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Chapter 21: Problem 72

In a certain region, the earth's magnetic field has a magnitude of \(5.4 \times 10^{-5} \mathrm{T}\) and is directed north at an angle of \(58^{\circ}\) below the horizontal. An electrically charged bullet is fired north and \(11^{\circ}\) above the horizontal, with a speed of \(670 \mathrm{m} / \mathrm{s}\). The magnetic force on the bullet is \(2.8 \times 10^{-10} \mathrm{N},\) directed due east. Determine the bullet's electric charge, including its algebraic sign \((+\) or \(-\) ).

Short Answer

Expert verified
The bullet's charge is \(+6.34 \times 10^{-10} \mathrm{C}\).

Step by step solution

01

Determine the Magnetic Field Components

First, resolve the earth's magnetic field into its horizontal \((B_{h})\) and vertical \((B_{v})\) components. Given: Magnitude of the magnetic field \(B = 5.4 \times 10^{-5} \mathrm{T}\), and the angle \(\theta = 58^{\circ}\) below the horizontal. The horizontal component is: \[B_{h} = B \cdot \cos(\theta) = 5.4 \times 10^{-5} \cdot \cos(58^{\circ})\]Calculate the horizontal component:\(B_{h} \approx 2.86 \times 10^{-5} \mathrm{T}\). The vertical component is:\[B_{v} = B \cdot \sin(\theta) = 5.4 \times 10^{-5} \cdot \sin(58^{\circ})\]Calculate the vertical component:\(B_{v} \approx 4.57 \times 10^{-5} \mathrm{T}\).
02

Determine Bullet's Velocity Direction Components

Resolve the bullet's velocity into its horizontal \((v_h)\) and vertical \((v_v)\) components. Given:Bullet's speed \(v = 670 \mathrm{m/s}\), and angle \(\phi = 11^{\circ}\) above the horizontal. The horizontal velocity is:\[v_h = v \cdot \cos(\phi) = 670 \cdot \cos(11^{\circ})\]Calculate the horizontal velocity:\(v_h \approx 658.12 \mathrm{m/s}\). The vertical velocity is:\[v_v = v \cdot \sin(\phi) = 670 \cdot \sin(11^{\circ})\]Calculate the vertical velocity:\(v_v \approx 127.78 \mathrm{m/s}\).
03

Calculate the Magnetic Force Using Magnitude and Direction

The magnetic force \(F\) on a moving charge can be given by:\[F = qvB\sin(\theta)\]Where:- \(F = 2.8 \times 10^{-10} \mathrm{N}\)- \(B\) is the horizontal component since magnetic force is due east, \(B = 2.86 \times 10^{-5} \mathrm{T}\)- \(v\) is the vertical component since the force direction is perpendicular, \(v = 127.78 \mathrm{m/s}\)- \(\theta = 90^{\circ}\) as the force is maximized perpendicular.Rearrange and solve for \(q\):\[q = \frac{F}{vB}\sin(\theta) = \frac{2.8 \times 10^{-10}}{127.78 \times 2.86 \times 10^{-5}}\sin(90^{\circ})\]Calculate charge \(q\):\(q \approx 6.34 \times 10^{-10} \mathrm{C}\).
04

Determine the Algebraic Sign of the Charge

Since the magnetic force is directed east and the initial velocity is northward, we use the right-hand rule, which predicts the force direction, to find that the force is consistent with a positive charge. Thus, the charge of the bullet is \(+6.34 \times 10^{-10} \mathrm{C}\).

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Key Concepts

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

Magnetic Field Components
Understanding how to resolve a magnetic field into its components is crucial when examining how it affects charged particles. The Earth's magnetic field, in this exercise, is given with a magnitude and an angle relative to the horizon. We can break it down into two components:
  • **Horizontal Component (\(B_h\)**): This part lies in the plane parallel to the Earth's surface. It's calculated using the cosine of the angle with the formula \(B_h = B \cdot \cos(\theta)\).

  • **Vertical Component (\(B_v\)**): This is the part that extends perpendicular to the horizontal plane, calculated using the sine of the angle with the formula \(B_v = B \cdot \sin(\theta)\).

