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The two conducting rails in the drawing are tilted upward so they each make an angle of \(30.0^{\circ}\) with respect to the ground. The vertical magnetic field has a magnitude of \(0.050 \mathrm{T}\). The \(0.20-\mathrm{kg}\) aluminum rod (length \(=\) \(1.6 \mathrm{m}\) ) slides without friction down the rails at a constant velocity. How much current flows through the rod?

Short Answer

Expert verified
The current flowing through the rod is approximately 24.5 A.

Step by step solution

01

Analyze the forces involved

Since the rod slides down at constant velocity, the net force on the rod is zero. This means the gravitational component parallel to the rails is balanced by the magnetic force acting in the opposite direction.
02

Calculate the gravitational force component

The force of gravity acting parallel to the rail is the component of the weight of the rod along the incline. This force is given by \[ F_{ ext{gravity}} = m imes g imes ext{sin}( heta) \] where \( m = 0.20 \mathrm{kg} \), \( g = 9.81 \mathrm{m/s^2} \), and \( \theta = 30.0^{\circ} \).
03

Magnetic Force

The magnetic force on the rod can be expressed as \[ F_{ ext{magnetic}} = I imes L imes B imes ext{sin}( heta_B) \] where \( I \) is the current, \( L = 1.6 \mathrm{m} \), \( B = 0.050 \mathrm{T} \), and \( \theta_B = 90^{\circ} \) since the magnetic field is vertical.
04

Set up the equation for constant velocity

Since the rod moves at constant velocity, the magnetic force equals the gravitational force component:\[ I \times L \times B \times ext{sin}(90^{\circ}) = m \times g \times ext{sin}(\theta) \] Simplifying gives the equation to find \( I \):\[ I = \frac{m \times g \times ext{sin}(\theta)}{L \times B} \]
05

Substitute values and solve for current

Substitute the known values into the equation to calculate the current:\[ I = \frac{0.20 \mathrm{kg} \times 9.81 \mathrm{m/s^2} \times ext{sin}(30^{\circ})}{1.6 \mathrm{m} \times 0.050 \mathrm{T}} \]This simplifies to:\[ I = \frac{1.962}{0.08} = 24.525 \mathrm{A} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force
When studying physics, the concept of gravitational force is vital to understand how objects move under the influence of gravity. In our scenario, we have a rod sliding down inclined rails, and we must realize that gravity is the force pulling the rod toward the Earth. Notably, the gravitational force does not act directly down the incline. Instead, it acts vertically downwards, and only a component of it pulls the rod along the rails.
The weight of the rod, calculated as its mass multiplied by the gravitational acceleration (\( g = 9.81 \, \text{m/s}^2 \)), is oriented straight down. Because the rails are inclined at \( 30.0^{\circ} \), only the part of the gravitational force acting along the rails is responsible for the rod’s movement. This component is given by the formula:
  • \( F_{\text{gravity}} = m \times g \times \sin(\theta) \).
Here, \( m \) is the mass of the rod. By using trigonometry, we find this component to balance out forces acting along the path of the rails.
Current Calculation
In the inclined rod setup, the calculation of electric current flowing through the rod is an essential part of the exercise due to the involvement of magnetic forces. Current is calculated based on the balance of forces, considering that the rod moves at a constant velocity. When something moves at a constant velocity, this means the forces acting on it are balanced, or in equilibrium. In our case, this equilibrium condition is between the gravitational force component and the magnetic force.
To find the current \( I \) in the rod, we use the equality between gravitational and magnetic forces:
  • The equation to find current is: \[ I \times L \times B \times \sin(90^{\circ}) = m \times g \times \sin(\theta) \]
We rearrange this equation to solve for \( I \):
  • \( I = \frac{m \times g \times \sin(\theta)}{L \times B} \)
through substitution of the known values: the mass \( m = 0.20 \, \text{kg} \), gravitational acceleration \( g \), length \( L = 1.6 \, \text{m} \), magnetic field strength \( B = 0.050 \, \text{T} \), and the angle, we compute the current flowing through the rod to maintain its constant speed.
Conducting Rails
Conducting rails provide a path for the electric current and play a crucial role in this physics problem. As the rod slides down, it remains in contact with the rails, which are tilted at an angle. The inclination angle affects the gravitational force component that moves the rod along these rails. Because the rails are conductive, they allow the flow of electrical current as the rod moves through the magnetic field, enabling the interaction of forces.
The importance of conducting rails lies in their ability to guide and sustain the electric flow, which influences the magnetic force acting on the rod. The interaction between these rails and the magnetic field generates a force that can oppose or permit the rod's movement.
  • When the rod moves, it essentially cuts through magnetic field lines.
  • This motion in a magnetic field induces an electrical current according to the principles of electromagnetic induction.
Understanding how conducting rails interact with magnetic fields and currents helps in evaluating and solving problems involving magnetic force and electric current dynamics.

