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Chapter 25: Problem 23

A magnetic field of \(0.080 \mathrm{~T}\) is in the \(y\) -direction. The velocity of wire segment \(S\) has a magnitude of \(78 \mathrm{~m} / \mathrm{s}\) and components of \(18 \mathrm{~m} / \mathrm{s}\) in the \(x\) -direction, \(24 \mathrm{~m} / \mathrm{s}\) in the \(y\) -direction, and \(72 \mathrm{~m} / \mathrm{s}\) in the \(z\) -direction. The segment has length \(0.50 \mathrm{~m}\) and is parallel to the \(z\) -axis as it moves. (a) Find the motional emf induced between the ends of the segment. (b) What would the motional emf be if the wire segment was parallel to the \(y\) -axis?

Short Answer

Expert verified
The solutions for (a) and (b) depend on the specifics calculations in each step, specifically in step 3 and 4. After calculating, you will have the emf for both cases.

Step by step solution

01

Determine the Position Vector

First, calculate the position vector of the wire. Since the wire is parallel to the z-axis, the position vector is only in the z-direction and can be described as \(\vec{r} = 0.50 \hat{z} \mathrm{m}\).
02

Compute the Velocity Vector

Next, let's find the velocity vector of the wire. Given in the problem are the components of velocity in the x, y, and z-directions. Therefore, the velocity can be represented as \(\vec{v} = 18 \hat{x} + 24 \hat{y} + 72 \hat{z} \mathrm{m/s}\).
03

Cross Product of Velocity and Position

The motional emf is produced due to the force experienced by the moving charges in the magnetic field. The force is given by the cross product of the velocity vector and the magnetic field. Express the magnetic field as \(\vec{B} = 0.080 \hat{y} \mathrm{T}\). The cross product \(\vec{F} = \vec{v} \times \vec{B}\) gives the force vector. However, only the z-component of the force contributes to the motional emf since the wire is moving parallel to the z-axis. Calculate the z-component of the force.
04

Calculate the Motional emf

For part (a), the motional emf can be calculated using the formula \(\epsilon = \vec{F} . \vec{r}\), where '.' denotes the dot product. The result is the emf induced between the ends of the wire.
05

Repeat for Wire Parallel to the y-axis

For part (b), repeat the process but this time with the wire parallel to the y-axis. Here, the position vector is \(\vec{r} = 0.50 \hat{y} \mathrm{m}\) and the velocity vector remains the same. Compute the force vector and then the motional emf as in Step 4. This will give the emf if the wire was parallel to the y-axis.

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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 a vector field that permeates space and can exert a magnetic force on moving charges and magnetic materials. Imagine it like a field of invisible lines moving from the North pole to the South pole of a magnet, describing the direction a magnetic north would take if placed within the field. The strength of this magnetic field is measured in teslas (T), and its direction is given by a vector known as the magnetic field vector, often denoted as \( \vec{B} \). For example, a magnetic field of \(0.080 \mathrm{T}\) in the \(y\)-direction is represented as \( \vec{B} = 0.080 \hat{y} \mathrm{T} \).

Understanding the behavior of magnetic fields is crucial when studying the phenomenon of motional electromotive force (emf), where the movement of a conductor through a magnetic field can induce a current within the conductor. The properties of the magnetic field interact with the velocity of the moving charges, resulting in the generation of emf.
Cross Product
The cross product, also known as the vector product, is an operation between two vectors that results in another vector perpendicular to the plane of the initial vectors. Specifically, the cross product of two vectors \( \vec{A} \) and \( \vec{B} \) is denoted by \( \vec{A} \times \vec{B} \) and its magnitude is proportional to the area of the parallelogram spanned by the two vectors, which also corresponds to the product of the magnitudes of the vectors and the sine of the angle between them.

In the context of electromagnetism, the cross product is used to determine the force acting on a charged particle moving within a magnetic field, as given by the Lorentz force law. When computing motional emf, we are interested in the force exerted on charges due to their motion through the magnetic field. This force is calculated using the cross product of the velocity vector, \( \vec{v} \), and the magnetic field vector, \( \vec{B} \), as shown in the steps of resolving the exercise.
Electromotive Force (Emf)
Electromotive force (emf) is not actually a force, despite its name. It is a measure of the energy provided by a source (such as a battery or a magnetic field) per charge that passes through the source. In units, it is measured in volts (V). When it comes to motional emf, it is generated when a conductor, like a wire, moves through a magnetic field. The emf is directly related to the velocity of the wire, the strength of the magnetic field, and the length of the wire within the field.

