Passage:Railgun

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A railgun is a form of gun that converts electrical energy into projectile kinetic energy rather than the more conventional means through chemical energy from an explosive propellant. Unlike gas pressure guns, rail guns are not limited by the laws of gas expansion and so they are capable of accelerating projectiles to extremely high speeds.


A railgun consists of two parallel metal rails connected to an electrical power supply. When a conductive projectile is inserted between the rails it completes the circuit. This flow of current makes the railgun act like an electromagnet, creating a powerful magnetic field which accelerates the projectile along the rails.


image:Passage_railgun_circuit.jpg


The magnitude of the force experienced by the projectile can be expressed mathematically in terms of the permeability constant (\mu_0), the radius of the rails (r), the distance between the rails (d) and the current in amps through the system (I) as follows:

F = \frac{\mu_0 I^2}{ \pi} \ln{ \frac{d-r}{r}}


If a very large power supply providing a million amperes or so of current is used, then the force on the projectile will be tremendous, and by the time it leaves the ends of the rails it can be travelling at many kilometres per second. 20 kilometers per second has been achieved with small projectiles explosively injected into the railgun. Although these speeds are theoretically possible, the heat generated from the propulsion of the object is enough to rapidly erode the rails. Such a railgun would require frequent replacement of the rails, or use a heat resistant material that would be conductive enough to produce the same effect.


1. What is the orientation of the magnetic field that the projectile feels in the schematic diagram of the railgun above?

into the page.
out of the page.
→ The orientation of the power source implies current flows counterclockwise in this circuit. By using the right hand rule, one can determine that the magnetic field wraps around the wires out of the page between the two wires but into the page above and below. The projectile thus feels the magnetic field going out of the page.
upwards, along the plane of the page.
→ Upwards and downwards can be ruled out because they are both parallel to the direction of current, which would imply no magnetic field is produced.
downwards, along the plane of the page.
→ Upwards and downwards can be ruled out because they are both parallel to the direction of current, which would imply no magnetic field is produced.

2. What limitations exist in the railgun apparatus?
          i) The rails experience strong repulsive forces against each other.
          ii) The projectile must be conductive.
          iii) The apparatus requires both strong magnetic and strong electric fields

i and iii.
i and ii.
ii, and iii.
i, ii, and iii.
→ Electric fields are not necessary in this apparatus as only the magnetic field generated from a current carrying wire is necessary.

3. According to the given mathematical expression of force experienced by the projectile, what must the units of the permeability constant, (\mu_0), be?

(N)/(A)
(Kg)(s)/(C2)
(C)(s)/(Kg)(m)
→ This false answer is the result of mixing up which terms go on the top and bottom when rearranging to isolate the constant.
(Kg)(m)/(C2)
→ The correct answer is (N)/(A)2. If one then replaces Amps with (C)(s) and simplifies, D becomes the only valid answer.

4. What is the magnitude of the force on a particle of charge -7nC moving westward at 10 Mm/s in a 1000 guass magnetic field aiming south?

7 x 10-5 N
7 x 10-3 N
→ Using F = qvB (sin can be ignored since they are perpendicular, and converting the units into N, m/s, and Tesla we can derive this answer easily.
→ = (7nC)(10 Mm/s)(1000 Gauss)
→ = (7 x 10-9)(10 x 106)(1000 Gauss x 1 Tesla/10000 Gauss)
→ = 7 x 10-3 N
7 x 103 N
7 x 105
→ If one fails to convert the units from nC to C, Mm/s to m/s, and Gauss to Tesla, this is the result.

5. Which of the following is true to the magnetic field, B, generated from a current carrying circular wire?

B is directly proportional to r2.
B is inversely proportional to r.
B is inversely proportional to I.
B is directly proportional to I2.

6. If, in the above railgun, a resistor were inserted in parallel with the rest of the system such that the new resistor had the same resistance as the projectile circuit, what would happen to the acceleration of the projectile rod?

Its acceleration will be halved.
Its acceleration will double.
Its acceleration will be one forth the original.
→ This false answer is the result of mixing up which terms go on the top and bottom when rearranging to isolate the constant.
Its acceleration will be quadrupled.
→ In answering this problem, one must first recognize that the Force equation given must be used. Secondly, one must recognize the I squared term, and replace it with V squared over R squared. Thus if R is halved, the overall value becomes a forth when squared. Therefore the force, and thus the acceleration, go down by a forth.
→ F = \frac{\mu_0 I^2}{ \pi} \ln{ \frac{d-r}{r}}
→ F = \frac{\mu_0 \left( V/R \right) ^2}{ \pi} \ln{ \frac{d-r}{r}}
→ Now, half the resistance because the new resistor went in parallel,
→ F = \frac{\mu_0 \left( V/\left(R/2 \right)\right) ^2}{ \pi} \ln{ \frac{d-r}{r}}
→ F = \frac{\mu_0 4 \left( V/R \right) ^2}{ \pi} \ln{ \frac{d-r}{r}}
→ 4 F = \frac{\mu_0 \left( V/R \right) ^2}{ \pi} \ln{ \frac{d-r}{r}}

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