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Chapter 21: Problem 79

A charge is moving perpendicular to a magnetic field and experiences a force whose magnitude is \(2.7 \times 10^{-3} \mathrm{N}\). If this same charge were to move at the same speed and the angle between its velocity and the same magnetic field were \(38^{\circ},\) what would be the magnitude of the magnetic force that the charge would experience?

Short Answer

Expert verified
1.6634 x 10^{-3} N.

Step by step solution

01

Understand the problem

We have a charge moving in a magnetic field, initially perpendicular to it, experiencing a force \(2.7 \times 10^{-3} \mathrm{N}\). We need to find the force when the angle between its velocity and the field changes to \(38^{\circ}\).
02

Recall the formula for magnetic force

The formula for the magnetic force experienced by a moving charge in a magnetic field is given by \( F = qvB\sin(\theta) \), 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.
03

Use given data for initial condition

Initially, \( \theta = 90^{\circ} \), so \(\sin(90^{\circ}) = 1\). Thus, the force is \( F = qvB \). We have \( F = 2.7 \times 10^{-3} \mathrm{N} \), which lets us express \( qvB = 2.7 \times 10^{-3} \) since \( F = qvB\).
04

Calculate the force for angle 38 degrees

Now with \( \theta = 38^{\circ} \), substitute into the formula: \( F = qvB\sin(38^{\circ}) \). Since \( qvB = 2.7 \times 10^{-3} \), we have \( F = 2.7 \times 10^{-3} \times \sin(38^{\circ}) \).
05

Evaluate the sine and solve

Calculate \( \sin(38^{\circ}) \approx 0.6157 \). Therefore, the force is \( F = 2.7 \times 10^{-3} \times 0.6157 = 1.6634 \times 10^{-3} ~\mathrm{N} \).

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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 an invisible force that surrounds magnets and electric currents. It affects how charged particles move. When a charged particle, like an electron, moves through a magnetic field, it experiences a force. This force can change the direction of the particle's motion but not its speed. The magnetic field is measured in teslas (T) and its strength affects how much force the charge experiences. In our example, the magnetic field's interaction with a moving charge is key to understanding the overall behavior of the force.
Velocity of Charge
The velocity of a charge is simply how fast the charge is moving and in what direction. When dealing with magnetic forces, both speed and direction matter. If a charged particle moves in a magnetic field, the force it experiences depends on its velocity. The faster it moves, the greater the potential force. Also, the path of movement relative to the magnetic field direction will determine how this force is applied. In the initial problem, the charge moves perpendicular to the field, leading to the first calculated force.
Angle Between Velocity and Magnetic Field
The angle between the velocity of a charge and the magnetic field direction is crucial for calculating the force the charge experiences. When the angle is 90 degrees, or perpendicular, the sine function (more on that soon) reaches its maximum value, leading to the maximum possible force. However, when the angle changes, the magnitude of the force adjusts accordingly. In our scenario, we shift the angle to 38 degrees, which decreases the force due to the mathematical nature of the sine function.
Sine Function in Physics
The sine function is a trigonometric function often used in physics to determine components of forces and other vectors. It is especially important in magnetic force equations. Here鈥檚 why: the force on a moving charge in a magnetic field depends on the sine of the angle between the charge's velocity and the field. With a sine function, \( \sin(\theta) \), when \( \theta = 90^{\circ} \), \( \sin(90^{\circ}) = 1\). This means full force is applied. However, at \( 38^{\circ} \), \( \sin(38^{\circ}) \approx 0.6157 \). This tells us that only about 61.57% of the maximum force is at play.

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Most popular questions from this chapter

In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5} \mathrm{T} .\) An electron is shot vertically straight up from the ground with a speed of \(2.1 \times 10^{6} \mathrm{m} / \mathrm{s} .\) What is the magnitude of the acceleration caused by the magnetic force? Ignore the gravitational force acting on the electron.

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 \(I ?\)

An electron is moving through a magnetic field whose magnitude is \(8.70 \times 10^{-4} \mathrm{T}\). The electron experiences only a magnetic force and has an acceleration of magnitude \(3.50 \times 10^{14} \mathrm{m} / \mathrm{s}^{2} .\) At a certain instant, it has a speed of \(6.80 \times 10^{6} \mathrm{m} / \mathrm{s}\). Determine the angle \(\theta\) (less than \(90^{\circ}\) ) between the electron's velocity and the magnetic field.

A charge of \(-8.3 \mu \mathrm{C}\) is traveling at a speed of \(7.4 \times 10^{6} \mathrm{m} / \mathrm{s}\) in a region of space where there is a magnetic field. The angle between the velocity of the charge and the field is \(52^{\circ} .\) A force of magnitude \(5.4 \times 10^{-3} \mathrm{N}\) acts on the charge. What is the magnitude of the magnetic field?

Two parallel rods are each \(0.50 \mathrm{m}\) in length. They are attached at their centers to either end of a spring (spring constant \(=150 \mathrm{N} / \mathrm{m}\) ) that is initially neither stretched nor compressed. When 950 A of current is in each rod in the same direction, the spring is observed to be compressed by \(2.0 \mathrm{cm} .\) Treat the rods as long, straight wires and find the separation between them when the current is present.
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