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Chapter 20: Problem 21

(III) A 3.40-g bullet moves with a speed of 155 m/s perpendicular to the Earth's magnetic field of 5.00 \(\times\) 10\(^{-5}\) T. If the bullet possesses a net charge of 18.5 \(\times\) 10\(^{-9}\) C, by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled 1.50 km?

Short Answer

Expert verified
The bullet is deflected approximately 1.98 micrometers from its path.

Step by step solution

01

Understand the Lorentz Force

The force experienced by a charged particle moving in a magnetic field is called the Lorentz force. It is given by the formula: \[ F = qvB \]where \( q \) is the charge, \( v \) is the velocity, and \( B \) is the magnetic field strength. This force will cause the bullet to experience a centripetal acceleration, altering its path.
02

Calculate the Lorentz Force

Substitute the known values into the Lorentz force equation to find the force:\[ F = (18.5 \times 10^{-9} \text{ C})(155 \text{ m/s})(5.00 \times 10^{-5} \text{ T}) \]Calculating this gives:\[ F = 1.44 \times 10^{-10} \text{ N} \]
03

Determine the Acceleration Due to the Force

Use Newton's second law, \( F = ma \), to compute the acceleration \( a \):\[ a = \frac{F}{m} = \frac{1.44 \times 10^{-10} \text{ N}}{3.40 \times 10^{-3} \text{ kg}} \]Calculate this to find:\[ a = 4.24 \times 10^{-8} \text{ m/s}^2 \]
04

Calculate the Time of Travel

Determine the time \( t \) it takes for the bullet to travel 1.50 km using the velocity:\[ t = \frac{d}{v} = \frac{1500 \text{ m}}{155 \text{ m/s}} \]This gives:\[ t = 9.68 \text{ s} \]
05

Determine the Deflection Distance

Use kinematics to calculate the deflection \( y \) experienced by the bullet due to the acceleration:\[ y = \frac{1}{2} a t^2 \]Substitute \( a = 4.24 \times 10^{-8} \text{ m/s}^2 \) and \( t = 9.68 \text{ s} \):\[ y = \frac{1}{2} (4.24 \times 10^{-8} \text{ m/s}^2 )(9.68 \text{ s})^2 \]Calculate this to find:\[ y \approx 1.98 \times 10^{-6} \text{ m} \] which is approximately 1.98 micro meters.

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Key Concepts

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

Magnetic Field
In physics, a magnetic field is a vector field that describes the magnetic influence on moving electric charges, electric currents, and magnetic materials. A charged particle moving through a magnetic field experiences a force perpendicular to both its velocity and the magnetic field direction.
This force influences the particle's motion, altering its path in a way unique to magnetic interactions.
  • The strength of a magnetic field is measured in teslas (T).
  • The magnetic field lines show the direction of the magnetic forces and are typically represented as going from the North to South pole.
  • When charged particles move across these lines, they feel a force that is perpendicular to both their velocity and the direction of the field.
This interaction between the charged bullet and the Earth's magnetic field is critical for understanding its change in path.
Charged Particle
A charged particle is an object with an electric charge, such as an electron or proton. In the context of the Lorentz force, a charged particle moving in a magnetic field is acted upon by a force proportional to the speed of the particle, the charge on the particle, and the strength of the magnetic field.
  • The charge (q) is often measured in coulombs (C), and it determines how strongly a particle will interact with electric and magnetic fields.
  • In this exercise, the 3.40-g bullet carries a net charge of 18.5\( \times 10^{-9} \) C, which allows the magnetic field to exert a force on it despite the lack of an overall electric field.
  • Due to its charge, this bullet experiences a continuous force as it moves through the Earth鈥檚 magnetic field, causing a change in its motion, observable as a deflection from its straight path.
The behavior of charged particles within magnetic fields forms the basis of many technologies, from accelerators to everyday electronic devices.
Centripetal Acceleration
Centripetal acceleration is the acceleration that occurs when an object moves in a circular path, always pointing towards the center of the circle. It arises due to the continuous change in the direction of the velocity of the object.
When a charged particle, such as the bullet in the exercise, moves perpendicularly across a magnetic field, this force causes it to undergo centripetal acceleration, altering its linear path into a curved trajectory.
  • The magnitude of centripetal acceleration ( a ) is determined by the equation \( a = \frac{v^2}{r} \), where v is the speed and r is the radius of the circular path.
  • In the problem, the field force acts perpendicular to the velocity, which does not change the speed of the bullet, but continuously changes its direction.
  • This concept is crucial to understanding how magnetic fields influence charged particles, making them spiral or bend rather than continue in straight lines.
Recognizing this motion explains why the bullet is deflected, producing a measurable distance from its original path.
Kinematics
Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion. It deals with variables like displacement, velocity, and time. In the context of this exercise, kinematics helps calculate the deflection of the bullet using its acceleration and travel time.
  • Displacement ( y ) due to acceleration can be calculated with the equation: \( y = \frac{1}{2}at^2 \), where a is the acceleration and t is the time.
  • In the given problem, the bullet travels a distance of 1.50 km and the velocity is 155 m/s, allowing calculation of travel time.
  • With known acceleration, determined from the Lorentz force equation, kinematics provides the final measure of how far off course the bullet strays.
Kinematics thus offers practical techniques to predict movement based on initial conditions and known forces acting on the object.

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Most popular questions from this chapter

(II) For a particle of mass \(m\) and charge \(q\) moving in a circular path in a magnetic field \(B\), (\(a\)) show that its kinetic energy is proportional to \(r^2\), the square of the radius of curvature of its path. (\(b\)) Show that its angular momentum is \(L = qBr^2\), around the center of the circle.

(I) A 240-m length of wire stretches between two towers and carries a 120-A current. Determine the magnitude of the force on the wire due to the Earth's magnetic field of 5.0 \(\times\) 10\(^{-5}\) T which makes an angle of 68\(^\circ\) with the wire.

Two long straight parallel wires are 15 cm apart. Wire A carries 2.0-A current. Wire B's current is 4.0 A in the same direction. (\(a\)) Determine the magnetic field magnitude due to wire A at the position of wire B. (\(b\)) Determine the magnetic field due to wire B at the position of wire A. (\(c\)) Are these two magnetic fields equal and opposite? Why or why not? (\(d\)) Determine the force on wire A due to wire B, and the force on wire B due to wire A. Are these two forces equal and opposite? Why or why not?

A uniform conducting rod of length \(\ell\) and mass \(m\) sits atop a fulcrum, which is placed a distance \(\ell/4\) from the rod's left-hand end and is immersed in a uniform magnetic field of magnitude \(B\) directed into the page (Fig. 20-69). An object whose mass \(M\) is 6.0 times greater than the rod's mass is hung from the rod's left-hand end.What current (direction and magnitude) should flow through the rod in order for it to be "balanced" (i.e., be at rest horizontally) on the fulcrum? (Flexible connecting wires which exert negligible force on the rod are not shown.)

(II) A long horizontal wire carries 24.0 A of current due north.What is the net magnetic field 20.0 cm due west of thewire if the Earth's field there points downward, 44\(^\circ\) below the horizontal, and has magnitude 5.0 \(\times\) 10\(^{-5}\) T?
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