/*! 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 12 A 0.250 m radius, 500-turn coil ... [FREE SOLUTION] | 91Ó°ÊÓ

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A 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in \(4.17 \mathrm{ms}\), originally having its plane perpendicular to a uniform magnetic field. (This is 60 rev/s.) Find the magnetic field strength needed to induce an average emf of \(10,000 \mathrm{V}\).

Short Answer

Expert verified
The magnetic field strength needed to induce an average emf of 10,000 V is \(48.9 \text{ T}\).

Step by step solution

01

Understand the Faraday's Law of Induction

Faraday's Law of Induction states that the induced electromotive force (emf) in any closed circuit is equal to the negative of the time rate of change of the magnetic flux through the circuit. This can be expressed as \(\text{emf} = -N\frac{\Delta\Phi_B}{\Delta t}\), where \(N\) is the number of turns, \(\Delta\Phi_B\) is the change in magnetic flux, and \(\Delta t\) is the time it takes for this change to occur.
02

Calculate the Change in Magnetic Flux \(\Delta\Phi_B\)

The magnetic flux \(\Phi_B\) is given by the product of the magnetic field strength \(B\), the area \(A\) of the coil, and the cosine of the angle \(\theta\) between the field and the normal to the coil's plane: \(\Phi_B = B \cdot A \cdot \cos(\theta)\). In this case, as the coil rotates one-fourth of a revolution from being perpendicular (\(\theta = 0^\circ\)) to the field, it ends at \(\theta = 90^\circ\), which means an initial \(\Phi_{Bi} = B \cdot A\) and a final \(\Phi_{Bf} = 0\), since \(\cos(90^\circ) = 0\). Therefore, \(\Delta\Phi_B = \Phi_{Bf} - \Phi_{Bi} = -B \cdot A\) since \(\Phi_{Bf} = 0\) and \(\Phi_{Bi} = B \cdot A\).
03

Calculate the Area \(A\) of the Coil

The area of the coil can be calculated using the formula for the area of a circle, \(A = \pi r^2\), where \(r\) is the radius. Substituting \(r = 0.250 \text{ m}\), we obtain \(A = \pi (0.250)^2 = 0.19635 \text{ m}^2\).
04

Determine the Time Interval \(\Delta t\)

The given time interval for the quarter turn rotation is \(4.17 \text{ ms} = 4.17 \times 10^{-3} \text{ s}\).
05

Solve for the Magnetic Field Strength \(B\)

Now that we have the change in flux \(\Delta\Phi_B = -B \cdot A\), the number of turns \(N = 500\), and the time interval \(\Delta t\), we can use Faraday's Law to find \(B\). Rewrite Faraday's law to solve for \(B\): \[ \text{emf} = -N\frac{\Delta\Phi_B}{\Delta t} = -N (\frac{-B \cdot A}{\Delta t}) \] Then, plug in the values for emf, N, A, and \(\Delta t\): \[ 10,000 \text{ V} = 500 (\frac{B \cdot 0.19635 \text{ m}^2}{4.17 \times 10^{-3} \text{ s}}) \] Solving for \(B\), we get: \[ B = \frac{10,000}{500 \cdot 0.19635 \cdot (4.17 \times 10^{-3})^{-1}} = 48.9 \text{ T} \]

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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 measure of the amount of magnetic field passing through a given area. Imagine you have a loop of wire exposed to a magnetic field; the extent to which the field lines penetrate the loop is described by magnetic flux. It is quantified as the product of the magnetic field strength, the area of the loop, and the cosine of the angle between the magnetic field direction and the normal (perpendicular) to the area:
\[\begin{equation}\Phi_B = B \times A \times \text{cos}(\theta)\end{equation}\]
Where
  • \(\Phi_B\) is the magnetic flux,
  • \(B\) is the strength of the magnetic field,
  • \(A\) is the area of the loop, and
  • \(\theta\) is the angle between the field and the normal to the loop's plane.
In our exercise, as the coil consists of 500 turns and it rotates one-fourth of a revolution, the magnetic flux linkages change, leading to the induction of an electromotive force (EMF).
Electromagnetic Force (EMF)
The electromotive force, or EMF, is the voltage generated by an external magnetic field when it changes the magnetic environment of a circuit. For instance, when you move a magnet towards a coil, the change in magnetic flux over time induces an EMF, pushing electrons around the circuit. In the case of our textbook problem, the induced EMF is created by rotating the coil within a magnetic field. Using Faraday's Law of Induction, it is determined by the negative rate of change of magnetic flux through the circuit. In equation form:
\[\begin{equation}\text{emf} = -N\frac{\Delta\Phi_B}{\Delta t}\end{equation}\]
Where
  • \(N\) is the number of turns in the coil,
  • \(\Delta\Phi_B\) is the change in magnetic flux, and
  • \(\Delta t\) is the time over which this change occurs.
In practice, to achieve a high induced EMF, either the number of turns can be increased, the magnetic field can be made more powerful, or the change can be made to occur more rapidly.
Magnetic Field Strength
Magnetic field strength, denoted by B, reflects how strong a magnetic field is at a point in space. It is a vector quantity and is measured in Tesla (T). The strength of the magnetic field plays a crucial role in Faraday's Law, as it directly affects the magnitude of the induced EMF in a coil or wire loop. The larger the magnetic field strength, the greater the potential for inducing a strong EMF when a conductor moves relative to the magnetic field. In our textbook example, to induce a specified EMF of 10,000 V by rotating the coil, the problem required finding the necessary magnetic field strength. The solution hinges upon calculating the rate of change of magnetic flux, and then applying Faraday's Law to back out the strength of the magnetic field.
Induction in Rotating Coils
When a coil rotates in a magnetic field, the magnetic flux linking with the coil changes. This is because the orientation of the coil with respect to the direction of the magnetic field is dynamic. For coils that rotate, such as in electrical generators, the flux changes cyclically, leading to the induction of an alternating EMF over time. Faraday's law precisely accounts for this phenomenon. In the exercise, the rapid rotation of the coil within the magnetic field—one-fourth revolution in 4.17 ms—leads to a significant change in magnetic flux, hence a large EMF. The faster the coil rotates, the higher the frequency of the changing magnetic field, and therefore, the higher the induced EMF will be. This principle underpins the operation of many electromagnetic devices, from electric guitars to power generators.

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Most popular questions from this chapter

(a) A particle and its antiparticle are at rest relative to an observer and annihilate (completely destroying both masses), creating two \(\gamma\) rays of equal energy. What is the characteristic \(\gamma\) -ray energy you would look for if searching for evidence of proton-antiproton annihilation? (The fact that such radiation is rarely observed is evidence that there is very little antimatter in the universe.) (b) How does this compare with the \(0.511-\mathrm{MeV}\) energy associated with electronpositron annihilation?

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The lowest frequency in the FM radio band is 88.0 MHz. (a) What inductance is needed to produce this resonant frequency if it is connected to a 2.50 pF capacitor? (b) The capacitor is variable, to allow the resonant frequency to be adjusted to as high as 108 MHz. What must the capacitance be at this frequency?

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