/*! 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 33 The loop \(A B C D\) is moving w... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The loop \(A B C D\) is moving with velocity ' \(v^{\prime}\) towards right. The magnetic field is \(4 \mathrm{~T}\). The loop is connected to a resistance of \(8 \mathrm{ohm}\). If steady current of \(2 \mathrm{~A}\) flows in the loop then value of \(v\) if loop has a resistance of 4 ohm, is : (Given \(A B=30 \mathrm{~cm}, A D=30 \mathrm{~cm}\) ) (a) \(\frac{50}{3} \mathrm{~m} / \mathrm{s}\) (b) \(20 \mathrm{~m} / \mathrm{s}\) (c) \(10 \mathrm{~m} / \mathrm{s}\) (d) \(\frac{100}{3} \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The velocity needed is 20 m/s, which corresponds to option (b).

Step by step solution

01

Understand the Problem

The problem involves a loop moving through a magnetic field, generating an electromotive force (EMF), which causes a current to flow. We're given specifics about the loop dimensions, the magnetic field, the resistances involved, and the current. We need to find the velocity at which the loop should move to maintain a steady current of 2 A.
02

Identify the Known Variables

We are given:- Magnetic field, \( B = 4 \mathrm{~T} \)- Current, \( I = 2 \mathrm{~A} \)- Total resistance in the loop (resistance of the loop plus connected resistance), \( R_{total} = 8 \Omega + 4 \Omega = 12 \Omega \)- The length of side \( AB = 0.3 \mathrm{~m} \).- The area perpendicular to the magnetic field, initially formed by sides \( AB \) and \( AD \).
03

Apply Faraday's Law of Electromagnetic Induction

According to Faraday's law, the EMF induced in the loop is given by:\[ \text{EMF} = B \cdot l \cdot v \]where \( l \) is the length of the loop moving through the magnetic field (\( AB = 0.3 \mathrm{~m} \)).
04

Relate EMF to Current

Ohm's Law states that \( \text{EMF} = I \cdot R_{total} \). Thus,\[ B \cdot l \cdot v = I \cdot R_{total} \]Substituting known values:\[ 4 \cdot 0.3 \cdot v = 2 \cdot 12 \].
05

Solve for Velocity \( v \)

Rearrange the equation to find \( v \):\[ v = \frac{2 \cdot 12}{4 \cdot 0.3} = \frac{24}{1.2} = 20 \mathrm{~m/s} \]
06

Conclusion and Select the Correct Option

Based on our calculations, the velocity \( v \) at which the loop needs to move is \( 20 \mathrm{~m/s} \). This 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
Faraday's Law of Electromagnetic Induction is a fundamental principle that forms the backbone of electromagnetism. It describes how a changing magnetic field creates an electromotive force (EMF), which in turn induces current flow in a closed circuit. This is the principle at work when our loop moves through the magnetic field.

In our exercise, the movement of the loop through a 4 Tesla magnetic field induces an EMF. This is because as the loop cuts through the magnetic field lines, it experiences a change in magnetic flux. The faster the loop moves, the greater the change in flux, and thus, the greater the EMF. The EMF is calculated using the formula:
  • \( \text{EMF} = B \cdot l \cdot v \)
where \( B \) is the magnetic field strength, \( l \) is the length of the loop section moving through the field (in this case, 0.3 meters), and \( v \) is the velocity of the loop.

Understanding Faraday's Law helps us calculate the necessary speed of the loop to maintain the desired current of 2 Amperes, by generating sufficient EMF. This exercise exemplifies the real-world applications of this law in technologies like electric generators and transformers.
Ohm's Law
Ohm’s Law is another crucial concept in understanding how circuits operate. It relates the current flowing through a conductor to the voltage across it and the resistance it encounters. The simple relationship states that:
  • \( \text{EMF} = I \cdot R \)
This indicates that the EMF (or voltage) is equal to the product of the current \( I \) and the total resistance \( R \) in the circuit.

In our problem, we need to balance the loop's induced EMF with the product of the desired steady current and the total resistance. The loop itself has a resistance of 4 ohms, and it’s connected to a resistor of 8 ohms, giving us a total resistance \( R_{total} \) of 12 ohms.

The problem requires solving for the speed \( v \) required to maintain a steady 2 Amperes current. By substituting the known values into Ohm's Law, we can determine the precise velocity that meets these conditions:
  • \( 4 \cdot 0.3 \cdot v = 2 \cdot 12 \)
  • \( v = \frac{24}{1.2} = 20 \; \text{m/s} \)
This relationship shows how critical Ohm's Law is to designing circuits that meet specific performance criteria.
Magnetic Field
The magnetic field is an invisible force field around magnetic materials, magnetically susceptible materials, and electrical currents. It is characterized by its strength and direction, denoted as the vector \( B \). In electromagnetic problems, the strength of the magnetic field significantly influences the level of interaction with electrical circuits.

In our exercise, the loop moves through a magnetic field of 4 Tesla, a relatively strong field, which directly affects the EMF generated. The strength of a magnetic field specifies how many lines of force are exerted in a particular area and, in our case, how the loop responds as it travels through this field.

When thinking about magnetic fields in terms of electromagnetic induction, it's helpful to consider how they interact with moving charged particles. As the loop travels through the magnetic field, its movement alters the magnetic flux, ensuring that an EMF is established. This induced voltage propels the current around the loop.
  • A stronger magnetic field results in higher EMF for a given speed and length of conductor in the field.
  • The direction of the field relative to the loop’s motion determines the polarity of the induced EMF and the resultant current direction.
These properties of the magnetic field and its orientation are vital in determining how the generated EMF can be controlled and harnessed effectively, as exemplified by this exercise.

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 flat horizontal coils are mounted as shown. Which one of the following actions will not cause the sensitive galvanometer \(G\) to deflect? (a) Coils stationary and coil 1 moves upwards with \(K_{1}\) and \(K_{2}\) closed (b) Both coils stationary, \(K_{2}\) closed and \(K_{1}\) switched on or off (c) With \(K_{1}\) and \(K_{2}\) closed, a variable resistance \(R\) is increased and decreased rapidly (d) Both coils stationary, \(K_{1}\) closed and \(K_{2}\) switched on or off

The energy stored in the magnetic field if current of \(5 \mathrm{~A}\) produces a magnetic flux of \(2 \times 10^{-3} \mathrm{~Wb}\) through a coil of 500 turns is : (a) \(2.5 \mathrm{~J}\) (b) \(0.25 \mathrm{~J}\) (c) \(250 \mathrm{~J}\) (d) \(1.5 \mathrm{~J}\)

Two coils are placed close to each other. The mutual inductance of the pair of coils depend upon : (a) the currents in the two coils (b) the rates at which currents are changing in the two coils (c) relative position and orientation of the two coils (d) the materials of the wires of the coils

A small circular loop of radius \(r\) is placed inside a circular loop of radius \(R(R \gg r)\). The loops are coplanar and their centres coincide. The mutual inductance of the system is proportional to: (a) \(r / R\) (b) \(r^{2} / R\) (c) \(r / R^{2}\) (d) \(r^{2} / R^{2}\)

An Indian ship with a vertical conducting mass navigates the Indian ocean in the latitude of magnetic equator. To induce the greatest emf in the mast, the ship should proceed : (a) northward (b) southward (c) eastward (d) none of these

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.