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A metal bar of length 25 cm is placed perpendicular to a uniform magnetic field of strength 3 T. (a) Determine the induced emf between the ends of the rod when it is not moving. (b) Determine the emf when the rod is moving perpendicular to its length and magnetic field with a speed of \(50 \mathrm{cm} / \mathrm{s}\).

Short Answer

Expert verified
The induced emf when the rod is stationary is 0 V, and when moving it is 0.375 V.

Step by step solution

01

Identify the Given Variables

The length of the metal bar \(L\) is given as 25 cm = 0.25 m. The strength of the magnetic field \(B\) is given as 3 T. The velocity \(v\) of the rod when moving is given as 50 cm/s = 0.5 m/s.
02

Calculate the Induced emf for the Stationary Rod

According to Faraday’s law, when a conducting rod is stationary in a magnetic field, there is no change in the magnetic field with respect to the rod, thus no emf is induced. So the induced emf is 0 V.
03

Calculate the Induced emf for the Moving Rod

For a conducting rod moving perpendicular to magnetic filed, the induced emf can be calculated using formula \(emf = BvL\). Subsituting given values, \(emf = 3 * 0.5 * 0.25 = 0.375\) V

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Key Concepts

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

Understanding Faraday's Law
Faraday's Law of electromagnetic induction is a fundamental principle that governs how electric current can be generated through the motion within a magnetic field. In essence, it states that an electromotive force (often abbreviated to emf) is induced in a circuit whenever there is a change in the magnetic flux linking that circuit. The 'change' could be due to altering the strength or direction of the magnetic field, moving the circuit within the field, or altering the area of the circuit.

According to Faraday's Law, the induced emf is directly proportional to the rate of change of the magnetic flux. Mathematically, it is expressed as: \[ emf = - \frac{d\text{\textPhi}_B}{dt} \] where \( \text{\textPhi}_B \) is the magnetic flux and \( \frac{d\text{\textPhi}_B}{dt} \) is its rate of change over time. The negative sign indicates the direction of the induced emf which is given by Lenz's Law. This dictates that the induced emf will act in a direction that opposes the change in flux that produced it—essentially, it's nature's way of maintaining equilibrium.

In the scenario of a metal rod placed in a uniform magnetic field, when the rod is stationary, there is no change in flux over time, and therefore by Faraday's Law, no emf is induced. When the rod is moved, it intersects magnetic field lines, creating a change in flux, and an emf is generated. It's important to understand that this induced emf can cause a current if there is a closed path in which charges can move.
The Role of a Uniform Magnetic Field
A uniform magnetic field is characterized by having the same strength and direction at every point within the field. This consistency is important because it provides a constant environment for experimental analysis and it simplifies calculations for induced emf, as seen in Faraday's Law applications.

In the given example of the textbook problem, a uniform magnetic field is crucial for ensuring an accurate calculation of the induced emf when the metal bar is moved. Since the field strength \( B \) is the same everywhere across the rod’s length, we can be sure that the entire rod experiences the same magnetic influence. In a non-uniform field, the emf would vary along the length of the rod, complicating the situation and necessitating a more complex analysis.

The existence of a uniform field means that for any straight-line motion of the rod perpendicular to both the rod's length and the magnetic field, the calculation of emf is straightforward. The concept of uniformity allows the formula \( emf = BvL \) to be valid, as it relies on the magnetic field strength being consistent throughout the motion of the rod.
Conducting Rod Motion in Electromagnetic Induction
The motion of a conducting rod in a magnetic field is at the heart of many electromagnetic applications, from electric generators to sensing devices. When the rod moves perpendicular to its length and to the magnetic field, it intersects magnetic field lines and induces a voltage across its ends. This phenomenon is explained by the Lorentz force, which describes the force experienced by a charged particle moving in a magnetic field.

In simpler terms, as the conducting rod moves through a magnetic field, free electrons within the metal experience a force that causes them to move along the length of the rod, resulting in an induced emf. The faster the rod moves, or the stronger the magnetic field (assuming uniformity), or the longer the rod, the greater the amount of induced emf, as described by the equation \( emf = BvL \).

This concept underpins the step-by-step solution where the induced emf for the moving metal bar is calculated. The bar's motion creates a situation where the conducting material 'cuts' through magnetic field lines, which by Faraday's Law, induces an emf. If the rod were moved parallel to the magnetic field lines or were not moving at all, no cutting action would occur, and no emf would be induced. The principle is elegantly captured in practical devices such as handheld dynamo torches, where the motion of a magnet or coil generates electric power.

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Most popular questions from this chapter

A long solenoid with 10 turns per centimeter is placed inside a copper ring such that both objects have the same central axis. The radius of the ring is \(10.0 \mathrm{cm},\) and the radius of the solenoid is \(5.0 \mathrm{cm}\). (a) What is the emf induced in the ring when the current \(I\) through the solenoid is 5.0 A and changing at a rate of \(100 \mathrm{A} / \mathrm{s}\) ? (b) What is the emf induced in the ring when \(I=2.0 \mathrm{A}\) and \(d I / d t=100 \mathrm{A} / \mathrm{s} ?(\mathrm{c}) \mathrm{What}\) is the electric field inside the ring for these two cases? (d) Suppose the ring is moved so that its central axis and the central axis of the solenoid are still parallel but no longer coincide. (You should assume that the solenoid is still inside the ring.) Now what is the emf induced in the ring? (e) Can you calculate the electric field in the ring as you did in part (c)?

Electromagnetic braking can be achieved by applying a strong magnetic field to a spinning metal disk attached to a shaft. (a) How can a magnetic field slow the spinning of a disk? (b) Would the brakes work if the disk was made of plastic instead of metal?

A 0.50-kg copper sheet drops through a uniform horizontal magnetic field of \(1.5 \mathrm{T}\), and it reaches a terminal velocity of \(2.0 \mathrm{m} / \mathrm{s}\). (a) What is the net magnetic force on the sheet after it reaches terminal velocity? (b) Describe the mechanism responsible for this force. (c) How much power is dissipated as Joule heating while the sheet moves at terminal velocity?

A current is induced in a circular loop of radius \(1.5 \mathrm{cm}\) between two poles of a horseshoe electromagnet when the current in the electromagnet is varied. The magnetic field in the area of the loop is perpendicular to the area and has a uniform magnitude. If the rate of change of magnetic field is \(10 \mathrm{T} / \mathrm{s}\), find the magnitude and direction of the induced current if resistance of the loop is \(25 \Omega\).

Discuss the factors determining the induced emf in a closed loop of wire.

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