
#1
Mar1512, 01:43 PM

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 IBsin90 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 



#2
Mar1512, 03:34 PM

Mentor
P: 11,415

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 yaxis" 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}## .




#3
Mar1512, 03:46 PM

P: 5

So i would do Vector Ix = vector Icos 90° 6.4x10^13 cos 90 =0 0.20 cos 90 =0 Sorry im confused? 



#4
Mar1512, 04:33 PM

Mentor
P: 11,415

Cross Product application problemLet: ##\vec{v} = 4.0x10^6 \hat{i}~m/s~~~~~and~~~~~ \vec{B} = 0.20 \hat{k}~T ## 



#5
Mar1612, 12:56 PM

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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