field using the given information. Since the electron has an initial velocity with an x component of zero, we can assume that there is no electric field in the x direction. This means that the acceleration in the x direction is also zero. In the y and z directions, the acceleration can be calculated using the given constant acceleration of 1.95 x 1012 m/s2. Therefore, the acceleration vector is a = (0, 1.95 x 1012, 1.95 x 1012) m/s2.

Next, we can calculate the cross product of the velocity and magnetic field. This can be done using the vector notation for cross product, v x B = (vyBz - vzBy, vzBx - vxBz, vxBx - vyBx). Plugging in the given values, we get v x B = (0, 378 µT(20.7 km/s), 378 µT(17.2 km/s)) = (0, 7.8 m/s, 6.5 m/s).

Finally, we can calculate the electric field using the equation E = (m/q)(a + v x B). Plugging in the calculated values, we get E = (9.11 x 10^-31 kg / -1.60 x 10^-19 C)(0, 1.95 x 1012, 1.95 x 1012) + (0, 7.8 m/s, 6.5 m/s) = (0, 1.14 x 10^13 V/m, 9.5 x 10^12 V/m). Therefore, the (a)x, (b)y, and (c)z components of the electric field are 0 V/m, 1.14 x 10^13 V/m, and