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(III) \(\mathrm{A} 3.40\) -g bullet moves with a speed of 155 \(\mathrm{m} / \mathrm{s}\) perpen- dicular to the Earth's magnetic field of \(5.00 \times 10^{-9} \mathrm{T}\) . If the bullet possesses a net charge of \(18.5 \times 10^{-9} \mathrm{C},\) by what distance will it be deflected from its path due to the Earth's magnetic field after it has traveled 1.00 \(\mathrm{km} ?\)

Short Answer

Expert verified
The bullet is deflected approximately \(8.78 \times 10^{-12}\) meters.

Step by step solution

01

Understanding the problem

A charged bullet moves through a magnetic field, which causes it to be deflected from its path. We need to calculate the distance of this deflection.
02

Identify the Relevant Formula

The magnetic force acting on a charge moving through a magnetic field is given by the formula: \[ F = qvB \sin(\theta) \]where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field direction. Since the bullet moves perpendicular to the magnetic field, \( \theta = 90^\circ \), so \( \sin(\theta) = 1 \). The formula simplifies to: \[ F = qvB \]
03

Calculate the Magnetic Force

Substitute the values into the formula for magnetic force: \[ F = (18.5 \times 10^{-9} \ \text{C}) \times (155 \ \text{m/s}) \times (5.00 \times 10^{-9} \ \text{T}) \]\[ F = 1.43375 \times 10^{-15} \ \text{N} \]
04

Determine the Mass in Kilograms

Convert the mass of the bullet from grams to kilograms: \[ m = 3.40 \ \text{g} = 0.00340 \ \text{kg} \]
05

Find the Acceleration due to Magnetic Force

Use Newton's second law \( F = ma \) to find the acceleration of the bullet:\[ a = \frac{F}{m} = \frac{1.43375 \times 10^{-15} \ \text{N}}{0.00340 \ \text{kg}} \]\[ a = 4.2166 \times 10^{-13} \ \text{m/s}^2 \]
06

Calculate the Deflection Distance

Use the kinematic equation for distance traveled under constant acceleration:\[ d = \frac{1}{2}at^2 \]First, find the time \( t \) it takes to travel 1 km:\[ t = \frac{1000 \ \text{m}}{155 \ \text{m/s}} \approx 6.4516 \ \text{s} \]Now calculate the deflection distance:\[ d = \frac{1}{2} \times 4.2166 \times 10^{-13} \ \text{m/s}^2 \times (6.4516 \ \text{s})^2 \]\[ d \approx 8.78 \times 10^{-12} \ \text{m} \]
07

Conclusion

The bullet's deflection from its original path after traveling 1.00 km due to Earth's magnetic field is approximately \( 8.78 \times 10^{-12} \) meters.

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Key Concepts

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

Lorentz force
When a charged particle, like a bullet with a net electric charge, moves through a magnetic field, it experiences a force known as the Lorentz force. This force is given by the formula:
  • \( F = qvB \sin(\theta) \)
where:
  • \( F \) is the magnetic force.
  • \( q \) is the charge of the particle.
  • \( v \) is the velocity of the particle.
  • \( B \) is the magnetic field strength.
  • \( \theta \) is the angle between the velocity and the direction of the magnetic field.
Since the bullet moves perpendicular to the Earth's magnetic field, the angle \( \theta \) is \( 90^\circ \), simplifying the equation to:
  • \( F = qvB \)
This makes the computation straightforward, assuming the charge, velocity, and magnetic field are known. The Lorentz force causes the charged particle to follow a curved path. The extent of this curvature depends on factors like the field strength and the particle's velocity.
Kinematic equations
The kinematic equations are used to predict the motion of objects when they are subjected to forces. In this case, once the Lorentz force is calculated, the next step is to find the resulting acceleration. Using Newton's second law:
  • \( F = ma \)
The acceleration \( a \) is:
  • \( a = \frac{F}{m} \)
where \( m \) is the mass of the bullet. With the acceleration known, kinematic equations help in finding the deflection. The kinematic equation for distance traveled under constant acceleration is:
  • \( d = \frac{1}{2}at^2 \)
Here, \( d \) is the deflection distance, \( a \) is the acceleration due to the magnetic force, and \( t \) is the time taken to travel a particular distance, like 1 km in this scenario. By understanding how acceleration due to an external force changes an object's path, these equations allow us to solve for the resulting movement.
Deflection due to magnetic field
When a charged particle such as a bullet travels through the Earth's magnetic field, it does not follow a straight-line path but instead gets slightly deflected. This deflection occurs because of the Lorentz force acting perpendicular to its motion. The path becomes slightly curved due to this sideway force. To determine how much the bullet deflects, it's important to recognize that the deflection distance depends largely on:
  • The magnetic force acting on the bullet.
  • The time the bullet spends in the magnetic field.
  • The bullet's initial velocity which dictates how quickly it covers ground.
All of these aspects can be calculated using the given physics equations. By combining the Lorentz force understanding with kinematic principles, the deflection over a given distance can be effectively predicted. Practically, this helps us anticipate how charged particles deviate from their path in magnetic fields, a principle applied in various fields like physics experiments and electronic devices.

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Most popular questions from this chapter

If the restoring spring of a galvanometer weakens by \(15 \%\) over the years, what current will give full-scale deflection if it originally required \(46 \mu \mathrm{A} ?\)

(II) One form of mass spectrometer accelerates ions by a voltage \(V\) before they enter a magnetic field \(B\) . The ions are assumed to start from rest. Show that the mass of an ion is \(m=q B^{2} R^{2} / 2 V,\) where \(R\) is the radius of the ions' path in the magnetic field and \(q\) is their charge.

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A proton moves through a region of space where there is a magnetic field \(\overrightarrow{\mathbf{B}}=(0.45 \hat{\mathbf{i}}+0.38 \hat{\mathbf{j}}) \mathrm{T}\) and an electric field \(\overrightarrow{\mathbf{E}}=(3.0 \hat{\mathbf{i}}-4.2 \hat{\mathbf{j}}) \times 10^{3} \mathrm{~V} / \mathrm{m}\). At a given instant, the proton's velocity is \(\overrightarrow{\mathbf{v}}=(6.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{j}}-5.0 \hat{\mathbf{k}}) \times 10^{3} \mathrm{~m} / \mathrm{s}\). Determine the components of the total force on the proton.

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