/*! 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 16 A circular loop of wire radius \... [FREE SOLUTION] | 91Ó°ÊÓ

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A circular loop of wire radius \(R\) rotates about \(z\) -axis with angular velocity \(\omega .\) The normal to the loop is perpendicular to z-axis. At \(t=0\) normal is parallel to \(y\) -axis. An external magnetic field \(\overrightarrow{\mathrm{B}}=B_{y} \hat{j}+B_{z} \hat{k}\) is applied. The emf induced in the coil will be: (a) \(\pi r^{2} \omega B_{y} \sin \omega t\) (b) \(\pi r^{2} \omega B_{z} \sin \omega t\) (c) \(\pi r^{2} \omega B_{z} \cos \omega t\) (d) \(\pi r^{2} \omega B_{y} \cos \omega t\)

Short Answer

Expert verified
The answer is (a) \( \pi r^2 \omega B_y \sin(\omega t) \).

Step by step solution

01

Understand the Problem and Set Up

We have a circular loop of wire with radius \( R \) rotating about the \( z \)-axis with angular velocity \( \omega \). At \( t=0 \), the normal to the loop is parallel to the \( y \)-axis. There is an external magnetic field given by \( \overrightarrow{\mathrm{B}} = B_{y} \hat{j} + B_{z} \hat{k} \). We need to find the induced emf in the coil as it rotates.
02

Determine the Magnetic Flux

Calculate the magnetic flux through the loop, \( \Phi \). Since the normal to the loop rotates with angular velocity \( \omega \), the angle \( \theta \) between the loop's normal and the \( y \)-axis changes with time: \( \theta = \omega t \). The magnetic flux is given by \( \Phi = \mathbf{B} \cdot \mathbf{A} = B_y A \cos \theta + B_z A \sin \theta \), where \( A = \pi R^2 \) is the area of the loop.
03

Differentiate Magnetic Flux to Find the EMF

The induced emf \( \varepsilon \) is given by Faraday's Law: \( \varepsilon = -\frac{d\Phi}{dt} \). Substitute \( \Phi = B_y A \cos(\omega t) + B_z A \sin(\omega t) \) into the formula and differentiate with respect to \( t \):\[ \varepsilon = -\frac{d}{dt}(B_y A \cos(\omega t) + B_z A \sin(\omega t)) \]This yields \( \varepsilon = -A \left(-B_y \omega \sin(\omega t) + B_z \omega \cos(\omega t) \right) \).
04

Simplify the Expression

Simplify the expression for \( \varepsilon \) and substitute back the area \( A = \pi R^2 \):\[ \varepsilon = \pi R^2 \omega (B_y \sin(\omega t) - B_z \cos(\omega t)) \]
05

Compare with Options and Select the Correct Answer

Based on the simplified expression, match the term to the given options. Recalling the structure of our expression, \( \varepsilon = \pi r^2 \omega B_y \sin(\omega t) + \pi r^2 \omega B_z \cos(\omega t) \), the correct match is option (a) \( \pi r^2 \omega B_y \sin(\omega t) \).

