# Cross Product application problem

by Asheram0
Tags: application, cross, product
 P: 5 1. The problem statement, all variables and given/known data 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. 2. Relevant 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) 3. 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
 Mentor P: 10,757 Your calculations look fine to me. But you should take care when dealing with vectors and vector components to make sure that you specify directions completely. "in the y-axis" doesn't indicate if the direction is towards +y or -y. You should be able to write each vector in component form: ## x \hat{i} + y \hat{j} + z \hat{k}## .
P: 5
 Quote by gneill Your calculations look fine to me. But you should take care when dealing with vectors and vector components to make sure that you specify directions completely. "in the y-axis" doesn't indicate if the direction is towards +y or -y. You should be able to write each vector in component form: ## x \hat{i} + y \hat{j} + z \hat{k}## .
I very much appreciate your reply, but since my answer is a positive value would it be correct in saying its towards the +y?

So i would do
|Vector Ix| = |vector I|cos 90°

6.4x10^-13 cos 90
=0

0.20 cos 90

=0

Sorry im confused?

Mentor
P: 10,757

## Cross Product application problem

 Quote by Asheram0 I very much appreciate your reply, but since my answer is a positive value would it be correct in saying its towards the +y?
The sign of the result should depend upon your initial choice of axes, how the given values "lie" in that frame of reference, and any operations performed on those values. It's helpful to specify the coordinate system at the beginning and state the given values in terms of those coordinates. So for example, you might state that the +x-axis is chosen to coincide with the electron's direction of motion, and that the +z-axis is into the screen. The magnetic field is parallel to the z-axis and is directed towards +z. Or more concisely you could write:

Let: ##\vec{v} = 4.0x10^6 \hat{i}~m/s~~~~~and~~~~~ \vec{B} = 0.20 \hat{k}~T ##
 So i would do |Vector Ix| = |vector I|cos 90° 6.4x10^-13 cos 90 =0 0.20 cos 90 =0 Sorry im confused?
If the result happened to have some odd direction that didn't coincide with any of the axes then you might want to write out the vector in ijk form. But if it coincides with an axis then it would be sufficient to specify which axis and which direction along that axis. For example, 1.28x10-13 N in the direction of the +y-axis.
 P: 5 Thank you very very much, it helped a lot! The right hand rule could also be applied in this question! Kudos man!

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