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Chapter 27: Problem 42

A rail gun accelerates a projectile from rest by using the magnetic force on a current-carrying wire. The wire has radius \(r=5.1 \cdot 10^{-4} \mathrm{~m}\) and is made of copper having a density of \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\). The gun consists of rails of length \(L=1.0 \mathrm{~m}\) in a constant magnetic field of magnitude \(B=2.0 \mathrm{~T}\) oriented perpendicular to the plane defined by the rails. The wire forms an electrical connection across the rails at one end of the rails. When triggered, a current of \(1.00 \cdot 10^{4}\) A flows through the wire, which accelerates the wire along the rails. Calculate the final speed of the wire as it leaves the rails. (Neglect friction.)

Short Answer

Expert verified
Answer: To find the final velocity of the wire when it leaves the rails, follow these steps: 1. Calculate the mass of the wire using its dimensions and the density of copper. 2. Determine the magnetic force on the wire using the given current and magnetic field strength. 3. Find the acceleration of the wire using the magnetic force and its mass. 4. Calculate the final velocity of the wire using the acceleration and distance it traveled along the rails. By following these steps, you can determine the final velocity of the copper wire as it leaves the rails in the rail gun.

Step by step solution

01

Calculate the mass of the wire

First, we need to find the volume and mass of the wire. We know the radius (r) and length (L) of the wire, so we can find the volume (V) using the formula for the volume of a cylinder: \(V = \pi r^2 L\) Then, we can calculate the mass (m) using the density (ρ) of copper: \(m = \rho \cdot V\)
02

Calculate the magnetic force on the wire

Next, we need to determine the magnetic force (F) acting on the wire due to the current (I) and the magnetic field (B). The magnetic force on a current-carrying wire can be found using the formula: \(F = I(LB)\)
03

Calculate the wire's acceleration

Now that we have the magnetic force acting on the wire, we can use Newton's second law of motion (F=ma) to find the acceleration (a) of the wire: \(a = \frac{F}{m}\)
04

Calculate the final velocity of the wire

Finally, we can find the final velocity (v) of the wire using the kinematic equation, knowing the initial velocity (u=0), the acceleration (a), and the distance (L): \(v^2 = u^2 + 2aL\) Since the initial velocity (u) is 0, we only need to compute the square root of the right-hand side to find v: \(v = \sqrt{2aL}\) Once all calculations are done, we'll have determined the final velocity of the wire as it leaves the rails.

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Most popular questions from this chapter

A high electron mobility transistor (HEMT) controls large currents by applying a small voltage to a thin sheet of electrons. The density and mobility of the electrons in the sheet are critical for the operation of the HEMT. HEMTs consisting of AlGaN/GaN/Si are being studied because they promise better performance at higher powers, temperatures, and frequencies than conventional silicon HEMTs can achieve. In one study, the Hall effect was used to measure the density of electrons in one of these new HEMTs. When a current of \(10.0 \mu\) A flows through the length of the electron sheet, which is \(1.00 \mathrm{~mm}\) long, \(0.300 \mathrm{~mm}\) wide, and \(10.0 \mathrm{nm}\) thick, a magnetic field of \(1.00 \mathrm{~T}\) perpendicular to the sheet produces a voltage of \(0.680 \mathrm{mV}\) across the width of the sheet. What is the density of electrons in the sheet?

A straight wire of length \(2.00 \mathrm{~m}\) carries a current of \(24.0 \mathrm{~A} .\) It is placed on a horizontal tabletop in a uniform horizontal magnetic field. The wire makes an angle of \(30.0^{\circ}\) with the magnetic field lines. If the magnitude of the force on the wire is \(0.500 \mathrm{~N}\), what is the magnitude of the magnetic field?

The figure shows a schematic diagram of a simple mass spectrometer, consisting of a velocity selector and a particle detector and being used to separate singly ionized atoms \(\left(q=+e=1.60 \cdot 10^{-19} \mathrm{C}\right)\) of gold \((\mathrm{Au})\) and molybdenum (Mo). The electric field inside the velocity selector has magnitude \(E=1.789 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\) and points toward the top of the page, and the magnetic field has magnitude \(B_{1}=1.00 \mathrm{~T}\) and points out of the page. a) Draw the electric force vector, \(\vec{F}_{E},\) and the magnetic force vector, \(\vec{F}_{B},\) acting on the ions inside the velocity selector. b) Calculate the velocity, \(v_{0}\), of the ions that make it through the velocity selector (those that travel in a straight line). Does \(v_{0}\) depend on the type of ion (gold versus molybdenum), or is it the same for both types of ions? c) Write the equation for the radius of the semicircular path of an ion in the particle detector: \(R=R\left(m, v_{0}, q, B_{2}\right)\). d) The gold ions (represented by the black circles) exit the particle detector at a distance \(d_{2}=40.00 \mathrm{~cm}\) from the entrance slit, while the molybdenum ions (represented by the gray circles) exit the particle detector at a distance \(d_{1}=19.81 \mathrm{~cm}\) from the entrance slit. The mass of a gold ion is \(m_{\text {gold }}=\) \(3.27 \cdot 10^{-25}\) kg. Calculate the mass of a molybdenum ion.

It would be mathematically possible, for a region with zero current density, to define a scalar magnetic potential analogous to the electrostatic potential: \(V_{B}(\vec{r})=-\int_{\vec{r}_{0}}^{\vec{r}} \vec{B} \cdot d \vec{s},\) or \(\vec{B}(\vec{r})=-\nabla V_{B}(\vec{r}) .\) However, this has not been done. Explain why not.

An electron is moving at \(v=6.00 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field. If the field strength is \(0.500 \cdot 10^{-4} \mathrm{~T}\), what is the radius of the electron's circular path?
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