For example, with \(B = 5.4 \times 10^{-5} \mathrm{T}\) and \(\theta = 58^{\circ}\), the horizontal and vertical components would be \(B_h \approx 2.86 \times 10^{-5} \mathrm{T}\) and \(B_v \approx 4.57 \times 10^{-5} \mathrm{T}\), respectively. These components allow us to understand how different aspects of the field interact with a moving charge.
Charged Particle Motion
When a charged particle moves through a magnetic field, its path can change depending on several factors. In this particular problem, we're analyzing a bullet as it travels through the Earth's magnetic field.

The bullet has an initial velocity with both vertical and horizontal components. The magnitude and direction of these components determine how the magnetic field influences its motion.
  • The horizontal velocity component is given by \(v_h = v \cdot \cos(\phi)\), and for this bullet, it's calculated as \(v_h \approx 658.12 \mathrm{m/s}\).

  • The vertical velocity component is \(v_v = v \cdot \sin(\phi)\), resulting in \(v_v \approx 127.78 \mathrm{m/s}\).
By understanding these components, we can predict how the perpendicular magnetic forces will influence the motion of the bullet as it travels north at an angle from the horizontal.
Right-Hand Rule
The right-hand rule is a fundamental technique used to determine the direction of the magnetic force acting on a moving charged particle. In this context, the rule helps establish the relationship between the velocity of the charge, the magnetic field direction, and the resulting force.

Here's how it works:
  • Point your thumb in the direction of the velocity of the charged particle (in this case, northward at an angle).

  • Extend your fingers in the direction of the magnetic field (in this case, a horizontal component directed north but tilted below the horizontal).

  • The force acting on a positive charge will be directed in the direction your palm pushes (in this problem, directed due east).
The right-hand rule shows that since the magnetic force is directed east, the charge must be positive to align with this predicted force direction. Thus, it confirms the bullet's charge is \(+6.34 \times 10^{-10} \mathrm{C}\).
Magnetic Force Calculation
To find the magnetic force acting on a charged particle moving through a field, we use the formula \(F = qvB\sin(\theta)\). Here:
  • \(F\) is the magnetic force, calculated as \(2.8 \times 10^{-10} \mathrm{N}\).

  • \(q\) represents the particle's charge, which we need to determine.

  • \(v\) is the component of velocity perpendicular to the magnetic field. In this case, it's the vertical velocity \(127.78 \mathrm{m/s}\).

  • \(B\) is the magnetic field's horizontal component \(2.86 \times 10^{-5} \mathrm{T}\).

  • \(\theta = 90^{\circ}\) since the force is perpendicular, making \(\sin(90^{\circ}) = 1\).
We then rearrange the formula to solve for \(q\): \[ q = \frac{F}{vB\sin(\theta)} \]Substituting the known values yields the charge: \(q \approx 6.34 \times 10^{-10} \mathrm{C}\), ensuring we've accounted for the direction and magnitude of the force just observed.

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Most popular questions from this chapter

A long solenoid has a length of \(0.65 \mathrm{m}\) and contains 1400 turns of wire. There is a current of \(4.7 \mathrm{A}\) in the wire. What is the magnitude of the magnetic field within the solenoid?

Two insulated wires, each \(2.40 \mathrm{m}\) long, are taped together to form a two-wire unit that is \(2.40 \mathrm{m}\) long. One wire carries a current of \(7.00 \mathrm{A}\); the other carries a smaller current \(I\) in the opposite direction. The twowire unit is placed at an angle of \(65.0^{\circ}\) relative to a magnetic field whose magnitude is \(0.360 \mathrm{T}\). The magnitude of the net magnetic force experienced by the two-wire unit is \(3.13 \mathrm{N}\). What is the current \(I ?\)

A wire has a length of \(7.00 \times 10^{-2} \mathrm{m}\) and is used to make a circular coil of one turn. There is a current of \(4.30 \mathrm{A}\) in the wire. In the presence of a \(2.50-\mathrm{T}\) magnetic field, what is the maximum torque that this coil can experience?

Suppose that a uniform magnetic field is everywhere perpendicular to this page. The field points directly upward toward you. A circular path is drawn on the page. Use Amp猫re's law to show that there can be no net current passing through the circular surface.

An ionized helium atom has a mass of \(6.6 \times 10^{-27} \mathrm{kg}\) and a speed of \(4.4 \times 10^{5} \mathrm{m} / \mathrm{s} .\) It moves perpendicular to a \(0.75-\mathrm{T}\) magnetic field on a circular path that has a 0.012-m radius. Determine whether the charge of the ionized atom is \(+e\) or \(+2 e\).
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