# Homework Help: Magnitude of magnetic field

1. Mar 24, 2016

### Koh Eng Kiat

1. The problem statement, all variables and given/known data

The boundary shown is that of a uniform magnetic field directed in the positive z direction. An electron enters the magnetic field with a velocity pointing along the x axis and exits 0.63 μs later at point A. What is the magnitude of the magnetic field?

2. Relevant equations

F = qvb
v = s/t
r = mv/qb

3. The attempt at a solution

I calculated v = 2m/0.63 x 10^-6s = 3174603m/s. But I am stuck after this next step. I know that q = 1.6 x 10^-19C. Is my approach to the question starting out right? I am confused and paralyzed, unable to proceed forward. Thanks if anyone could offer some hints.

2. Mar 24, 2016

### phyzguy

I think you calculated v correctly. Now look at your last equation. You know m, v, and q, and you want to find b. So you need to know r. What does r need to be so that an electron that enters on the left exits at point A? I think you are supposed to assume that the electron enters at the origin, where the x-axis and y-axis cross..

3. Mar 24, 2016

### Staff: Mentor

What is the shape of the path that the electron takes from its point of entry to its point of exit? What is the length of that path?

4. Mar 24, 2016

### Koh Eng Kiat

Guys, thanks for replying! But I do not understand. What is the path that the electron takes? Why is my velocity correct? Isnt it displacement over time? I googled and know that the radius of a square is from the center to any of the four vertices, which I calculated to be 1.41m.

Last edited: Mar 24, 2016
5. Mar 25, 2016

### Staff: Mentor

When you calculated the velocity, why did you take the distance traveled to be 2 m? What is the shape of the trajectory of a charged particle moving in a uniform magnetic field that is perpendicular to the velocity vector?

Can you draw the trajectory?

6. Mar 25, 2016

### phyzguy

Yes, please ignore my earlier post stating that you have calculated the velocity correctly. gneill's suggestion is the correct one.

7. Mar 25, 2016

### Koh Eng Kiat

I now know that the trajectory of the electron is curved upwards. The reason is that the velocity direction is perpendicular to the direction of the uniform magnetic field. Hence, the centripedal force will result in a curved motion. Therefore, it is the top half of the diagram. For my question, the radius is 2m. and velocity is calculated by taking displacement over time. The displacement can be found using pythagoras theorem. After getting the radius and velocity, I just use r = mv/qb and I get 13 micro T, which is about 14 micro T. I guess the difference is due to accuracy and rounding up issues. So, is my line of thinking correct? Can anyone verify this for me? Thanks for the help!

8. Mar 25, 2016

### Staff: Mentor

You don't want the average velocity here. Consider a case where an electron travels a full circle, arriving back to where it started over some time period t. Would you use a velocity of zero to compute the radius of curvature of the trajectory?

You want to use the length of the path actually taken by the electron from the origin to point A.

9. Mar 25, 2016

### Koh Eng Kiat

So I should use the circumference of the segment of the circle? The distance would be (2 x pi x 2m) / (4) = pi. Using pi / time to find velocity? Sorry guys. I have been trying quite hard on this question. What would be the ideal solution then?

Last edited: Mar 25, 2016
10. Mar 25, 2016

### Staff: Mentor

Yes, use the quarter circle's path length.

11. Mar 25, 2016

### Koh Eng Kiat

a.18 μT
b.14 μT
c.28 μT
d.34 μT
e.227 μT

The answer is b, according to my answer given. I get a much larger answer using the quarter circle's path length. Oh man...

12. Mar 25, 2016

### Staff: Mentor

13. Mar 25, 2016

### Koh Eng Kiat

r = mv/qb

v = 3.142 / 0.63 x 10^-6s = 4986655 m/s

2 = (9.11 x 10^-31kg)(4986655m/s)/(1.6 x 10^-19C)(b)
2 = (4.54 x 10^-24)/(1.6 x 10^-19C)(b)
b = (4.52 x 10^-24)/(3.2 x 10^-19C) = 14 micro T.

The answer is correct.Thanks for the help. It was calculation mistake on my part. However, isnt velocity = displacement over time, where displacement is the shortest distance from initial to final point. Shouldnt the displacement used be a straight line from (0,0) to (2,2) than the quarter circle's path length? I mean, now I have gotten the answer I wanted, but I do not understand the reasoning for using the quarter circle path length despite what you said. I apologize for the lengthy post.

14. Mar 25, 2016

### Staff: Mentor

The force on the moving charge depends upon the instantaneous velocity of the charge, not the average velocity over the whole path. total displacement over total time gives you the average velocity for the trip, not the instantaneous velocity at a particular point along the way.

The length of the path that the electron traveled is not the same as the net displacement, as I pointed out in my example where the electron travels a complete circle. There the displacement would be zero and the average velocity would be zero, yet the electron certainly had some non-zero speed throughout its travels.

15. Mar 26, 2016

Thanks!