What is the acceleration of a rod in a rail gun?

AI Thread Summary
To determine the acceleration of the conducting rod in a rail gun setup, the discussion highlights the need to calculate the current (I) using Ohm's law, where I = V/R. The magnetic force acting on the rod can be derived from the equation F = I(L × B), where L is the length of the rod and B is the magnetic field strength. To find the acceleration, Newton's second law (F = ma) is applied. Additionally, it is suggested to incorporate Faraday's law by formulating a differential equation that accounts for the back electromotive force (emf) generated as the rod moves. This approach integrates the effects of voltage, magnetic field, resistance, and mass to derive the rod's acceleration.
matt747
Messages
1
Reaction score
0
Hi I'm trying to come up with an expression for the accleration of the rod in Rail Gun.

The conducting rod is free to slide on two parallel rails with negligible friction. At the right end of the rails, a voltage source of strength "V" in series with a resistor of resistance "R" makes a closed circuit together with the rails and the rod. The rails and the rod are taken to be perfect conductors. The rails extend to infinity on the left.

There is a uniform magnetic field of magnitude B, pervading all space, perpendicular to the plane of rod and rails. The rod is released from rest, and it is observed that it accelerates to the left.

I'm trying to find the acceleration based on the following variables:

V (voltage)
B (magnetic field)
v (velocity of the rod)
L (Length of the rod)
R (Resistance)
m (mass)

http://mp.pearsoncmg.com/probhtml/16467_a.jpg

Thanks,

Matt
 
Last edited by a moderator:
Physics news on Phys.org
You need to find the current ##I## in the rail using Ohm's law and then find the magnetic force on the rail using ##\vec F=I\vec L\times \vec B##. Finally, Newton's law will give you the acceleration. If you have seen Faraday's law, you need to write a differential equation taking the back emf into account.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top