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Key Concepts

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

Faraday's Law of Electromagnetic Induction
Faraday's Law of Electromagnetic Induction is a fundamental principle that helps us understand how electrical currents are generated within a circuit in the presence of a changing magnetic field. This concept is named after Michael Faraday, who discovered that changing magnetic fields can induce an electromotive force (EMF) and subsequently electric currents.
In its simplest form, Faraday's Law states that the induced EMF in a closed loop is directly proportional to the negative rate of change of magnetic flux through the loop. Mathematically, it is expressed as:
  • \( \varepsilon = -\frac{d\Phi}{dt} \)
Here, \( \varepsilon \) represents the induced EMF and \( \Phi \) is the magnetic flux.
Faraday's Law is critical in understanding how generators and transformers work. In our current scenario, it helps calculate the induced EMF in a rotating loop subjected to a magnetic field, allowing us to deduce how changes in flux generate electricity.
Magnetic Flux
Magnetic Flux is a measure of the quantity of magnetism, considering the strength and extent of the magnetic field. It is essential in understanding how magnetic fields interact with physical objects like loops or coils of wire.
The formula to calculate magnetic flux \( \Phi \) through a given area \( A \) is:
  • \( \Phi = \mathbf{B} \cdot \mathbf{A} = BA \cos(\theta) \)
Where:
  • \( \mathbf{B} \) is the magnetic field vector
  • \( A \) is the area vector perpendicular to the field lines
  • \( \theta \) is the angle between \( \mathbf{B} \) and \( A \)
In the rotating loop scenario from the exercise, the loop changes orientation over time, hence the angle \( \theta \) becomes time-dependent \( \theta = \omega t \). This affects how the magnetic flux through the loop changes over time and helps in determining the induced EMF through Faraday's Law.
Angular Velocity
Angular Velocity is a measure of the rate of rotation of an object relative to an axis. It's represented by the symbol \( \omega \) and is usually expressed in radians per second. Angular velocity provides insight into how quickly the object is rotating.
In the exercise, the circular loop of wire rotates about the z-axis with angular velocity \( \omega \). At time \( t = 0 \), the normal to the loop is aligned with the y-axis. The angular velocity determines how quickly the orientation of the coil changes relative to the external magnetic field components. This rotational attribute further influences the calculation of magnetic flux and thus the induced EMF based on how rapidly the angle \( \theta = \omega t \) changes as time progresses, affecting the trigonometric components of the magnetic flux equation.
Electromagnetic Force (EMF)
Electromagnetic Force, often denoted as EMF, refers to the voltage generated by any source of electrical energy such as a battery or a rotating loop in the presence of a magnetic field. Despite its name, EMF is not a force but a potential difference.
The calculation of the EMF in the exercise revolves around the rotation of a loop in a magnetic field. EMF is induced due to changes in the magnetic flux as the normal of the loop rotates away from the y-axis. Utilizing Faraday's Law, the induced EMF \( \varepsilon \) in this case is given by differentiating the magnetic flux with respect to time. After simplification, the expression for EMF becomes:
  • \( \varepsilon = \pi R^2 \omega (B_y \sin(\omega t) - B_z \cos(\omega t)) \)
This expression shows that the EMF is influenced by the components of the magnetic field \( B_y \) and \( B_z \), the area of the loop \( \pi R^2 \), and the angular velocity \( \omega \). Understanding EMF is key to harnessing electrical energy from mechanical systems, as it relates directly to how electricity can be generated in this context.

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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

A closed circuit consists of a source of \(\operatorname{emf} E\) and an inductor coil of inductance \(L\), connected in series. The active resistance of whole circuit is \(R\). At the moment \(t=0\), inductance of coil abruptly decreased to \(L / n\). Then current in the circuit immediately after, is : (a) zero (b) \(E / R\) (c) \(\frac{n E}{R}\) (d) \(\frac{E}{n R}\)

A bicycle wheel of radius \(0.5 \mathrm{~m}\) has 32 spokes. It is rotating at the rate of 120 revolutions per minute, perpendicular to the horizontal component of earth's magnetic field \(B_{H}=4 \times 10^{-5} \mathrm{~T}\). The emf induced between the rim and the centre of the wheel will be : (a) \(6.28 \times 10^{-5} \mathrm{~V}\) (b) \(4.8 \times 10^{-5} \mathrm{~V}\) (c) \(6.0 \times 10^{-5} \mathrm{~V}\) (d) \(1.6 \times 10^{-5} \mathrm{~V}\)

A system \(S\) consists of two coils \(A\) and \(B\). The coil \(A\) have a steady current \(I\) while the coils \(B\) is su \(_{1}\) - nded near by as shown in figure. Now the system is hc:,\(\quad\) (a) the two coils show attraction (b) the two coils show repulsion (c) there is no change in the position of the two coils (d) induced currents are not possible in coil \(B\)

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) \(\frac{100}{3} \mathrm{~m} / \mathrm{s}\)

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