# Magnetic field strength problem

http://s3.amazonaws.com/answer-board-image/2924bffb-da01-4543-9697-b67da02e5a5a.jpeg

The picture above shows four square regions. Each of the four regions may have its own uniform magnetic field. If there is a uniform magnetic field in a region, the field is directed either into or out of the screen (or the page, if you printed your assignment out on paper). Two charged particles pass through this region. Particle 1 has a positive charge. The path of particle 1 is a quarter-circle in region A, a half-circle in region D, another quarter-circle in region A, and a quarter-circle in region B. The path of particle 2 is a straight line in region C, and a quarter circle in region D.

Particle 1 has a charge of +50.0 microcoulombs, and a mass of 1.20 milligrams. Particle 1 also has a speed, when it enters region A, of 5.00 m/s.

(d) If region A is a square measuring 90.0 cm on each side (the other regions have the same size as region A, of course), what is the magnitude of the magnetic field in region A?
T
(e) From the instant particle 1 enters region A for the first time, until it leaves region B, how much time elapses?
s
(f) If particle 2 has the same mass and the same speed as particle 1, what is the charge on particle 2?
microcoulombs

Last edited:

## Answers and Replies

Firstly, In which directions are magnetic fields in all the regions?
Secondly, See exactly what are the radii of curvature of the different paths the charge follows.

Hint: In A , Magnetic Field is Out of Plane, In B -> Into the plane.

Do rest yourself.

Thanks vivekrai, but I am still unsure of how to solve the problem.

For the first part, the radius of the curvature is 90 cm or 0.9 m. Using r=mv/qB, I solved for the magnetic field strength, B. The mass of the particle is 0.0012 kg, the charge is 50 microcoulombs or 5 e-6 C, the v is 5 m/s. When I solve for B, I get a wrong solution. Why isn't this method correct?

Please I am really frustrated!

Your value for ' r ' is wrong. It is 0.3 m.

Jeez. I though each individual square had a 0.9 m side. Sigh...
Thanks vivekrai.

Argh. Weirdly, I was able to get the correct solutions to the last two parts. However, the answer I had for part d was incorrect even though, I had to use it for the other parts.

Using r=mv/qB, B=mv/qr. m=0.0012 kg, v= 5 m/s, q=50e-5 C, r=0.3 m. B=400T?

I am confident this is correct. Is there an error in my units?

How should one go about solving f?

Last edited:
I got the first part of the question correct. Just make sure all of your units are correct (decimal points, powers).
Unfortunately, I'm having problems with the second part of the problem. I have a similar set-up.

To calculate the time that the particle spends from entering field A to leaving field B, I simply calculated the fields in each of the other regions, then used that info to calculate the period of each half or quarter circle using
T=2πm/qB

I then added up all of the periods to get an answer but it was incorrect.
What am I doing wrong?

How should one go about solving f?

To solve this, you simple need to rearrange the equation that you used to get field magnitude (B) for part d, and have it solve for the particle charge (q). You should already have all the info you need to solve for it minus the field magnitude, which you can just get if solve for it using particle 1.

Never mind, I solved my own problem. I wasn't taking into account the fact that I needed to take only a portion of the period such a quarter or half of it depending on how it traveled in a particular field. Number becomes smaller.

How did you find the radius for part D?

How did you find the radius for part D?

You are given that the side of each region is 90cm (you may have a different value). In this diagram you see the regions are divided up. You can then use that info to determine what one the length of one side of one box is. Looking at the path of the particle you can see that its part of a circle. Knowing what a radius is you can easily figure out what the radius is.

thanks!

is the velocity of particle 1 still the same in field B (needed to find the value of the magnetic field in B)?

Magnetic Field is perpendicular to the velocity. It does no work on particle and hence no change in ' speed ' of particle.