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emf in a Bullet. At the equator, the earth's magnetic field is approximately horizontal, is directed toward the north, and has a value of \(8 \times 10^{-5} \mathrm{T}\) . (a) Estimate the emf induced between the top and bottom of a bullet shot horizontally at a target on the equator if the bullet is shot toward the east. Assume the bullet has a length of 1 \(\mathrm{cm}\) and a diameter of 0.4 \(\mathrm{cm}\) and is traveling at 300 \(\mathrm{m} / \mathrm{s} .\) Which is at higher potential: the top or bottom of the bullet? (b) What is the emf if the bullet travels south? (c) What is the emf induced between the front and back of the bullet for any horizontal velocity?

Short Answer

Expert verified
(a) EMF is 9.6 × 10^−5 V. (b) & (c) requires directions.

Step by step solution

01

Understand the Problem

We have a bullet traveling in Earth's magnetic field and need to find the emf induced under different conditions. We will use the formula for motional emf, which is given by \(\text{emf} = B \cdot v \cdot L\), where \(B\) is the magnetic field, \(v\) is the velocity of the bullet, and \(L\) is the length of the part of the bullet perpendicular to the velocity and magnetic field.
02

Calculate EMF for Eastward Motion (a)

For part (a), the bullet is traveling east at the equator. The magnetic field is horizontal and directed north. The emf induced is based on the length in the vertical direction, which is the bullet's diameter. Using \(B = 8 \times 10^{-5} \, \text{T}\), \(v = 300 \, \text{m/s}\), and \(L = 0.004 \, \text{m}\) (diameter in meters), the equation becomes:\[\text{emf} = B \cdot v \cdot L = 8 \times 10^{-5} \times 300 \times 0.004 = 9.6 \times 10^{-5} \, \text{V}\]

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Key Concepts

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

Magnetic Fields
Magnetic fields are regions around a magnet or electric current in which magnetic forces can be felt. These fields influence the movement of charged particles and can induce electromotive force (emf) under certain conditions. The Earth's magnetic field is a perfect real-world example. It is relatively weak but extensive, influencing countless natural processes and human-made devices.

At the equator, the Earth's magnetic field is nearly horizontal and directed from south to north. This specific orientation is crucial for problems involving motional emf, a type of electromagnetic induction, as it determines how the magnetic field interacts with moving objects, like bullets, in our scenario. Understanding this interaction forms the foundation for calculating induced emf.
Lorentz Force
The Lorentz force is a fundamental principle that determines how charged particles move within a magnetic and electric field. It helps in understanding the direction and magnitude of forces exerted on objects within magnetic fields.

The force acts perpendicular to both the velocity of the particle and the magnetic field direction, which we observe as the emf induced when the bullet travels across the Earth's magnetic field.
It can be calculated using the formula: \( F = q(abla \times B) × v\), where \( F \) is the force, \( q \) is the charge, \( v \) is velocity, and \( B \) is the magnetic field.

Lorentz Force is instrumental in practical applications of physics, including explaining how motional emf helps generate voltage in generators or affects radio waves in communication technologies.
Electromagnetic Induction
Electromagnetic induction is the creation of voltage (emf) in a conductor situated within a changing magnetic field. It is a vital concept for addressing how energy conversion works in many technologies, from motors to transformers.

In our case, the bullet shot horizontally cuts through the Earth's magnetic field lines, creating this induced voltage due to its motion. This phenomenon, explained by Faraday’s Law, gives us the foundational formula used in the exercise: \( \text{emf} = B \cdot v \cdot L\). Here, \( B \) is the magnetic field strength, \( v \) is the relative velocity, and \( L \) is the length of the object across the fields.

Understanding electromagnetic induction equips students with the knowledge to solve practical problems of voltage generation in various real-world scenarios.
Physics Problem Solving
Physics problem solving involves dissecting given conditions, applying relevant physical laws, and performing accurate calculations to find a solution. It requires both analytical thinking and a practical approach.

In tackling motional emf problems, you start by clearly understanding the physical setting and forces at play. The next step involves identifying the pertinent formulas – in this case, the motional emf formula – which then guides your calculations. It's crucial to recognize elements like directional movement and object dimensions that influence the final result.

Effective problem solvers use diagrams to visualize the scenario, organize known values, and methodically work through calculations, making sure units are consistent and logical. This systematic approach ensures a comprehensive grasp and the ability to apply it in various circumstances.

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Most popular questions from this chapter

A long, thin solenoid has 400 turns per meter and radius 1.10 \(\mathrm{cm} .\) The current in the solenoid is increasing at a uniform rate \(d i / d t .\) The induced electric field at a point near the center of the solenoid and 3.50 \(\mathrm{cm}\) from its axis is \(8.00 \times 10^{-6} \mathrm{V} / \mathrm{m}\) . Calculate di/dt.

A rectangle measuring 30.0 \(\mathrm{cm}\) by 40.0 \(\mathrm{cm}\) is located inside a region of a spatially uniform magnetic field of 1.25 \(\mathrm{T}\) , with the field perpendicular to the plane of the coil (Fig. E29.24). The coil is pulled out at a steady rate of 2.00 \(\mathrm{cm} / \mathrm{s}\) traveling perpendicular to the field lines. The region of the field ends abruptly as shown. Find the emf induced in this coil when it is (a) all inside the field; (b) partly inside the field; (c) all outside the field.

A closely wound rectangular coil of 80 turns has dimensions of 25.0 \(\mathrm{cm}\) by 40.0 \(\mathrm{cm}\) . The coll is rotated from a position where it makes an angle of \(37.0^{\circ}\) with a magnetic field of 1.10 \(\mathrm{T}\) to a position perpendicular to the field. The rotation takes 0.0600 s. What is the average emf induced in the coil?

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 12 \(\mathrm{cm}^{2}\) is rotated in 0.040 s from a position where its plane is perpendicular to the earth's magnetic field to a position where its plane is parallel to the field. The earth's magnetic field at the lab location is \(6.0 \times 10^{-5} \mathrm{T}\) . (a) What is the total magnetic flux through the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?

Terminal Speed. A conducting rod with length \(L\) mass \(m,\) and resistance \(R\) moves without friction on metal rails as shown in Fig. \(29.11 .\) A uniform magnetic field \(\vec{B}\) is directed into the plane of the figure. The rod starts from rest and is acted on by a constant force \(\vec{\boldsymbol{F}}\) directed to the right. The rails are infinitely long and have negligible resistance. (a) Graph the speed of the rod as a function of time. (b) Find an expression for the terminal speed (the speed when the acceleration of the rod is zero).

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