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Ch 29: The Magnetic Field
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 29: The Magnetic FieldProblem 76
Chapter 29, Problem 76

An electromagnetic rail gun uses magnetic forces to launch projectiles. FIGURE P29.76 shows a 10-cm-long, 10 g metal wire that can slide without friction along 1.0-m-long horizontal rails. The rails are connected to a 300 V source, and a 0.10 T magnetic field fills the space between the rails. Each rail has linear resistivity ⋋ = 0.10 Ω/m, which means that the resistance is ⋋ multiplied by the length of rail through which current flows. Assume that the sliding wire and the left end, where the voltage source is, have zero resistance. The wire is initially placed at x₀ = 5.0 cm then the switch is closed. What is the wire's speed as it leaves the rails?

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Step 1: Understand the forces acting on the wire. The wire experiences a magnetic force due to the current flowing through it and the magnetic field. This force is given by the Lorentz force formula: F = I × L × B, where I is the current, L is the length of the wire, and B is the magnetic field strength.
Step 2: Calculate the current in the circuit. The total resistance in the circuit includes the resistance of the rails, which depends on their length and linear resistivity (λ). The resistance of the rails is R = λ × L_rails, where L_rails is the length of the rails. Using Ohm's law, the current can be calculated as I = V / R, where V is the voltage of the source.
Step 3: Determine the magnetic force acting on the wire. Substitute the values of I, L, and B into the Lorentz force formula to find the force acting on the wire. This force will accelerate the wire along the rails.
Step 4: Use Newton's second law to find the acceleration of the wire. The net force acting on the wire is equal to its mass times its acceleration: F = m × a. Rearrange to solve for acceleration: a = F / m, where m is the mass of the wire.
Step 5: Calculate the final speed of the wire as it leaves the rails. Use the kinematic equation v² = u² + 2 × a × d, where u is the initial speed (0 m/s), a is the acceleration, and d is the distance traveled by the wire (from x₀ to the end of the rails). Solve for v, the final speed.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electromagnetic Induction

Electromagnetic induction is the process by which a changing magnetic field within a closed loop induces an electromotive force (EMF) in the wire. In the context of the rail gun, as the wire moves through the magnetic field, it experiences a change in magnetic flux, which generates a current due to Faraday's law of induction. This induced current interacts with the magnetic field, producing a force that propels the wire forward.
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Lorentz Force

The Lorentz force is the force experienced by a charged particle moving through a magnetic field. It is given by the equation F = q(v × B), where F is the force, q is the charge, v is the velocity of the particle, and B is the magnetic field. In the rail gun scenario, the current flowing through the wire creates a magnetic field that interacts with the external magnetic field, resulting in a force that accelerates the wire along the rails.
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Ohm's Law and Resistance

Ohm's Law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. In this case, the resistance of the rails is determined by their resistivity and length, which affects the current flowing through the circuit when the switch is closed. Understanding the relationship between voltage, current, and resistance is crucial for calculating the wire's speed as it exits the rails.
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Related Practice
Textbook Question

It is shown in more advanced courses that charged particles in circular orbits radiate electromagnetic waves, called cyclotron radiation. As a result, a particle undergoing cyclotron motion with speed v is actually losing kinetic energy at the ratedKdt=−(μ0q46πcm2)B2v2\(\frac{dK}{dt}\) = - \(\left\)( \(\frac{\mu_0 q^4}{6\pi c m^2}\) \(\right\)) B^2 v^2

How long does it take (a) an electron and (b) a proton to radiate away half its energy while spiraling in a 2.0 T magnetic field?

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Textbook Question

A scientist measuring the resistivity of a new metal alloy left her ammeter in another lab, but she does have a magnetic field probe. So she creates a 6.5-m-long, 2.0-mm-diameter wire of the material, connects it to a 1.5 V battery, and measures a 3.0 mT magnetic field 1.0 mm from the surface of the wire. What is the material's resistivity?

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Textbook Question

A wire along the x-axis carries current I in the negative x-direction through the magnetic field B⃗={B0xlk^0≤x≤l0elsewhere\(\vec{B}\)= \(\begin{cases}\) B_0\(\dfrac{x}{l}\]\hat{k}\) & 0 \(\leq\) x \(\leq\) l \\ 0 & \(\text{elsewhere}\) \(\end{cases}\). Find an expression for the net torque on the wire about the point x = 0.

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Textbook Question

FIGURE CP29.79 is an edge view of a 2.0 kg square loop, 2.5 m on each side, with its lower edge resting on a frictionless, horizontal surface. A 25 A current is flowing around the loop in the direction shown. What is the strength of a uniform, horizontal magnetic field for which the loop is in static equilibrium at the angle shown?

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Textbook Question

A proton moves in the uniform fields E = 2500 k V/m and B = 0.50 k T. At t = 0 s the proton is moving in a 1.0-cm-diameter circle in the xy-plane. How many revolutions will the proton have made during this time interval?

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Textbook Question

In FIGURE P29.75, a long, straight, current-carrying wire of linear mass density μ is suspended by threads. A magnetic field perpendicular to the wire exerts a horizontal force that deflects the wire to an equilibrium angle θ. Find an expression for the strength and direction of the magnetic field B.

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