An Electromagnetic Rail Gun. A conducting bar with mass m...

An Electromagnetic Rail Gun. A conducting bar with mass $m$ and length $L$ slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current $I$ in the rails and bar, and a constant, uniform, vertical magnetic field $\overrightarrow{\boldsymbol{B}}$ fills the region between the rails (Fig. 27.63$)$ . (a) Find the magnitude and direction of the net force on the conducting bar Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass $m$ , find the distance $d$ that the bar must move along the rails from rest to attain speed $v$ . (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth $(11.2 \mathrm{km} / \mathrm{s}) .$ Let $B=0.50 \mathrm{T}, \quad I=2.0 \times 10^{3} \mathrm{A}$ $m=25 \mathrm{kg},$ and $L=50 \mathrm{cm} .$ For simplicity assume the net force on the object is equal to the magnetic force, as in parts $(a)$ and $(b)$ , even though gravity plays an important role in an actual launch in space.

An Electromagnetic Rail Gun. A conducting bar with mass $m$ and length $L$ slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current $I$ in the rails and bar, and a constant, uniform, vertical magnetic field $\overrightarrow{\boldsymbol{B}}$ fills the region between the rails (Fig. 27.63$)$ . (a) Find the magnitude and direction of the net force on the conducting bar Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass $m$ , find the distance $d$ that the bar must move along the rails from rest to attain speed $v$ . (c) It
has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth $(11.2 \mathrm{km} / \mathrm{s}) .$ Let $B=0.50 \mathrm{T}, \quad I=2.0 \times 10^{3} \mathrm{A}$ $m=25 \mathrm{kg},$ and $L=50 \mathrm{cm} .$ For simplicity assume the net force on the object is equal to the magnetic force, as in parts $(a)$ and $(b)$ , even though gravity plays an important role in an actual launch in space.

Key Concepts

Escape Velocity

Escape velocity is the minimum speed required for an object to overcome a gravitational field without further propulsion. This concept is pertinent when assessing the capabilities and limitations of railgun-based systems for launching payloads into space.

Kinematics in Uniform Acceleration

The kinematics equations describe the relationship between displacement, initial velocity, acceleration, and time for bodies under constant acceleration. These equations are essential for calculating the distance the bar travels in achieving a specified speed, as in the railgun scenario.

Work-Energy Principle

The work done by a force over a displacement is equal to the change in kinetic energy of an object. In the context of the railgun, this principle is used to determine how the work done by the magnetic force translates into the kinetic energy (and thus speed) of the moving bar.

Lorentz Force

This principle states that a current-carrying conductor in a magnetic field experiences a force given by the cross product F = I L × B. The direction of the force is determined by the right-hand rule. This concept is fundamental in understanding how electromagnetic railguns work by converting electrical energy into kinetic energy.

Newton's Second Law of Motion

This law, expressed as F = ma, relates the net force applied to an object to its mass and the resulting acceleration. It is crucial when analyzing the motion of the railgun’s conducting bar as it accelerates under the magnetic force, linking force to the resultant acceleration.

Transcript

00:01 In this problem, we have an electromagnetic rail gun, which consists of two parallel metal rails, with a metal rod crossing across them, which is free to move left and right.

00:14 This rod has length l and mass m.

00:26 We're told that between the rails, there is a constant b field, or a constant magnetic field, pointing in the vertical direction, which in our case, since we're looking down on the rails, would be pointing out of the page.

00:40 Here is our magnetic field.

00:44 And then finally we have a constant current going through the rails and through the rod, which we're going to say is going in the clockwise direction.

00:54 That's i.

00:57 Now in part one, they want us to find the magnetic force on the rod.

01:01 So you'll remember that for a point charge, the force is the charge of the particle times its velocity crossed with the magnetic field.

01:10 Now here we have many charges and so we have a continuous current and that turns out to be for a continuous current of charge to be the length of the rod or the wire or whatever it is the current is going through times the current crossed with the magnetic field now in most textbooks they actually write this as i times l and then l is the vector crossed with b where l points in the direction of i.

01:46 So it seems to make more sense to me to just say i is the vector, because i is actually what points in some direction where l is just a length.

01:54 But just a heads up that it's generally written the other way in textbooks.

01:59 Now the magnitude of the force is just going to be the magnitude of this cross product, which is l, i, b, and then times the sign of the angle between the current and the b field.

02:14 Now in our case the current through the rod is going straight up in this drawing and the b field points out of the page so these two are perpendicular so sign of 90 is 1 so that means that the force or the magnitude of the force is just l ib and now we just have to find the direction and to do that we use the right -hand rule and that means you take your right hand you point your fingers along the first vector which means you point your fingers along i which is straight up in our schematic here and then curl them towards the second vector which would be you curl them out of the page right curl them towards b and your thumb will point in the direction of f and that turns out to be to the right to the right and that's the answer to part one for part two let's give us some room here they want us to find the velocity, or actually they want us to find the distance that the rod will have to travel in order to get some velocity.

03:33 So distance as a function of velocity.

03:35 Now you'll notice that this here is a constant force.

03:38 So you could actually just solve kinematic equations for constant acceleration and find the distance as a function of velocity.

03:49 But there's actually an easier way to do this, which is to look at the work done on the rod.

03:54 So work is equal to the force dotted with the displacement and then integrated, which is equal to the change in kinetic energy.

04:12 Now we're starting from rest, so that means the kinetic energy will just be one half the mass of the rod times its velocity squared and that will equal the force.

04:22 So f.

04:22 D .r, this is a dot product, so we take the magnitude of the force, which is just l -i -b and then times dr...

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