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Chapter 19: Problem 41

A horizontal metal bar of length \(L\) falls vertically through a horizontal magnetic field of strength \(B\). (a) Show that if the bar falls with speed \(v,\) the induced emf between its ends is \(\mathcal{E}=B L v\) (b) Suppose \(B=0.25 \mathrm{~T}\) and \(L=0.50 \mathrm{~m} .\) What's the induced emf \(1.0 \mathrm{~s}\) after the bar is dropped from rest?

Short Answer

Expert verified
The induced emf after 1 second is approximately 1.23 V.

Step by step solution

01

Understanding the formula for induced emf

To find the induced electromotive force (emf), we utilize Faraday's Law in a moving conductor. According to Faraday's Law, the induced emf (\( \mathcal{E} \)) in a conductor moving in a magnetic field is given by the formula \( \mathcal{E} = B \cdot L \cdot v \), where \( B \) is the magnetic field strength, \( L \) is the length of the conductor, and \( v \) is the speed at which the conductor is moving perpendicular to the field.
02

Derivation using Faraday's Law

When the metal bar falls vertically through a horizontal magnetic field, it experiences a change in magnetic flux. Since the bar is moving perpendicular to the field, the velocity \( v \) is included in the calculation of emf. Thus, the induced emf is directly given by the expression \( \mathcal{E} = B \cdot L \cdot v \). This formula is derived from the basic principle of electromagnetic induction, focusing on the rate of change of magnetic flux through the loop formed by the falling bar.
03

Calculating motion of falling bar

When the bar is released from rest, it accelerates due to gravity. After \( t = 1.0 \, \text{s} \), its velocity \( v \) can be calculated using the formula for uniformly accelerated motion: \( v = g \cdot t \), where \( g \approx 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity. Thus, after \( 1.0 \, \text{s} \), the velocity is \( v = 9.81 \, \text{m/s}^2 \times 1.0 \, \text{s} = 9.81 \, \text{m/s} \).
04

Calculate the induced emf after 1 second

Now, substitute the values into the emf formula: \( \mathcal{E} = B \cdot L \cdot v \). With \( B = 0.25 \, \text{T} \), \( L = 0.50 \, \text{m} \), and \( v = 9.81 \, \text{m/s} \), calculate \( \mathcal{E} = 0.25 \, \text{T} \times 0.50 \, \text{m} \times 9.81 \, \text{m/s} = 1.22625 \, \text{V} \). Therefore, the induced emf is approximately \( 1.23 \, \text{V} \).

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Key Concepts

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

Induced EMF
When a conductor such as a metal bar moves through a magnetic field, it experiences a change in magnetic flux. This results in the induction of an electromotive force, commonly known as emf, within the conductor. The idea of induced emf is at the heart of Faraday's Law of Induction. The formula
  • \( \mathcal{E} = B \cdot L \cdot v \)
helps us quantify this electrical output, where:
  • \(B\) is the magnetic field strength in Tesla,
  • \(L\) is the length of the conductor in meters,
  • \(v\) is the velocity of the conductor in meters per second as it moves perpendicularly through the magnetic field.
Thus, the induced emf is directly proportional to the product of these three quantities. The more dynamically the conductor intersects the magnetic field lines, the greater the induced emf.
Electromagnetic Induction
Electromagnetic induction refers to the process by which a voltage or emf is generated in a conductor due to its interaction with a changing magnetic field. It's a principle also discovered by Michael Faraday and based on his laws of induction. Essentially, when the metal bar falls through the magnetic field:
  • The movement translates into a change in magnetic flux through the bar's loop.
  • Induction is the result of the conductor cutting across magnetic field lines, creating current within it.
  • The relationship is linear, as showcased by the formula for induced emf.
This principle is foundational to many electrical systems, including generators and transformers, which exploit induction for the conversion between mechanical and electrical energy.
Magnetic Field Interaction
Magnetic field interaction involves the complex interplay between moving conductive materials and magnetic field lines. In the exercise, a horizontal magnetic field interacts with a vertical falling bar. This interaction is what causes the induction of emf in the conductor. Key aspects of magnetic field interaction:
  • The field exerts a force on the electrons within the metal bar.
  • It splits charge distribution, causing one end of the bar to become more positive and the other more negative.
  • This charge separation induces an electric potential difference, or emf, across the bar.
Understanding how magnetic fields influence and interact with conductive bodies is crucial for designing systems that involve induction, like electric motors and sensors.
Uniformly Accelerated Motion
Uniformly accelerated motion describes the movement of an object experiencing constant acceleration, such as gravity. In the context of the exercise, this comes into play because the bar accelerates under the influence of Earth's gravitational force.Under this constant acceleration:
  • The velocity of the bar increases linearly over time, calculated using the formula \( v = g \cdot t \), where \( g \approx 9.81 \, \text{m/s}^2 \) represents gravitational acceleration.
  • Applying this motion formula, we find that after one second, the bar reaches a velocity of approximately \( 9.81 \, \text{m/s} \).
  • This velocity subsequently influences the calculation of induced emf, revealing a direct connection between mechanical motion and electrical phenomena.
Thus, understanding uniformly accelerated motion is crucial for predicting how quickly the bar will move鈥攁nd consequently, the magnitude of the induced emf.

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Most popular questions from this chapter

A resistor connected across an AC power supply has a current given by \(I=(1.20 \mathrm{~A}) \cos (300 t)\) when connected to a power supply with emf \(100 \mathrm{~V} \mathrm{rms}\). Find (a) the rms current, (b) the resistance, and (c) the average power delivered to the resistor.

A 50 -turn circular coil has diameter \(6.2 \mathrm{~cm}\) and resistance \(0.75 \Omega\). A magnetic field perpendicular to the coil is changing at \(0.50 \mathrm{~T} / \mathrm{s} .\) The induced current in the coil is (a) \(10 \mathrm{~mA} ;\) (b) \(30 \mathrm{~mA}\) (c) \(100 \mathrm{~mA} ;\) (d) \(300 \mathrm{~mA}\).

An AC power supply operates with peak emf \(340 \mathrm{~V}\) and frequency \(120 \mathrm{~Hz} .\) A \(3.50-\mathrm{k} \Omega\) resistor is connected across the supply. Find (a) the rms current and (b) the average power dissipated in the resistor. (c) Graph the potential difference across and current through the resistor, both as functions of time.

In \(48-\mathrm{V}\) battery is in series with a switch, a \(0.50-\mathrm{H}\) inductor, and a \(10-\Omega\) resistor. (a) What's the maximum current in this circuit? (b) What's the current at \(t=0.10 \mathrm{~s}\) after the switch is closed? (c) When does the current reach half the maximum value? (d) What's the current at time \(t=2 \tau ?\)

A \(240-\mathrm{V}\) (rms), 50 -Hz power supply is connected to a \(4700-\mu \mathrm{F}\) capacitor. Find (a) the capacitor's reactance and (b) the peak current. (c) What's the peak current if the power supply's frequency is changed to \(60 \mathrm{~Hz} ?\)
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