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Asheram0
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Calculating magnitude and direction of an magnetic force on an electron
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.
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)
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
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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