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Chapter 19: Problem 6

A charge of \(0.250 \mathrm{C}\) moves vertically in a field of \(0.500 \mathrm{~T}\) that is oriented some angle from the vertical. If the charge's speed is \(2.0 \times 10^{2} \mathrm{~m} / \mathrm{s}\), what field angle(s) will ensure that the force acting on the charge is \(5.0 \mathrm{~N}\) ?

Short Answer

Expert verified
The angle(s) are approximately 11.54掳 and 168.46掳.

Step by step solution

01

Review of the Magnetic Force Formula

The formula for the magnetic force acting on a moving charge is given by \\[ F = qvB\sin\theta \] \where \( F \) is the magnetic force, \( q \) is the charge, \( v \) is the velocity of the charge, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity vector and the magnetic field vector.
02

Substitute in Known Values

Substitute the known values into the formula: \\[ 5.0 = 0.250 \times 200 \times 0.500 \times \sin\theta \] \This simplifies to: \\[ 5.0 = 25 \times \sin\theta \]
03

Solve for \( \sin\theta \)

Divide both sides of the equation by 25 to solve for \( \sin\theta \): \\[ \sin\theta = \frac{5.0}{25} = 0.2 \]
04

Find the Angle \( \theta \)

Now, use the inverse sine function to find \( \theta \): \\[ \theta = \sin^{-1}(0.2) \] \Calculating this gives \( \theta \approx 11.54^\circ \). \Since a sine value of 0.2 can correspond to angles in the first and second quadrants, the possible angles are \( \theta \approx 11.54^\circ \) and \( \theta \approx 180^\circ - 11.54^\circ = 168.46^\circ \).

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Key Concepts

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

Magnetic Field
A magnetic field is a region in space where a magnetic force can be felt by moving charges or magnetic materials. Its strength is measured in Tesla (T), which expresses how strong the magnetic influence is.
Magnetic fields are often depicted by field lines. These lines illustrate the direction and strength of the field. The closer the lines are to each other, the stronger the field in that region.
In the exercise given, the magnetic field strength is denoted as 0.500 T. This tells us how intense the magnetic field is around the moving charge, which in combination with the charge's velocity and direction, will determine the force applied on the charge.
Trigonometry in Physics
Trigonometry plays a crucial role in physics, especially when dealing with vectors like velocity and force. Since these vectors often don't line up perfectly, trigonometric functions help describe their interactions.
In the context of magnetic forces, the angle between the velocity of a charge and the magnetic field is vital. This angle, denoted as \( \theta \), is used in the equation \( F = qvB\sin\theta \).
This equation incorporates \( \sin\theta \), which determines how much of the force is actually aligned with the movement of the particle. If \( \theta \) is 0 or 180 degrees, \( \sin\theta \) becomes 0, giving no force, while at 90 degrees, the force is maximized.
  • Angles between 0 and 180 degrees can thus yield different force strengths, as seen with the given possible angles \( 11.54^\circ \) and \( 168.46^\circ \).
Charge Velocity
Velocity is a vector quantity, meaning it has both magnitude and direction. When dealing with charges in a magnetic field, the direction of movement is critical to determining how the magnetic field influences it.
The magnitude of the velocity is straightforward鈥攊n this case, 200 meters per second (m/s). This describes how fast the charge moves.
Combining this with the angle provided, we can determine how the direction of the charge's velocity vector impacts the force experienced. A charge moving directly with or against the field line experiences different forces compared to a charge moving at an angle.
Coulombs and Tesla
Coulombs (C) and Tesla (T) are units used to measure charge and magnetic field strength, respectively. Understanding these units is crucial in the context of magnetic force calculations.
The Coulomb measures how much electric charge is present. In our example, the charge is 0.250 C. A higher charge results in a stronger interaction with the magnetic field.
Tesla, on the other hand, conveys the intensity of a magnetic field. A field of 0.500 T indicates a moderate strength.
  • These values are plugged into the magnetic force formula and directly affect the force's magnitude. More Coulombs or a stronger Tesla value increases the resultant force.

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Most popular questions from this chapter

An ionized deuteron (a bound proton-neutron system with a net \(+e\) charge) passes through a velocity selector whose perpendicular magnetic and electric fields have magnitudes of \(40 \mathrm{mT}\) and \(8.0 \mathrm{kV} / \mathrm{m}\), respectively. Find the speed of the ion.

An electron travels at a speed of \(2.0 \times 10^{4} \mathrm{~m} / \mathrm{s}\) through a uniform magnetic field whose magnitude is \(1.2 \times 10^{-3} \mathrm{~T}\). What is the magnitude of the magnetic force on the electron if its velocity and the magnetic field (a) are perpendicular, (b) make an angle of \(45^{\circ},(\mathrm{c})\) are parallel, and (d) are exactly opposite?

A rectangular wire loop with a cross-sectional area of \(0.20 \mathrm{~m}^{2}\) carries a current of \(0.25 \mathrm{~A}\). The loop is free to rotate about an axis that is perpendicular to a uniform magnetic field with strength \(0.30 \mathrm{~T}\). The plane of the loop is at an angle of \(30^{\circ}\) to the direction of the magnetic field. (a) What is the magnitude of the torque on the loop? (b) How would you change the magnetic field to double the magnitude of the torque in part (a)? (c) How could you change only the current to double the torque in part (a)? (d) If you wanted to double the torque by changing only the loop area, what would the new area have to be? (e) Could you double the torque in part (a) by changing only the angle?

A beam of protons exits from a particle accelerator due east at a speed of \(3.0 \times 10^{5} \mathrm{~m} / \mathrm{s}\). The protons then enter a uniform magnetic field of magnitude \(0.50 \mathrm{~T}\) that is oriented at \(37^{\circ}\) above the horizontal relative to the beam direction. (a) What is the initial acceleration of a proton as it enters the field? (b) What if the magnetic field were angled at \(37^{\circ}\) below the horizontal instead? (c) If the beam were made of electrons traveling at the same speed rather than protons and the field were angled upward at \(37^{\circ},\) would there be any difference in the force on the electrons compared to the protons? Explain. (d) In part (c), what would be the ratio of the acceleration of an electron to that of a proton?

1 lies on the \(x\) -axis and its north end is at \(x=+1.0 \mathrm{~cm},\) while its south end is at鈥� # Two bar very narrow magnets are located in the \(x\) -y \(y\) plane. Magnet #1 lies on the \(x\) -axis and its north end is at \(x=+1.0 \mathrm{~cm},\) while its south end is at \(x=+5.0 \mathrm{~cm}\) Magnet \(\\# 2\) lies on the \(y\) -axis and its north end is at \(y=+1.0 \mathrm{~cm},\) while its south end is at \(y=+5.0 \mathrm{~cm}\) Magnet #2 produces a magnetic field that is only onehalf the magnitude of magnet #1. (a) In what direction would a compass point if it were located at the origin? (b) Repeat part (a) for the situation where magnet #1 is reversed in polarity.
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