/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 The two rails of a railway track... [FREE SOLUTION] | 91Ó°ÊÓ

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The two rails of a railway track insulated from each other and the ground are connected to a milli-voltmeter. What is the reading of the \(\mathrm{mV}\), when a train travels at a speed of \(180 \mathrm{kmh}^{-1}\) along the track, given that the horizontal component of earth's magnetic field is \(0.2 \times 10^{-4} \mathrm{Wbm}^{-2}\) and the rails are separated by \(1 \mathrm{~m}\).(a) \(10^{-2} \mathrm{mV}\) (b) \(10 \mathrm{mV}\) (c) \(10^{2} \mathrm{mV}\) (d) \(1 \mathrm{mV}\)

Short Answer

Expert verified
The reading of the milli-voltmeter is (d) 1 mV.

Step by step solution

01

Understanding the concept of electromagnetic induction

When a train moves along the tracks, which act as a loop with the earth's magnetic field, it experiences electromagnetic induction. This phenomenon generates an electromotive force (EMF) across the rails.
02

Identifying the formula for induced EMF

The EMF \( \varepsilon \) generated by a moving conductor in a magnetic field is given by the formula: \( \varepsilon = B \cdot l \cdot v\), where \( B \) is the magnetic field strength, \( l \) is the separation between the rails (length of the conductor), and \( v \) is the velocity of the train.
03

Converting train speed to meters per second

The train speed is given as \( 180 \ \mathrm{kmh}^{-1} \). Convert this speed to meters per second by using the conversion factor: \( 1 \ \mathrm{kmh}^{-1} = \frac{5}{18} \ \mathrm{ms}^{-1} \). Therefore, \( v = 180 \times \frac{5}{18} = 50 \ \mathrm{ms}^{-1} \).
04

Substituting values into the formula

Substitute the known values into the formula: \[ \varepsilon = (0.2 \times 10^{-4} \ \mathrm{Wbm}^{-2}) \times (1\ \mathrm{m}) \times (50\ \mathrm{ms}^{-1}) = 10^{-3} \ \mathrm{V} = 1 \ \mathrm{mV} \].
05

Conclusion and selecting the correct option

The calculated EMF is \( 1 \ \mathrm{mV} \). Therefore, the correct answer is (d) \( 1 \ \mathrm{mV} \).

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Key Concepts

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

Induced Electromotive Force
Induced electromotive force (EMF) is a fascinating aspect of physics, particularly relevant in the study of electromagnetic induction. This phenomenon occurs when a conductor experiences a change in magnetic field, resulting in the generation of an electrical voltage, known as EMF. The **principle of electromagnetic induction** states that any change in the magnetic environment of a coil of wire will cause an induced EMF in the coil. In the example of a train moving along the rails, the train itself acts as a moving conductor within Earth's magnetic field. As the train travels, the magnetic field interacts with the conductive structure of the train and the rails, generating an EMF across the rails.

This EMF can be calculated using the formula: \( \varepsilon = B \cdot l \cdot v \), where:
  • \( B \) = magnetic field strength
  • \( l \) = separation between the rails (length of the conductor)
  • \( v \) = velocity of the train
This formula helps in determining the voltage induced due to the movement of the train along the rails in a magnetic field.
Earth's Magnetic Field
The Earth's magnetic field plays a crucial role in many natural processes, as well as in various technological applications. It exerts a force that affects moving charged particles and is responsible for phenomena such as the Northern and Southern Lights. In the context of electromagnetic induction, the horizontal component of Earth's magnetic field is of particular interest. This component is often denoted as \( B \) and measured in units like Weber per square meter (\( \mathrm{Wb/m^2} \)).

For the train and railway example, the Earth's magnetic field component of \( 0.2 \times 10^{-4} \ \mathrm{Wb/m^2} \) acts horizontally and interacts with the train moving along the tracks. This interaction is essential for the generation of induced EMF while the train moves, highlighting a practical application of Earth's magnetic field.
Conversion of Units
Unit conversion is a fundamental skill in physics, essential for performing accurate calculations. In many physics problems, quantities are often given in different units, and it's necessary to convert them into consistent units to apply formulas correctly.

In the railway track problem, the train's speed is initially given as \( 180 \ \mathrm{km/h} \). Since the formula for induced EMF involves speed in meters per second (\( \mathrm{m/s} \)), it is crucial to convert the speed using the conversion factor:
  • \( 1 \ \mathrm{km/h} = \frac{5}{18} \ \mathrm{m/s} \)
Applying this factor, the speed of \( 180 \ \mathrm{km/h} \) converts to \( 50 \ \mathrm{m/s} \). This step ensures the accuracy of the calculations when using the induced EMF formula.

Understanding and implementing unit conversion effectively helps prevent errors and enhances clarity in problem-solving.
Velocity in Magnetic Field
The velocity of a conductor within a magnetic field is a key factor that influences the magnitude of the induced EMF. **Velocity** refers to the speed and direction with which the conductor moves in the magnetic field. In electromagnetic induction problems, like the train example, the velocity directly affects the amount of EMF generated.

For the given problem, the train's converted speed of \( 50 \ \mathrm{m/s} \) is used in the formula for EMF calculation. The motion of the train within the Earth's magnetic field sets up a condition where the moving metallic structure cuts through magnetic field lines, leading to the production of an induced EMF.
  • This scenario highlights how increased velocity can lead to a higher induced EMF, provided the magnetic field strength and the length of the conductor remain constant.
  • It emphasizes the interdependency of velocity, magnetic field, and distance in understanding electromagnetic induction.
Recognizing the impacts of velocity in such contexts broadens the comprehension of how effectively moving conductors generate electricity in a magnetic field.

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Most popular questions from this chapter

A 110 volt \(\mathrm{AC}\) is connected to a transformer of ratio 10\. If resistance of secondary is \(550 \Omega\), current through secondary will be (a) \(10 \mathrm{~A}\) (b) \(2 \mathrm{~A}\) [c) zero (d) \(55 \mathrm{~A}\)

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