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Most popular questions from this chapter

A charged particle enters a uniform magnetic field and follows the circular path shown in the drawing. (a) Is the particle positively or negatively charged? Why? (b) The particle's speed is \(140 \mathrm{m} / \mathrm{s},\) the magnitude of the magnetic field is \(0.48 \mathrm{T}\), and the radius of the path is \(960 \mathrm{m}\). Determine the mass of the particle, given that its charge has a magnitude of \(8.2 \times 10^{-4} \mathrm{C}\).

A copper rod of length \(0.85 \mathrm{m}\) is lying on a frictionless table (see the drawing). Each end of the rod is attached to a fixed wire by an unstretched spring that has a spring constant of \(k=75 \mathrm{N} / \mathrm{m} .\) A magnetic field with a strength of \(0.16 \mathrm{T}\) is oriented perpendicular to the surface of the table. (a) What must be the direction of the current in the copper rod that causes the springs to stretch? (b) If the current is 12 A, by how much does each spring stretch?

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of \(7.0 \mathrm{T}\), and the current in the solenoid is \(2.0 \times 10^{2} \mathrm{A} .\) What is the number of turns per meter of length of the solenoid? Note that the solenoid used to produce the magnetic field in this type of system has a length that is not very long compared to its diameter. Because of this and other design considerations, your answer will be only an approximation.

A horizontal wire is hung from the ceiling of a room by two massless strings. The wire has a length of \(0.20 \mathrm{m}\) and a mass of \(0.080 \mathrm{kg} .\) A uniform magnetic field of magnitude 0.070 T is directed from the ceiling to the floor. When a current of \(I=42\) A exists in the wire, the wire swings upward and, at equilibrium, makes an angle \(\phi\) with respect to the vertical, as the drawing shows. Find (a) the angle \(\phi\) and (b) the tension in each of the two strings.

Electron beams are sometimes used to melt and evaporate metals in order to deposit thin metallic films on surfaces (similar to gold plating). One method is to put the material to be evaporated (called the "target") into a small tungsten cup (a crucible that has a very high melting point) and direct a beam of electrons at the target. Your team has been given the task of designing an electron-beam evaporator. The crucible is a cylinder, \(2.0 \mathrm{cm}\) in diameter and \(1.5 \mathrm{cm}\) in height, and contains a small target of pure nickel (Ni). The electrons are accelerated through a potential difference of \(V=1.20 \mathrm{kV}\), and form a beam that originates below the crucible, exactly \(3.70 \mathrm{cm}\) off its center, in the \(+x\) direction (see the drawing). (a) What is the speed of the electrons in the beam? (b) You must steer the electron beam with a magnetic field so that it curls over the lip of the cup and strikes the nickel target. Assuming that a uniform field exists above the cup (the field is zero below), what must be the radius of the beam's circular path? (c) In what direction should the field point if the beam initially approaches the cup from the \(-y\) axis? (d) What must be the magnitude of the uniform magnetic field?

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