/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 Due to friction with the air, an... [FREE SOLUTION] | 91Ó°ÊÓ

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Due to friction with the air, an airplane has acquired a net charge of \(1.70 \times 10^{-5} \mathrm{C}\) The airplane travels with a speed of \(2.80 \times 10^{2} \mathrm{~m} / \mathrm{s}\) at an angle \(\theta\) with respect to the earth's magnetic field, the magnitude of which is \(5.00 \times 10^{-5} \mathrm{~T}\). The magnetic force on the airplane has a magnitude of \(2.30 \times 10^{-7} \mathrm{~N}\). Find the angle \(\theta\) (There are two possible angles.)

Short Answer

Expert verified
The possible angles are approximately \(75.11^\circ\) and \(104.89^\circ\).

Step by step solution

01

Identify the Equation for Magnetic Force

The magnetic force on a charged particle moving through a magnetic field is given by the equation \( F = qvB \sin(\theta) \), where \( F \) is the magnetic force, \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field, and \( \theta \) is the angle between the velocity and the magnetic field.
02

Rearrange for the Angle

We need to find the angle \( \theta \). To do this, rearrange the equation to solve for \( \sin(\theta) \): \( \sin(\theta) = \frac{F}{qvB} \).
03

Substitute Known Values

Substitute the given values into the equation: \( F = 2.30 \times 10^{-7} \, \text{N} \), \( q = 1.70 \times 10^{-5} \, \text{C} \), \( v = 2.80 \times 10^{2} \, \text{m/s} \), \( B = 5.00 \times 10^{-5} \, \text{T} \). Thus, \( \sin(\theta) = \frac{2.30 \times 10^{-7}}{(1.70 \times 10^{-5})(2.80 \times 10^{2})(5.00 \times 10^{-5})} \).
04

Calculate \( \sin(\theta) \)

Perform the calculation: \( \sin(\theta) = \frac{2.30 \times 10^{-7}}{2.38 \times 10^{-7}} \), yielding \( \sin(\theta) \approx 0.966 \).
05

Find Angles using Inverse Sine

Use the inverse sine function to find the angles. Thus, \( \theta_1 = \sin^{-1}(0.966) \approx 75.11^\circ \). Since sine is positive in the first and second quadrants, the other possible angle is \( \theta_2 = 180^\circ - 75.11^\circ \approx 104.89^\circ \).

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Key Concepts

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

charged particle motion
When considering the motion of a charged particle, it is essential to understand its behavior in the presence of various forces. In an environment with a magnetic field, this motion becomes particularly intriguing. A charged particle will move differently than if only gravitational or frictional forces were acting on it.
Unlike other forces, magnetic force acts perpendicular to the velocity of the charged particle, altering its path without changing its speed. For a plane like the one in our exercise, which has become charged due to friction, this means the airplane's motion is affected by Earth's magnetic field. This influences the plane's trajectory without slowing it down or speeding it up.
The key takeaway here is that the velocity, charge, and path of motion all interact with the magnetic field to generate a force that dictates the plane's behavior.
magnetic field interaction
Magnetic fields are invisible forces that exert influence on charged particles. The Earth's magnetic field is an example, subtly impacting anything with a charge moving through it.
When a charged particle like our airplane enters a magnetic field, it experiences a force that is perpendicular to both its velocity and the magnetic field lines. This scenario can be visualized as the airplane interacting with an unseen grid of magnetic lines, causing a deflection.
This interaction is quantified by the magnetic force equation: \[ F = qvB \sin{(\theta)} \] where:
  • \( F \) is the magnetic force,
  • \( q \) is the charged object's electric charge,
  • \( v \) is the velocity,
  • \( B \) represents the magnetic field strength,
  • and \( \theta \) is the angle between velocity and field lines.
This formula shows the importance of all these factors in predicting how strongly the airplane is affected.
trigonometry in physics
Trigonometry is a branch of mathematics dealing with triangles, and it plays a pivotal role in physics, especially in understanding forces and motions in different directions.
In the context of our problem, trigonometry helps determine the relationship between the magnetic force, the direction of the magnetic field, and the charged airplane's velocity. Specifically, the angle \( \theta \) between the velocity vector of the airplane and the magnetic field vector is crucial. This angle dictates the component of the velocity that effectively interacts with the magnetic field, which is captured using the sine function.
By rearranging the formula for magnetic force, we use trigonometric principles to isolate and calculate \( \sin{(\theta)} \). Then, through inverse trigonometric functions, we determine the potential angles the aircraft's velocity vector makes with the magnetic field.
sine function application
The sine function is a fundamental trigonometric function used to relate angles to their respective side ratios in right triangles. Its application extends beyond pure geometry into physics, where it helps determine variables like the angle of interaction between a charged particle's velocity and a magnetic field.
In our specific case, the sine function is used to isolate the angle \( \theta \) in the magnetic force equation. Because the sine of this angle determines how much of the velocity vector contributes to the magnetic force, it becomes essential in understanding the plane's trajectory.
Once we have the expression for \( \sin{(\theta)} \), we can employ the inverse sine function to find the exact angles. It's important to note that sine shows symmetry between two angles within 0 to 180 degrees, leading to two possible scenarios for \( \theta \). Using a calculator, we find the primary angle and its supplement, providing a complete picture of possible orientations of the plane's path with respect to the Earth's magnetic field.

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Most popular questions from this chapter

The electrons in the beam of a television tube have a kinetic energy of \(2.40 \times 10^{-15} \mathrm{~J}\). Initially, the electrons move horizontally from west to east. The vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of \(2.00 \times 10^{-5} \mathrm{~T}\). (a) In what direction are the electrons deflected by this field component? (b) What is the acceleration of an electron in part (a)?

What must be the radius of a circular loop of wire so the magnetic field at its center is \(1.8 \times 10^{-4} \mathrm{~T}\) when the loop carries a current of \(12 \mathrm{~A} ?\)

A particle that has an \(8.2-\mu C\) charge moves with a velocity of magnitude \(5.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\). When the velocity points along the \(+x\) axis, the particle experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge could experience has a magnitude of \(0.48 \mathrm{~N}\). Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field. Make sure that your answers are consistent with your answers to the Concept Questions.

A positively charged particle of mass \(7.2 \times 10^{-8} \mathrm{~kg}\) is traveling due east with a speed of \(85 \mathrm{~m} / \mathrm{s}\) and enters a 0.31 -T uniform magnetic field. The particle moves through onequarter of a circle in a time of \(2.2 \times 10^{-3} \mathrm{~s}\), at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.

At a certain location, the horizontal component of the earth's magnetic field is \(2.5 \times 10^{-5} \mathrm{~T}\), due north. A proton moves eastward with just the right speed, so the magnetic force on it balances its weight. Find the speed of the proton.

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