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## Homework Statement

An electron with velocity ##\vec{v}_0=8.7*10^4(m/s)## (in the î direction) passing through an area with a uniform magnetic field ##\vec{B}=0.80 T## (in the negative k̂ direction). There's also a uniform electric field in this area.

What is the magnitude and direction of this electric field if the electron travels through this area without being deflected? [see picture below]

## Homework Equations

The Lorentz force ##\vec{F}=q(\vec{E}+\vec{v}×\vec{B})##

Right Hand Rule

## The Attempt at a Solution

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The speed is low in comparison to that of the speed of light and the Lorentz factor ##γ=1.000000042## so we won't deal with special relativity in this problem.

So first let's use the RHR to decide the direction of the magnetic force on the electron. The crossproduct ##(\vec{v}×\vec{B})## multiplied by the

__negative__scalar ##q## (since we're dealing with an electron) gives the magnetic force to point downward. The electric field that we seek has to point upwards and be

__equal__to the magnetic force to make the electron stay fixed in its y-position.

The two parts of the Lorentz force

__has to be equal__.

##\vec{F}_B=\vec{F}_E## and since we're dealing with uniform fields and the magnetic field and the velocity are perpendicular to eachother the ##sin(θ)## is just ##1##.

We also know the direction of all the fields (the direction ##E##-field follows from definition and the wanted result in this problem) so we can skip the vector signs and simply have:

##qvB=qE## and ##E=vB##.

**So**the electric field has to point upward in the picture and have a magnitude of:

##E=(8.7*10^4 (m/s))(0.80 T)=6.09*10^3 (N/C)##

Did I get it right and is my reasoning on point?