What is the magnetic field produced by an electron beam in a cathodic ray tube?

[fishingspree2]

Homework Statement

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

Homework 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 I am 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.

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.