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A square loop of wire is held in a uniform 0.24 T magnetic field directed perpendicular to the plane of the loop. The length of each side of the square is decreasing at a constant rate of \(5.0 \mathrm{~cm} / \mathrm{s}\). What emf is induced in the loop when the length is \(12 \mathrm{~cm} ?\)

Short Answer

Expert verified
The induced emf is 2.88 mV.

Step by step solution

01

Understand the Problem

We have a square loop placed in a uniform magnetic field. We're asked to find the induced electromotive force (emf) when the side length is 12 cm, given that it decreases at 5 cm/s.
02

Calculate Initial Parameters

The area of a square loop with side length \(s\) is \(s^2\). For \(s = 12\text{ cm}\), the area \(A = 12^2 = 144\,\text{cm}^2\), or converting to square meters, \(A = 0.0144\,\text{m}^2\). The rate of change of side length is \(-0.05\, \text{m/s}\) (since it's decreasing).
03

Use Faraday's Law of Induction

Faraday's Law states \(\text{emf} = -\frac{d\Phi}{dt}\), where \(\Phi = B \cdot A\) is the magnetic flux. Calculate the rate of change of the magnetic flux \(\Phi\).
04

Differentiate the Magnetic Flux

\(\Phi = B \cdot A = B \cdot s^2\). Differentiate \(\Phi\) with respect to time: \(\frac{d\Phi}{dt} = B \cdot 2s \cdot \frac{ds}{dt}\).
05

Calculate the Induced EMF

Substitute \(B = 0.24\,\text{T}\), \(s = 0.12\,\text{m}\), and \(\frac{ds}{dt} = -0.05\,\text{m/s}\) into the differentiated formula: \(\frac{d\Phi}{dt} = 0.24 \cdot 2 \cdot 0.12 \cdot (-0.05)\). So, \(\text{emf} = - (0.24 \cdot 2 \cdot 0.12 \cdot (-0.05))\).
06

Compute the Result

Calculate the emf: \(\text{emf} = 0.24 \times 2 \times 0.12 \times 0.05 = 0.00288\, \text{V}\). The emf is 2.88 mV.

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Key Concepts

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

Magnetic Flux
Magnetic flux is a fundamental concept when understanding electromagnetic phenomena like induction. Magnetic flux, represented by the symbol \( \Phi \), is essentially the total magnetic field passing through a surface. It is calculated using the formula: \[ \Phi = B \cdot A \cdot \cos(\theta) \]where:
  • \( B \) is the magnetic field strength.
  • \( A \) is the area the magnetic field lines pass through.
  • \( \theta \) is the angle between the field lines and the normal to the surface.
In our exercise, the magnetic field is perpendicular to the loop, so \( \theta = 0^\circ \) and \( \cos(0) = 1 \). Thus, the magnetic flux simplifies to \( \Phi = B \cdot A \). The time-dependent change in the magnetic flux leads to an electromotive force (emf). By knowing how the area changes over time—thanks to the changing side length of the square loop—we can determine the rate of flux change.
Electromotive Force
Electromotive force, or emf, is a fascinating consequence of changing magnetic environments. It's not a force in the traditional sense but rather represents a potential difference. When magnetic flux through a loop changes over time, Faraday's Law of Induction springs into action: \[ \text{emf} = -\frac{d\Phi}{dt} \]Here, the negative sign follows Lenz's Law, indicating that the induced emf generates a current opposing the change in flux.
In the exercise, as the square loop shrinks, its area decreases, directly affecting the magnetic flux. This results in a time-varying flux and, subsequently, the induction of an emf. The rate at which the side length of the loop decreases helps us find \( \frac{d\Phi}{dt} \). Understanding how this process works unleashes the power to predict and control electric currents in various applications, from electric generators to sensors.
Uniform Magnetic Field
In this context, a uniform magnetic field is one where the magnetic field strength \( B \) has the same magnitude and direction at every point within the field. Such fields are often used in physics problems because they simplify calculations, ensuring that the magnetic field is constant across the entire area of interest.
In the exercise provided, the uniform magnetic field strength is 0.24 T (teslas), which remains constant as the loop's size changes. This consistency allows for straightforward calculation of the magnetic flux through the loop as a simple product of field strength and loop area. When working with non-uniform fields, calculations become more complex, requiring integration over the area to account for variations. Understanding uniform magnetic fields is crucial in many applications, such as in MRI machines, where consistent magnetic fields ensure operational precision.

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Most popular questions from this chapter

A loop antenna of area \(2.00 \mathrm{~cm}^{2}\) and resistance \(5.21 \mu \Omega\) is perpendicular to a uniform magnetic field of magnitude \(17.0 \mu \mathrm{T}\). The field magnitude drops to zero in \(2.96 \mathrm{~ms} .\) How much thermal energy is produced in the loop by the change in field?

A small circular loop of area \(2.00 \mathrm{~cm}^{2}\) is placed in the plane of, and concentric with, a large circular loop of radius \(1.00 \mathrm{~m} .\) The current in the large loop is changed at a constant rate from \(200 \mathrm{~A}\) to -200 A (a change in direction) in a time of \(1.00 \mathrm{~s}\), starting at \(t=0 .\) What is the magnitude of the magnetic field \(\vec{B}\) at the center of the small loop due to the current in the large loop at (a) \(t=0\), (b) \(t=0.500 \mathrm{~s},\) and (c) \(t=1.00 \mathrm{~s} ?\) (d) From \(t=0\) to \(t=1.00 \mathrm{~s},\) is \(\vec{B}\) reversed? Because the inner loop is small, assume \(\vec{B}\) is uniform over its area. (e) What emf is induced in the small loop at \(t=0.500 \mathrm{~s} ?\)

A circular coil has a \(10.0 \mathrm{~cm}\) radius and consists of 30.0 closely wound turns of wire. An externally produced magnetic field of magnitude \(2.60 \mathrm{mT}\) is perpendicular to the coil. (a) If no current is in the coil, what magnetic flux links its turns? (b) When the current in the coil is \(3.80 \mathrm{~A}\) in a certain direction, the net flux through the coil is found to vanish. What is the inductance of the coil?

At \(t=0,\) a battery is connected to a series arrangement of a resistor and an inductor. At what multiple of the inductive time constant will the energy stored in the inductor's magnetic field be 0.500 its steady-state value?

A coil with an inductance of \(2.0 \mathrm{H}\) and a resistance of \(10 \Omega\) is suddenly connected to an ideal battery with \(\mathscr{E}=100 \mathrm{~V}\). At \(0.10 \mathrm{~s}\) after the connection is made, what is the rate at which (a) energy is being stored in the magnetic field, (b) thermal energy is appearing in the resistance, and (c) energy is being delivered by the battery?

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