Electric and magnetic fields with electron

[cas111]

Homework Statement

at time t1, an electron is sent along the positive direction of an x axis, through both an electric field E and a magnetic field B, with E directed parallel to the y axis. Figure 28-34 gives the y component Fnet, y of the net force of the electron due to the two fields, as a function of the electron's speed v at time t1. the x and z components of the net force are zero at t1. Assuming Bx=0, find the magnitude E and B in unit-vector notion.

the graph has the y-axis set as the net force in the y direction and the x-axis as the velocity of the electron. the y-axis goes from -2 through 2 and the x-axis goes from 0 to 100. the x-axis is zero at -2 on the y axis, and 100 at 2 on the y axis. the graph is a straight line through those two points.

Homework Equations

im not really sure

The Attempt at a Solution

to be honest i have no idea where to start on this one, any help would be appreciated.

 

[phyjen08]

I think I'm in your class. I don't know if this is right but I think I have the idea.

Ok, becuase B is always perpendicular to v and E, B should be pointing in the page in the k direction. The formla you should use is Fnet=q(E+VxB) E,V,B are vectors. Becuase V and B are parallel their cross product is vB so now Fnet=q(E+vB). Your vectors will be stright lines when Fnet=0 so 0=q(E+vB). Obviously the q goes away and you are left with B=E/v, where E is the only vector. From here it gets a little trick. You have to use the KE formula for velocity. This is as far as I have gotten becuase you have to use voltage in the KE formula and I can't find a relation in the problem to get velocity for anything. If you figure the rest out could you post it so I can finish? I couldn't find the problm in the soultions manual unless it is number 7. 7 doesn't have a part b and it doesn't tell you how to find E. It uses V/m to find E, and again I have to clue to to find V.

 

[Moetasim]

I can provide some guidance on how to approach this problem. First, we need to understand the concept of electric and magnetic fields and how they interact with charged particles like electrons. Electric fields are created by stationary charges, while magnetic fields are created by moving charges. Both fields can exert a force on charged particles.

In this problem, we have an electron moving along the positive direction of the x-axis, which means its velocity is in the same direction as the electric and magnetic fields. The electric field is directed parallel to the y-axis, meaning it has no effect on the electron's motion in the x-direction. The magnetic field is directed in the z-direction, but since Bx=0, it also has no effect on the electron's motion in the x-direction.

To find the magnitude of the electric and magnetic fields, we can use the equation for the net force on a charged particle in an electric and magnetic field:

Fnet = q(E + v x B)

Where q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field. Since we know that the x and z components of the net force are zero, we can simplify this equation to only include the y component:

Fnet,y = q(Ey + vxBy)

We are given the net force in the y-direction as a function of the electron's velocity, so we can plug in the values for Fnet,y and v and solve for Ey and By. This will give us the magnitudes of the electric and magnetic fields in the y-direction.

Ey = Fnet,y/q
By = Fnet,y/(qv)

Since the graph is a straight line, we can choose any two points and use the slope formula to find the values for Ey and By. For example, at v=0, we can see that Fnet,y=0, so both Ey and By must also be zero. At v=100, we can see that Fnet,y=2, so using the above equations we can find the magnitudes of Ey and By.

Ey = 2/q
By = 2/(100q)

Now we just need to convert these values to unit-vector notation. Since Ey is in the y-direction, it can be written as Ey = Ey j, where j is the unit vector in the y-direction. Similarly, By can be written as By = By k, where k is the