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ssm In New England, the horizontal component of the earth's magnetic field has a magnitude of \(1.6 \times 10^{-5}\) 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.

Short Answer

Expert verified
The acceleration is approximately \( 5.9 \times 10^{10} \text{ m/s}^2 \).

Step by step solution

01

Identify the Known Values

The horizontal component of the Earth's magnetic field \( B = 1.6 \times 10^{-5} \text{ T} \) and the speed of the electron \( v = 2.1 \times 10^{6} \text{ m/s} \) are given. Also, the charge of an electron \( q = -1.6 \times 10^{-19} \text{ C} \) and the mass of an electron \( m = 9.11 \times 10^{-31} \text{ kg} \).
02

Calculate the Magnetic Force

The magnetic force on a charged particle moving in a magnetic field is calculated using the equation: \( F = qvB \sin\theta \). Here, \( \theta = 90^\circ \) because the electron is moving perpendicular to the magnetic field, so \( \sin\theta = 1 \). Thus, \( F = qvB \). Substitute the known values: \( F = (1.6 \times 10^{-19} \text{ C})(2.1 \times 10^{6} \text{ m/s})(1.6 \times 10^{-5} \text{ T}) \).
03

Compute the Acceleration

Using Newton's second law \( F = ma \), solve for the acceleration of the electron: \( a = \frac{F}{m} \). Substitute the expression for \( F \) from the previous step and the mass of the electron: \( a = \frac{(1.6 \times 10^{-19} \text{ C})(2.1 \times 10^{6} \text{ m/s})(1.6 \times 10^{-5} \text{ T})}{9.11 \times 10^{-31} \text{ kg}} \).
04

Perform the Calculation

After substituting the values, compute the numerical result to find \( a \). It involves multiplying the charge, velocity, and magnetic field together, then dividing by the mass of the electron. Calculate: \( a \approx 5.9 \times 10^{10} \text{ m/s}^2 \).

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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 around a magnetic material or a moving electric charge within which the force of magnetism acts. It is measured in teslas (T), and in this exercise, we're focusing on the horizontal component of the Earth's magnetic field. This component is given as a tiny force, specifically \(1.6 \times 10^{-5}\, \text{T}\), yet it can exert significant forces on moving charged particles like electrons.
The force experienced by a charged particle in a magnetic field depends on several factors: the strength of the magnetic field, the charge of the particle, its velocity, and the angle at which it meets the field. Importantly, the force is maximized when the motion of the particle is perpendicular to the magnetic field, as is the case in this exercise.
Understanding how magnetic fields interact with charged particles is crucial in fields ranging from physics to engineering. It explains phenomena seen in technologies such as cathode ray tubes and the behavior of particles in space.
Electron velocity
The velocity of an electron determines how it interacts with a magnetic field. In this scenario, the electron has an initial velocity of \(2.1 \times 10^6 \text{ m/s}\), directed vertically upwards. This high speed is crucial because the magnetic force exerted on the electron is directly proportional to its velocity.
When a charged particle moves through a magnetic field, its path can curve depending on the field's direction and the speed of the particle. This happens because the magnetic force acts perpendicular to both the velocity and the magnetic field.
The high velocity in this exercise means that the interaction between the electron and the magnetic field is robust. Understanding velocity's role in magnetic forces brings insight into designing technologies like particle accelerators and understanding how solar winds interact with the Earth's magnetic field.
Acceleration calculation
Calculating the acceleration of an electron under a magnetic force involves applying known physics principles and equations. The initial step is to determine the magnetic force acting on the electron. This force \( F \) is calculated using the formula: \[ F = qvB \sin \theta \] where:
  • \( q \) is the charge of the electron (\(-1.6 \times 10^{-19} \text{ C}\))
  • \( v \) is the velocity of the electron (\(2.1 \times 10^6 \text{ m/s}\))
  • \( B \) is the magnetic field (\(1.6 \times 10^{-5} \text{ T}\))
  • \( \theta \) is the angle between the velocity and the magnetic field, which is \(90^\circ\), making \(\sin\theta = 1\)
This results in the necessary force to further calculate acceleration.
Next, using Newton's second law, the acceleration \( a \) can be determined by rearranging the formula \( F = ma \) to \( a = \frac{F}{m} \). Substituting the mass of the electron \(9.11 \times 10^{-31} \text{ kg}\), we find that the acceleration is approximately \(5.9 \times 10^{10} \text{ m/s}^2\). This precise calculation reveals the remarkable effects even a tiny magnetic field can have on subatomic particles.
Newton's second law
Newton's second law of motion is fundamental in physics, stating that the force acting on an object is equal to the mass of that object multiplied by its acceleration: \[ F = ma \] This principle acts as a cornerstone for understanding how forces affect motion. In this exercise, it's essential for transforming a force value into something more meaningful: the electron's acceleration.
Once the magnetic force on the electron is calculated, applying Newton's second law allows us to find the acceleration by dividing the force by the electron's mass. The concept highlights the proportionality between force and acceleration and inversely with mass, explaining why less massive objects like electrons can have significant acceleration under relatively small forces.
This principle not only assists in academic exercises but also provides insight into dynamics in everyday life, from the motion of planets to the functioning of engines.

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Most popular questions from this chapter

Multiple-Concept Example 7 discusses how problems like this one can be solved. A \(+6.00 \mu \mathrm{C}\) charge is moving with a speed of \(7.50 \times 10^{4} \mathrm{m} / \mathrm{s}\) parallel to a very long, straight wire. The wire is 5.00 \(\mathrm{cm}\) from the charge and carries a current of 67.0 \(\mathrm{A}\) in a direction opposite to that of the moving charge. Find the magnitude and direction of the force on the charge.

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 magnitude of the magnetic force that the charge would experience?

You have a wire of length \(L=1.00 \mathrm{m}\) from which to make the square coil of a dc motor. The current in the coil is \(I=1.7 \mathrm{A},\) and the magnetic field of the motor has a magnitude of \(B=0.34 \mathrm{T}\) . Find the maximum torque exerted on the coil when the wire is used to make a single-turn square coil and a two-turn square coil.

ssm When beryllium-7 ions \(\left(m=11.65 \times 10^{-27} \mathrm{kg}\right)\) pass through a mass spectrometer, a uniform magnetic field of 0.283 \(\mathrm{T}\) curves their path directly to the center of the detector (see Figure 21.14\() .\) For the same accelerating potential difference, what magnetic field should be used to send beryllium-10 ions \(\left(m=16.63 \times 10^{-27} \mathrm{kg}\right)\) to the same location in the detector? Both types of ions are singly ionized \((q=+e)\) .

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 \(0.31-\mathrm{T}\) uniform magnetic field. The particle moves through one- quarter 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.

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