Question

A 6.70 −μC particle moves through a region of space where an electric field of magnitude 1250 N/C points in the positive x direction, and a magnetic field of magnitude 1.03 T points in the positive z direction.

.If the net force acting on the particle is 6.22×10^{−3}
N in the positive *x* direction, find the components of the
particle's velocity. Assume the particle's velocity is in the
*x*-*y* plane?

Answer #1

First, you write your variables:

Q = 6.70 x 10^-6 C

E = 1250 N/C --->+x direction

B = 1.03 T ------>+z direction

Fnet = 6.22x 10^-3 N ----->+x direction

Fnet = Fe - Fb

=qE - qvB

Solve for v so v = (1/B)(E - (Fnet/q))

v= 312.27 m/s

once you get the value for velocity. You need to know which direction it's going.

You know +x direction for E, +z direction for B and +x for
Fnet.

Do the right hand rule where:

your right thumb goes toward the Fnet, then your index finger
points to B (z direction) Then curl your middle,ring, and pink 90
angle. This shows where v is going which is -y direction.

Answer: vx, vy,vz = 0, -312.27, 0

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positive z direction. If the net force acting on the particle is
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Express your answer using two significant figures.
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I seem to be getting...

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