Question

please show work when possible so that I may better understand. Thank you.

x=L The rail gun consists of two thick conducting rails connected to a power supply and there is a magnetic field assumed to be uniform and constant in the direction shown. A short conducting bar that is to be (or carry) the projectile is placed across the bars at x -0. The current flows through the bars as shown. The magnetic force on the bar causes it to slide down the rails picking up speed until they leave the rails where x -L. The goal is to launch the bar at a desired fast speed. Simplifying assumptions: No friction There is some kind of guide to keep the projectile in contact with the rails and sliding only in the +x direction. Electrical resistance is zero in all parts except in the projectile itself. The bar has a resistance of R. The power supply is able to maintain a constant current during the launch. The source of the magnetic field is able to maintain a uniform and constant magnetic . . field everywhere between the rails during the launch. 2000[A] SIs 3000LA] 30.0 [m] SLS 40.0 [m] . 1.00[A] R 1.001[A] Allowable range for each design parameter 0.500[T]s B s 1.500[T] 40.0 [cm] s W s 60.0 [cm] 20.0 [kg] S m s 30.0 [kg] The goals are to choose the values of the parameters that will give you the fastest possible launch velocity and to find the launch velocity. There are also some questions about some aspects of the operation of the device. Fill out the following answer sheet. Attach your work on separate paper labeled "rail gun" Rail Gun Answer Sheet 1) Choose values for your design parameters: m- 2) Write an expression (in terms of the design parameters) for the x-component of the force on the projectile bar as it travels down the rails. Then solve for the force using your design. For your design F- Formula: F 3) Using the work energy theorem, write an expression (in terms of the design parameters and x) for the speed of the projectile bar when it is passing arbitrary location "x" on the rails. It started at rest at x-0. Then solve for the launch speed using your design. (x L at the launch.) Launch speed for your design v(L) = Formula: v(x) 4) Consider the motion of the bar from an arbitrary location x to x+dx along the rails. Using your formula for force, write a new expression (in terms of the design variables, x) for the rate that work is done on the bar during the differential move. Power is the rate of work done, P dW/dt. Formula: P(x)- 5) During the same differential move from x to dx, there is induced EMF in the circuit due to the changing magnetic flux that penetrates the loop of the circuit. Write an expression (in terms of the design variables and x) for the induced EMF during the differential move. I'll call it EMFs since it's due to the magnetic field. Formula: EMFB In order to maintain the constant current "l"', the power supply is going to have to adjust its EMF (I'l call it EMFrs). It has to push the charges in the circuit hard enough to move against the changing induced EMFs and still have enough energy left to get through the resistance of the bar. Using your expression for EMFs and the constant EMF needed to overcome the bar resistance (to make AV-IR in the bar), write an expression (in terms of the design variables and x) for the total EMFrs of the power supply when the bar is at location x Formula: EMFPS

Answer 1

Anse Given thot →from the above d

es-tect rm So dv WB p2 七

(Bw) Now the X-camponent d -th ^orcc on -the Α-n to.crion oLon c ComDonnt rojeti own Then

hore w o. 一-) 꽈 started atyt.ct at č 0 So The launch sped mula.

1 2 しょ40 Nouw, ao 26000 2 a0 o300 Laench Specd 103.qm.i· 八ourns V ('L

on C X to Xtdx alonq the vails. Her C, ay Hom an o Considuy So oimuxo magne HC

NImu eYt LO dct 03 MF bar ,s at-location χ Heit rs