/*! 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 14 A loop of wire has the shape sho... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 14

A loop of wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius \(r=0.20 \mathrm{m} .\) The normal to the plane of the loop is parallel to a constant magnetic field \(\left(\phi=0^{\circ}\right)\) of magnitude 0.75 T. What is the change \(\Delta \Phi\) in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolution?

Short Answer

Expert verified
The change in magnetic flux is approximately \(-0.0471 \text{ Wb}.\)

Step by step solution

01

Calculate Initial Magnetic Flux

First, compute the initial magnetic flux \( \Phi_1 \) through the loop. The formula for magnetic flux through a surface is given by \[ \Phi = B \cdot A \cdot \cos(\phi), \]where \( B = 0.75 \text{ T} \) is the magnetic field strength, \( A \) is the area the magnetic field passes through, and \( \phi = 0^\circ \) is the angle between the normal to the loop surface and the magnetic field. Since \( \phi = 0^\circ \), \( \cos(0^\circ) = 1 \).The area of the semicircle is \[ A = \frac{1}{2} \pi r^2, \]where \( r = 0.20 \text{ m} \). So,\[ A = \frac{1}{2} \pi (0.20)^2 \approx 0.0628 \text{ m}^2. \]Therefore, the initial magnetic flux is\[ \Phi_1 = 0.75 \times 0.0628 \times 1 \approx 0.0471 \text{ Wb}. \]
02

Calculate Final Magnetic Flux

After the semicircle is rotated through half a revolution, the plane of the loop is perpendicular to the magnetic field. Now \( \phi = 90^\circ \), and \( \cos(90^\circ) = 0 \). Thus, the final magnetic flux \( \Phi_2 \) is:\[ \Phi_2 = B \cdot A \cdot \cos(90^\circ) = 0.75 \times 0.0628 \times 0 = 0 \text{ Wb}. \]
03

Calculate Change in Magnetic Flux

The change in magnetic flux \( \Delta \Phi \) is the difference between the initial and final magnetic flux:\[ \Delta \Phi = \Phi_2 - \Phi_1 = 0 - 0.0471 \approx -0.0471 \text{ Wb}. \]

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Key Concepts

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

Magnetic Field
A magnetic field is the region around a magnet where the force of magnetism acts.
It is represented by the symbol \( B \) and measured in teslas (T). The magnetic field in the problem has a strength of 0.75 T.
A magnetic field can exert a force on moving charges, which is how it interacts with loops or coils of wire.
In the given exercise, understanding the direction and strength of the magnetic field is key, as it affects the magnetic flux through any surface within its influence. To visualize a magnetic field, think of field lines that emerge from the north pole of a magnet and loop around to the south pole.
The density of these lines represents the strength of the field; denser lines mean a stronger field.
In our problem, these field lines are uniform, indicating the field鈥檚 constant nature across the loop.
Semicircle
A semicircle is simply half of a full circle In mathematics, the area of a semicircle can be calculated using the formula: \[ A_\text{semi} = \frac{1}{2} \pi r^2 \] where \( r \) is the radius of the semicircle.
For the exercise, the radius is 0.20 m, giving us an area of approximately \( 0.0628 \text{ m}^2 \). In practical applications like our problem, a loop or wire may be bent into more specific shapes like a semicircle.
This affects the total area exposed to a magnetic field, hence impacting magnetic flux.
Despite being half the area of a full circle, the calculation and implications remain important, especially within magnetic fields.
Magnetic Flux Change
Magnetic flux quantifies the total magnetic field passing through a specific area. It is determined by:
  • The strength of the magnetic field \(B\)
  • The area \(A\) it penetrates
  • The angle \(\phi\) between the field lines and the perpendicular to the surface
The formula for magnetic flux is given as: \[ \Phi = B \cdot A \cdot \cos(\phi) \]
Initially, the loop's surface is completely aligned with the field, yielding maximum flux (\(\cos(0^\circ) = 1\)).
After rotation, \(\phi\) becomes \(90^\circ\), and \(\cos(90^\circ) = 0\), meaning no flux passes through, causing a change. In the exercise, the change in magnetic flux \(\Delta \Phi\) is noted by the shift from \(0.0471 \text{ Wb}\) to 0 as a result of re-orientation.
This concept is fundamental in electromagnetic applications, influencing how devices generate electricity.
Rotation in Magnetic Fields
Rotating an object in a magnetic field significantly affects the magnetic flux it experiences.
With rotation, the angle \(\phi\) changes, altering the effective area through which the magnetic field passes. When the loop in the problem is rotated, it goes from being directly aligned with the field to perpendicular to it.
Initially, the field lines pass completely through the loop鈥檚 surface, but after a half-revolution, they do not penetrate at all. This alteration is crucial in devices such as electric generators, where rotation within a magnetic field leads to fluctuating flux, inducing voltage (Faraday's Law). Understanding how rotation influences the magnetic field interaction assists in grasping the principles of electromagnetism.

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Most popular questions from this chapter

Suppose there are two transformers between your house and the high-voltage transmission line that distributes the power. In addition, assume that your house is the only one using electric power. At a substation the primary coil of a step-down transformer (turns ratio \(=1: 29\) ) receives the voltage from the high-voltage transmission line. Because of your usage, a current of \(48 \mathrm{mA}\) exists in the primary coil of this transformer. The secondary coil is connected to the primary of another step-down transformer (turns ratio \(=1: 32\) ) somewhere near your house, perhaps up on a telephone pole. The secondary coil of this transformer delivers a \(240-\mathrm{V}\) emf to your house. How much power is your house using? Remember that the current and voltage given in this problem are rms values.

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is \(7.0 \times 10^{-3}\) Wb. What is the flux that passes through the circular loop?

A magnetic field is passing through a loop of wire whose area is \(0.018 \mathrm{m}^{2} .\) The direction of the magnetic field is parallel to the normal to the loop, and the magnitude of the field is increasing at the rate of \(0.20 \mathrm{T} / \mathrm{s}\) (a) Determine the magnitude of the emf induced in the loop. (b) Suppose that the area of the loop can be enlarged or shrunk. If the magnetic field is increasing as in part (a), at what rate (in \(\mathrm{m}^{2} / \mathrm{s}\) ) should the area be changed at the instant when \(B=1.8 \mathrm{T}\) if the induced emf is to be zero? Explain whether the area is to be enlarged or shrunk.

Coil 1 is a flat circular coil that has \(N_{1}\) turns and a radius \(R_{1}\). At its center is a much smaller flat, circular coil that has \(N_{2}\) turns and radius \(R_{2}\). The planes of the coils are parallel. Assume that coil 2 is so small that the magnetic field due to coil 1 has nearly the same value at all points covered by the area of coil \(2 .\) Determine an expression for the mutual inductance between these two coils in terms of \(\mu_{0}, N_{1}, R_{1}, N_{2},\) and \(R_{2}\)

Indicate the direction of the electric field between the plates of the parallel plate capacitor shown in the drawing if the magnetic field is decreasing in time. Give your reasoning.
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