/*! 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 50 A horizontal straight wire \(20 ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A horizontal straight wire \(20 \mathrm{~m}\) long extending from east to west is falling with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\), at right angles to the horizontal component of the earth's magnetic field0.30 \(\times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\). The instantaneous value of the emf induced in the wire will be [AIFE 2011] (a) \(6.0 \mathrm{mV}\) (b) \(3 \mathrm{mV}\) (c) \(4.5 \mathrm{mV}\) (d) \(1.5 \mathrm{mV}\)

Short Answer

Expert verified
The instantaneous value of the emf induced in the wire is 3 mV (option b).

Step by step solution

01

Understand the Context

We need to find the emf (electromotive force) induced in a wire that is moving through a magnetic field. The wire is 20 meters long, and the speed with which it moves perpendicular to the magnetic field is 5 m/s. The magnetic field strength is given as \(0.30 \times 10^{-4} \, \text{Wb/m}^2 \) (Webers per square meter), which is also \( ext{Tesla} \).
02

Identify the Formula

The emf induced in a wire moving through a magnetic field can be calculated using the formula: \( \text{emf} = B \times L \times v \), where \( B \) is the magnetic field strength, \( L \) is the length of the wire, and \( v \) is the speed of the wire.
03

Substitute the Values

Insert the given values into the formula: \( \text{emf} = (0.30 \times 10^{-4} \, \text{T}) \times (20 \, \text{m}) \times (5 \, \text{m/s}) \).
04

Perform the Calculation

Calculate the product: \( 0.30 \times 10^{-4} \times 20 \times 5 \). First, multiply the values: \( 0.30 \times 20 = 6 \) and then, \( 6 \times 5 = 30 \). Finally, multiply \( 30 \times 10^{-4} = 3.0 \times 10^{-3} \text{V} \). Convert to millivolts: \( 3.0 \times 10^{-3} \text{V} = 3 ext{mV} \).
05

Verify the Answer

Check if the calculated emf matches any of the given options. We found \( 3 ext{mV} \), which corresponds to option (b).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Faraday's Law of Induction
The heart of electromagnetic induction lies in Faraday's Law of Induction, a fundamental principle of electromagnetism. This law explains how an electromotive force (emf) is induced in a conductor, like our wire, when it experiences a changing magnetic field. Simply put, when a conductor moves perpendicular to the magnetic field lines, or when the magnetic field through a stationary coil changes, an emf is produced.

This phenomenon occurs due to the magnetic flux, which is the product of the magnetic field, the area through which it passes, and the cosine of the angle between the field and the perpendicular to the area. Mathematically, Faraday's Law is given by:
  • \( \text{emf} = - \frac{d\Phi}{dt} \)
where \( \Phi \) is the magnetic flux. In our problem, because the movement is perpendicular, the angle between the field and motion is 90 degrees, and the cosine is zero, making the change in magnetic flux straightforward to calculate.

Ultimately, Faraday's Law helps us understand why the wire moving through Earth's magnetic field generates a measurable emf.
Magnetic Fields
Magnetic fields are invisible forces that exert influence on moving charges and magnetic materials. They originate from magnets and electric currents, and they can be visualized as lines that flow from the north to the south pole of a magnet.

In this exercise, Earth's magnetic field plays a central role. Although it is weak compared to artificial magnets, the field is always present. It's measured in webers per square meter (Wb/m²) or teslas (T). In our case, the "horizontal component" of Earth's magnetic field is specified as \(0.30 \times 10^{-4} \text{ T}\).

These magnetic field lines interact with the wire as it moves, creating a change in the magnetic environment around the wire. This movement essentially cuts across the magnetic field lines, causing the induction of emf, as articulated by Faraday's Law. Understanding the nature of how magnetic fields interact with conductors is crucial for predicting and calculating the behaviors observed in electrodynamic systems.
Electromotive Force
Electromotive force, commonly abbreviated as emf, is pivotal in the study of electromagnetism and circuits. It is the "electric potential" generated in a circuit due to a changing magnetic field. While it sounds like a force, it is in fact measured in volts (V) and essentially defines the work done to drive charge around a closed loop.

When the wire in the exercise moves through Earth's magnetic field, an emf is induced in the wire. This emf can be calculated with the equation:
  • \( \text{emf} = B \times L \times v \)
where:
  • \( B \) is the magnetic field strength in teslas (T),
  • \( L \) is the length of the wire in meters (m),
  • \( v \) is the speed of the wire in meters per second (m/s).
In our scenario, substituting the values gives us \(3 \text{ mV} \), an earthy but useful voltage that clearly illustrates the process of induction. This concept is essential for generating electricity, from small experiments like this, to large-scale power stations where turbines cut through magnetic fields, generating electric power. Understanding emf opens the gateway to mastering how we produce and manage electrical energy efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two short bar magnets of length \(1 \mathrm{~cm}\) each have magnetic moments \(1.20 \mathrm{Am}^{2}\) and \(1.00 \mathrm{Am}^{2}\) respectively. They are placed on a horizontal table parallel to each other with their \(N\) poles pointing towards the south. They have a common magnetic equator and are separated by a distance of \(20.0 \mathrm{~cm}\). The value of the resultant horizontal magnetic induction at the mid-point \(O\) of the line joining their centres is elose to (Horizontal component of the earth's magnetic induction is \(3.6 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\) ) IJEE Main 2013| (a) \(3.6 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\) (b) \(2.56 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (c) \(3.50 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\) (d) \(5.80 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\)

The magnetie flux linked with the coil varies with time as \(\phi=3 t^{2}+4 t+9\). The magnitude of the induced emf at 2 s is (a) \(9 \mathrm{~V}\) (b) \(16 \mathrm{~V}\) (c) \(3 \mathrm{~V}\) (d) \(4 \mathrm{~V}\)

Assertion-Reason type. Each of these contains two Statements: Statement 1 (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion. (c) If Assertion is true but Reason is false. (d) If Assertion is false but the Reason is true. Assertion An iron bar magnet is heated to \(1000^{\circ} \mathrm{C}\) and then cooled in a magnetic field in free space. It does not retain its magnetism. Reason Only ferro-magnets can show hysteresis.

At a certain location in Africa, a compass points \(12^{\circ}\) west of the geographie north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points \(60^{\circ}\) above the horizontal. The horizontal component of the earth's field is measured to be \(0.16\) G. Specify the direction and magnitude of the earth's field at the location. (a) \(32^{\circ}\) west of geographical meridian and \(3.2 \times 10^{-4} \mathrm{~T}\) (b) \(12^{\circ}\) west of geographical meridian and \(0.32 \times 10^{-4} \mathrm{~T}\) (c) \(12^{2}\) east of geographical meridian and \(0.32 \times 10^{-4} \mathrm{~T}\) (d) 32 ' cast of geographical meridian and \(3.2 \times 10^{-4} \mathrm{~T}\)

In a permanent magnet at room temperature [NCERT Exemplar] (a) magnetic moment of each molecule is zero (b) the individual molecules have non-zero magnetic moment which are all perfectly aligned (c) domains are partially aligned (d) domains are all perfectly aligned

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.