A long rod, radius R, that has uniform positive charge density p is spun around its central axis with angular velocity w, as shown below, so that the charges follow circular paths centered on the axis. Current is the flow of charge, so we have rings of current of varying radius r. You may want to review the following relationships: Current density J = dI/dA= pv (from chapter 24!) and v=wr (from PHY 204!) This means that the rotating rod of charge is equivalent to a non-uniform current density J = pwr,
A long rod, radius R, that has uniform positive charge density p is spun around its central axis with angular velocity w, as shown below, so that the charges follow circular paths centered on the axis. Current is the flow of charge, so we have rings of current of varying radius r. You may want to review the following relationships: Current density J = dI/dA= pv (from chapter 24!) and v=wr (from PHY 204!) This means that the rotating rod of charge is equivalent to a non-uniform current density J = pwr,
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Transcribed Image Text:A long rod, radius R, that has uniform positive charge density p is spun around its central axis with
angular velocity w, as shown below, so that the charges follow circular paths centered on the axis.
Current is the flow of charge, so we have rings of current of varying radius r.
You may want to review the following relationships:
Current density J = dI/dA= pv (from chapter 24!) and v=wr (from PHY 204!)
This means that the rotating rod of charge is equivalent to a non-uniform current density J = pwr,
where the current is circulating around the axis and r is the radial distance from the axis. Note this
has the same symmetry as the solenoid (current flowing in circles, field lines straight and parallel to
the axis). The difference is that this is like having many nested solenoids, and as we head out from
the axis, we see more and more current.
R
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I will walk you through the steps for using Ampere's law to find the magnetic field inside the rod.
(a) Draw a diagram of a cross section of the rod, cut across the diameter, and the appropriate test
loop to use in applying Ampere's law in finding the magnetic field at a distance r from the axis.
Be very careful with your diagram: the test loop is just like the one for the solenoid, so be honest
about what part of our source is inside the test loop!
(b) Use part (a) to find an expression for how much current is enclosed in (pokes through) the test
loop (in terms of p, w, r, R). Be sure to reference your diagram, and in particular note what part
of cylinder is inside the test loop! Note also that since the current density is not uniform, you
DO need to integrate. It will help to draw a slice of the loop with current dI and label it with
the distances r (the test point location, and location of one edge of the test loop,) R (surface of
the rod,) and y (the distance from the axis to the slice of current.)
(c) You now have the current enclosed by the test loop for the right hand side of Ampere's law.
The other side (field times length) of the law is the same as in the original Ampere's law for a
solenoid. Solve Ampere's law for the field as a function of r.
(d) The field should be strongest at points on the axis (at r = 0) and decrease as we move away,
going to zero when r = R. Why (physics argument) is this true? Check (math) that this is
consistent with the implications of your answer to part (d). If it is not, try to find your mistake!
Transcribed Image Text:(e) Let's focus on one of the charges inside the rod. It now is a moving charge in a magnetic field!
What is the direction of the force on it?
Problem 4
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