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Chapter 29: Problem 56

A wire of length \(\ell=10.0 \mathrm{~cm}\) is moving with constant velocity in the \(x y\) -plane; the wire is parallel to the \(y\) -axis and moving along the \(x\) -axis. If a magnetic field of magnitude \(1.00 \mathrm{~T}\) is pointing along the positive \(z\) -axis, what must the velocity of the wire be in order to induce a potential difference of \(2.00 \mathrm{~V}\) across it?

Short Answer

Expert verified
Answer: The required velocity of the wire is 20.0 m/s.

Step by step solution

01

Write down the formula for induced EMF

The formula to find the induced electromotive force or EMF in a moving wire (conductor) inside a magnetic field is given by Faraday's law of electromagnetic induction, which states: \(EMF = B * l * v\) Where, - \(EMF\) (Volts) is the induced electromotive force across the conductor, - \(B\) (Tesla) is the magnitude of the magnetic field, - \(l\) (meters) is the length of the conductor, and - \(v\) (meters/second) is the velocity of the conductor. In our case, we are given the potential difference (EMF), magnetic field strength, and the length of the wire. We need to determine the required velocity (\(v\)).
02

Plug in the known values

We are given the following values: 1. Length of the wire, \(l = 10.0\,\text{cm} = 0.1\,\text{m}\). 2. Magnetic field strength, \(B = 1.00\,\text{T}\). 3. Induced potential difference (EMF), \(EMF = 2.00\,\text{V}\). Plugging these values into the EMF formula: \(2.00 = (1.00) * (0.1) * v\)
03

Solve for the velocity

Now we need to solve for the velocity (\(v\)). Using algebra, we can isolate the velocity by dividing both sides of the equation by the product of the magnetic field strength and the length of the wire: \(v = \frac{2.00}{1.00 * 0.1}\) \(v = 20.0\,\text{m/s}\) Hence, the required velocity of the wire in order to induce a potential difference of 2.00 V across it is \(20.0\,\text{m/s}\).

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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
Understanding Faraday's Law of Electromagnetic Induction is essential when dealing with the principles of electromagnetism. This fundamental law explains how electric current can be produced by varying a magnetic field. Faraday's Law 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. In simpler terms, it tells us that a changing magnetic field will produce an electric current in a conductor.

But how does this translate into real-world applications? Consider an electric generator: As the coil within the generator moves through a magnetic field, an electric current is induced in the coil. This principle is what allows us to generate electricity on a large scale. And within your textbook problem, when we move a conductor through a magnetic field, an EMF is induced, which can be calculated using the straightforward formula EMF = B * l * v, accounting for magnetic field strength (B), conductor's length (l), and its velocity (v).
Magnetic Field Strength
Magnetic field strength, denoted as 'B' and measured in Tesla (T), is a quantitative expression of the intensity of a magnetic field at a point in space. A single Tesla is quite strong — it's about 10,000 times the strength of Earth's magnetic field!

The magnetic field strength plays a crucial role in inducing an EMF in a moving conductor. According to Faraday's Law, the magnitude of the induced EMF is directly proportional to the strength of the magnetic field: the stronger the field, the greater the induced EMF. That's why in MRI machines, which rely on very strong magnetic fields, a high magnetic field strength is crucial for creating detailed images of the inside of the human body.
Electromotive Force (EMF)
Electromotive Force (EMF), often measured in volts (V), is not a 'force' as the name suggests, but a potential difference — it's the voltage generated by a battery or by the magnetic force according to Faraday's Law. This induced EMF is what drives the current through the circuit, and it's fundamentally what allows electrical devices to function.

In the context of the exercise you're working on, an EMF of 2.00 V is induced across a wire moving in a magnetic field. It's important to note that this induced EMF is the reason electrons flow through the wire, and thereby, an electric current is generated. This is the same principle that allows for the operation of electric motors and generators.
Velocity of a Conductor
The velocity of a conductor refers to the speed and direction at which the conductor moves through a magnetic field. It's a vector quantity, meaning it has both magnitude and direction. An interesting aspect of velocity in electromagnetic induction is that only the component of the velocity perpendicular to the magnetic field lines contributes to the induction of EMF.

When solving for induced EMF, if the conductor moves faster, more lines of magnetic field are cut in a shorter period, which increases the rate of change of magnetic flux, thus inducing a larger EMF. It clarifies why, in your textbook problem, the induction of a 2.00 V potential difference across the wire is achievable only with a particular velocity. The formula used in the problem is an expression derived from Faraday's Law that is specific to this scenario of a straight conductor moving at a constant velocity in a uniform magnetic field.

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Most popular questions from this chapter

A respiration monitor has a flexible loop of copper wire, which wraps about the chest. As the wearer breathes, the radius of the loop of wire increases and decreases. When a person in the Earth's magnetic field (assume \(\left.0.426 \cdot 10^{-4} \mathrm{~T}\right)\) inhales, what is the average current in the loop, assuming that it has a resistance of \(30.0 \Omega\) and increases in radius from \(20.0 \mathrm{~cm}\) to \(25.0 \mathrm{~cm}\) over \(1.00 \mathrm{~s}\) ? Assume that the magnetic field is perpendicular to the plane of the loop.

A circular conducting loop with radius \(a\) and resistance \(R_{2}\) is concentric with a circular conducting loop with radius \(b \gg a(b\) much greater than \(a\) ) and resistance \(R_{1}\). A time-dependent voltage is applied to the larger loop, having a slow sinusoidal variation in time given by \(V(t)=V_{0} \sin \omega t\) where \(V_{0}\) and \(\omega\) are constants with dimensions of voltage and inverse time, respectively. Assuming that the magnetic field throughout the inner loop is uniform (constant in space) and equal to the field at the center of the loop, derive expressions for the potential difference induced in the inner loop and the current \(i\) through that loop.

An 8 -turn coil has square loops measuring \(0.200 \mathrm{~m}\) along a side and a resistance of \(3.00 \Omega\). It is placed in a magnetic field that makes an angle of \(40.0^{\circ}\) with the plane of each loop. The magnitude of this field varies with time according to \(B=1.50 t^{3}\), where \(t\) is measured in seconds and \(B\) in teslas. What is the induced current in the coil at \(t=2.00 \mathrm{~s} ?\)

A solenoid with 200 turns and a cross-sectional area of \(60 \mathrm{~cm}^{2}\) has a magnetic field of \(0.60 \mathrm{~T}\) along its axis. If the field is confined within the solenoid and changes at a rate of \(0.20 \mathrm{~T} / \mathrm{s}\), the magnitude of the induced potential difference in the solenoid will be a) \(0.0020 \mathrm{~V}\). b) \(0.02 \mathrm{~V}\). c) \(0.001 \mathrm{~V}\). d) \(0.24 \mathrm{~V}\).

A student wearing a 15.0 -g gold band with radius \(0.750 \mathrm{~cm}\) (and with a resistance of \(61.9 \mu \Omega\) and a specific heat capacity of \(c=129 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C}\) ) on her finger moves her finger from a region having a magnetic field of \(0.0800 \mathrm{~T}\), pointing along her finger, to a region with zero magnetic field in \(40.0 \mathrm{~ms}\). As a result of this action, thermal energy is added to the band due to the induced current, which raises the temperature of the band. Calculate the temperature rise in the band, assuming all the energy produced is used in raising the temperature.
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