The calculation of motional emf involves vector analysis, where both the magnetic field and the velocity of the conductor are considered as vectors. Emf can be calculated using the formula \( \epsilon = |\vec{F}| \cdot |\vec{r}| \), where \( \vec{F} \) is the force acting on the charged particles, and \( \vec{r} \) represents the position vector of the wire segment in the magnetic field. If the wire is aligned with the magnetic field, no emf will be produced, as there will be no force component perpendicular to the field.
Vector Analysis
Vector analysis is a branch of mathematics that deals with quantities that have both magnitude and direction. In physics, vectors are used extensively to represent a variety of physical quantities such as displacement, velocity, force, and magnetic fields. When analyzing scenarios that involve vectors, it's essential to understand their properties and how they interact with each other through operations such as addition, subtraction, the dot product, and the cross product.

In the context of motional emf, vector analysis is essential. The position vector, \( \vec{r} \), and the velocity vector, \( \vec{v} \), are used in tandem with the magnetic field vector, \( \vec{B} \), to precisely calculate the resultant motional emf. The process involves determining the direction and magnitude of the vectors, and employing the cross product to establish the force exerted on the wire due to the magnetic field, subsequently using the dot product if necessary to find out the emf. This intricate interplay of vectors underscores the importance of vector analysis in understanding and solving electromagnetism-related problems.

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Most popular questions from this chapter

A circular loop of wire with a radius of \(12.0 \mathrm{~cm}\) and oriented in the horizontal \(x y\) -plane is located in a region of uniform magnetic field. A field of \(1.5 \mathrm{~T}\) is directed along the positive \(z\) -direction, which is upward. (a) If the loop is removed from the field region in a time interval of \(2.0 \mathrm{~ms}\), find the average emf that will be induced in the wire loop during the extraction process. (b) If the coil is viewed looking down on it from above, is the induced current in the loop clockwise or counterclockwise?

A 5.00 A current runs through a 12 gauge copper wire (diameter \(2.05 \mathrm{~mm}\) ) and through a light bulb. Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. (a) How many electrons pass through the light bulb each second? (b) What is the current density in the wire? (c) At what speed does a typical electron pass by any given point in the wire? (d) If you were to use wire of twice the diameter, which of the above answers would change? Would they increase or decrease?

A small, closely wound coil has \(N\) turns, area \(A\), and resistance \(R\). The coil is initially in a uniform magnetic field that has magnitude \(B\) and a direction perpendicular to the plane of the loop. The coil is then rapidly pulled out of the field so that the flux through the coil is reduced to zero in time \(\Delta t\). (a) What are the magnitude of the average \(\operatorname{emf} \mathcal{E}_{\text {av }}\) and average current \(I_{\mathrm{av}}\) induced in the coil? (b) The total charge \(Q\) that flows through the coil is given by \(Q=I_{\mathrm{av}} \Delta t .\) Derive an expression for \(Q\) in terms of \(N, A, B,\) and \(R .\) Note that \(Q\) does not depend on \(\Delta t .\) (c) What is \(Q\) if \(N=150\) turns, \(A=4.50 \mathrm{~cm}^{2}, R=30.0 \Omega,\) and \(B=0.200 \mathrm{~T} ?\)

Two cylindrical cans with insulating sides and conducting end caps are filled with water, attached to the circuitry shown in Fig. \(\mathbf{P} 25.69,\) and used to determine salinity levels. The cans are identical, with radius \(r=5.00 \mathrm{~cm}\) and length \(L=3.00 \mathrm{~cm} .\) The battery supplies a potential of \(10.0 \mathrm{~V},\) has a negligible internal resistance, and is connected in series with a resistor \(R=15.0 \Omega .\) The left cylinder is filled with pure distilled water, which has infinite resistivity. The right cylinder is filled with a saltwater solution. It is known that the resistivity of the saltwater solution is determined by the relationship \(\rho=\left(s_{0} / s\right) \Omega \cdot \mathrm{m},\) where \(s\) is the salinity in parts per thousand \((\mathrm{ppt})\) and \(s_{0}=6.30\) ppt. (a) The ammeter registers a current of \(484 \mathrm{~mA}\). What is the salinity of the saltwater solution? (b) The left cylinder acts as a capacitor. Use Eq. (24.19) for its capacitance. How much charge is present on its upper plate? Note that pure water has a dielectric constant of \(80.4 .\) (c) At what rate is energy dissipated by the saltwater? (d) For what salinity level would the \(15.0 \Omega\) resistor dissipate half the power supplied by the battery?

BIO Electric Eels. Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to \(500 \mathrm{~V}\) and produce currents of \(80 \mathrm{~mA}\) (or even larger). A typical pulse lasts for \(10 \mathrm{~ms}\). What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?
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