# Magnetic field problem

#### fishingspree2

1. The problem statement, all variables and given/known data
In a cathodic ray tube, the canon launches a circular electron beam on the screen. The beam has a 0.22 mm diameter, the electrons have a kinetic energy of 25 keV and 5.6*10^14 electrons reach the screen each second. Find the magnetic field produced by the beam on a point located at a 1.5mm distance of the beam axis.

The answer is 12 nT

2. Relevant equations
I know that the magnetic field of a circular arc is
B = (u*i*angle)/(4*Pi*distance) but in this case I don't have an angle so im pretty sure this is not the correct formula

Also we could integrate using the biot savart rule but we don't have the limits of integration.

3. The attempt at a solution
I have found the current by using the fact that 5.6*10^14 electrons reach the screen each second. By converting electrons to coulombs, I get 8.97 * 10^-5 Amperes. Then I am stuck, I have no idea how to continue

thank you

#### AEM

You should review Ampere's law. Particularly the integral form. By the way, do you remember the shape of the magnetic field line in the vicinity of a current flowing in a straight line?

#### fishingspree2

A circle I believe?

#### fishingspree2

I still can't do it =( I don't understand what to do with the kinetic energy... I can find the electrons velocity but it would only be useful in the formula F = q v x B and I don't think that would apply here

#### AEM

A circle I believe?
Yes, the magnetic field lines take the shape of a circle going around a current i that is traveling in a straight line.

Now do you know Ampere's law? You have already found the magnitude of the current. Ampere's law involves a line integral taken over a suitable path. This actually is a very simple problem once you check your textbook (or Wikipedia). You will have to simplify a dot product, but that should be no problem.

#### AEM

I still can't do it =( I don't understand what to do with the kinetic energy... I can find the electrons velocity but it would only be useful in the formula F = q v x B and I don't think that would apply here
The kinetic energy is a red herring and F = qv X B is useless in this context.

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