A particle has a charge of q = +4.8 ?C and is located at the origin. As the draw

Question

A particle has a charge of q = +4.8 ?C and is located at the origin. As the drawing shows, an electric field of Ex = +202 N/Cexists along the +x axis. A magnetic field also exists, and its x and y components are Bx = +1.6 T and By = +1.9 T. Calculate the force (magnitude and direction) exerted on the particle by each of the three fields when it is (a) stationary, (b) moving along the +x axis at a speed of 345 m/s, and (c) moving along the +z axis at a speed of 345 m/s.

For this answer, I need the force in each direction (Fe, Fbx, and Fby) for each of the three magnetic fields, and the resulting direction of the force. SO, there should be 9 answers total with the resulting direction of each! Thanks!

Explanation / Answer

PART-A

when stationary there is no force due to the magnetic field, so

F = E*q

F= 202*4.8x10^-6

F= 9.69x10^-4N at 0 degee (along the + x axis)

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PART-2

The force form the electric doesn't change but the force from the magnetic field is

F= q*v*B*sin(?)

. Only the y component of the field will create a force on a charge moving along the x axis

FB = 4.8x10^-6*345*1.9

FB = 3.1464x10^-3N in the +z direction

F = sqrt((9.69x10^-4N)^2 +(3.1464x10^-3)^2)

F= 3.292x10^-3N .

The direction is in the x-z plane at an angle

arctan(3.1464x10^-3/9.69x10^-4)

The direction is in the x-z plane at an angle= 72.88 to the x axis

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PART-C

Again FE = 19.69x10^-4N

The X component of the magnetic force = 4.8x10^-6*345*1.6 = 2.6496x10^-3N  but its direction is in the -x

The Y component of the magnetic force = 4.8x10^-6*345*1.9 = 3.1464x10^-3N,

So net force is Fx = 9.69x10^-4N- 2.6496x10^-3N = -1.6806x10^-3N and

the y component Fy =3.1464x10^-3N

So F = sqrt((-1.6806x10^-3)^2 + (3.1464x10^-3)^2) = 3.567x10^-3N

the angle = arctan(3.1464x10^-3/(-1.6806x10^-3)) = -61.88 degree but since x is< 0 + 180

so theta = -61.88 + 180 = 118.11

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