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**Calculating magnitude and direction of an magnetic force on an electron**

## Homework Statement

Okay so the question says that the magnetic force (Vector)FM on a particle which is in a magnetic field is found by (Vector)FM = Vector I x Vector B.

Vector I = charge multiplied by the velocity of a charged particle

Vector B= the strength of the magnetic field (Measured in Tesla(T) ).

An electron accelerated from REST to the RIGHT, in a horizontal directed electric field.

The electron then leaves the electric field at a speed of 4.0 x 10^6, entering the magnetic field of magnitude 0.20 Tesla that is directed into the screen.

From the given information calculate the magnitude and direction of the magnetic force on the electron given that the charge on an electron is q= 1.6 x 10^-19.

## Homework Equations

All right, since we have to solve this in the cross product method i'm quite lost.

I take it we want to end up with something like this: (ax, ay, az) and b = (bx, by, bz)

giving us a x b = = (aybz − azby , azbx − axbz , axby − aybx)

## The Attempt at a Solution

So far i analyzed that

Vector I = charge of an electron (q = 1.6 x 10^-19) x velocity of a charged particle (4.06 x 10^6) (Since it says electron accelerates from rest to the right horizontally, i assume vector I is the x component?)

and for

Vector B = strength of the magnetic field in Tesla (0.20 T). (And since it says its directed into the screen, i assume its the z component? Right?)

So now we have

Vector I = 1.6x10^-19(4.0 x 10^6) = 6.4 x 10^-13

Vector B = 0.20

|I||B|sin90

|6.4x10^-13| x |0.20| sin 90 = 1.28 x 10^-13

So the magnitude is 1.28 x 10^-13 and the direction is in the y axis?

I defenitley think i hugely messed up somewhere, please help

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