Analyzing Angular Momentum of Solenoid & Charge

[Tarikura]

Homework Statement

Given a long thin solenoid with a point charge of e. Initial conditions, e is at rest, the solenoid is on, i.e., there is current running through it, and the magnetic field outside is essientally 0 due to the length of the solenoid. Final conditions: the solenoid is turned off, i.e. current is zero, the charge then gets a kick so that its velocity is not zero. The magnetic flux of the induction coil is zero. Is the angular momentum conserved?

Homework Equations

L = r X mv

The Attempt at a Solution

Well, I'm supposed to "analyze" this problem from the points of view of the Newtonian (ma type of equations) and the Hamiltonian formulas. Anyone know how I should get started, this problem kinda confuses me a bit.

 

[Jhero]

First, let's define some variables for the problem:

- e = point charge
- v = initial velocity of the point charge (before the kick)
- r = distance from the center of the solenoid to the point charge
- m = mass of the point charge
- I = current running through the solenoid
- B = magnetic field inside the solenoid
- L = angular momentum of the point charge

Now, let's look at this problem from a Newtonian perspective. We can use the equation F = ma to analyze the motion of the point charge. In this case, the only force acting on the point charge is the magnetic force, given by F = qvB. Since the magnetic field outside the solenoid is essentially zero, we can assume that there is no force acting on the point charge after the solenoid is turned off. Therefore, the point charge will continue to move in a straight line with constant velocity. This means that the angular momentum of the point charge, L, will remain constant, since there are no external forces causing it to change.

Now, let's look at this problem from a Hamiltonian perspective. The Hamiltonian is defined as H = T + V, where T is the kinetic energy and V is the potential energy. In this case, the potential energy is zero since the magnetic field outside the solenoid is essentially zero. The kinetic energy of the point charge can be written as T = 1/2mv^2. When the solenoid is turned off, the point charge will receive a kick, giving it an initial velocity v. This means that the kinetic energy of the point charge will increase, and therefore the Hamiltonian will also increase. However, since there are no external forces acting on the point charge, the total energy (H) will remain constant. This means that the angular momentum, L, will also remain constant.

In conclusion, from both a Newtonian and Hamiltonian perspective, the angular momentum of the point charge will be conserved in this scenario.