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Chapter 20: Problem 7

\(\bullet \mathrm{A}^{9}\) Be nucleus containing four protons and five neutrons has a mass of \(1.50 \times 10^{-26} \mathrm{kg}\) and is traveling vertically upward at 1.35 \(\mathrm{km} / \mathrm{s} .\) If this particle suddenly enters a horizontal magnetic field of 1.12 T pointing from west to east, find the magnitude and direction of its acceleration vector the instant after it enters the field.

Short Answer

Expert verified
The acceleration is \( 6.42 \times 10^{10} \) m/s虏 directed north.

Step by step solution

01

Identify Given Values

Start by identifying the values provided in the question. The beryllium-9 nucleus has a mass \( m = 1.50 \times 10^{-26} \) kg, it has a speed \( v = 1.35 \) km/s = \( 1350 \) m/s, and it enters a magnetic field \( B = 1.12 \) T. Also, we know it has 4 protons, so the elementary charge \( q \) of a proton is \( 1.60 \times 10^{-19} \) C.
02

Calculate Charge of Nucleus

The charge of the nucleus is equal to the charge of 4 protons since it has 4 protons. So, \( q = 4 \times 1.60 \times 10^{-19} \) C. This gives \( q = 6.40 \times 10^{-19} \) C.
03

Apply Lorentz Force Equation

The Lorentz force acting on a charged particle moving in a magnetic field is given by \( F = qvB \sin \theta \), where \( \theta \) is the angle between velocity and the magnetic field. Here, \( \theta = 90^{\circ} \), thus \( \sin \theta = 1 \). Therefore, \( F = qvB \). Substitute the values: \( F = (6.40 \times 10^{-19})(1350)(1.12) \).
04

Calculate the Force

Calculate the force using the values from the previous step: \( F = 9.63 \times 10^{-16} \) N.
05

Determine the Acceleration

To find the acceleration, use Newton's second law, \( F = ma \). Rearrange this to find acceleration: \( a = \frac{F}{m} \). Substitute the known values: \( a = \frac{9.63 \times 10^{-16}}{1.50 \times 10^{-26}} \).
06

Calculate the Acceleration

Compute the acceleration: \( a = 6.42 \times 10^{10} \) m/s虏.
07

Determine the Direction of Acceleration

Use the right-hand rule to determine the direction: Point your thumb upward in the direction of velocity (upwards), and extend your fingers in the direction of the magnetic field (west to east), the force (and hence the acceleration) is directed towards the palm which is towards the north.

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Key Concepts

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

Lorentz Force
The Lorentz force is a fundamental concept in electromagnetism. It describes the force exerted on a charged particle moving through a magnetic and/or electric field. This force influences the particle鈥檚 trajectory and can be calculated using the formula:
  • \( F = qvB \sin \theta \)
Where:
  • \( F \) is the Lorentz force, measured in Newtons (N)
  • \( q \) is the charge of the particle, measured in Coulombs (C)
  • \( v \) is the velocity of the particle, measured in meters per second (m/s)
  • \( B \) is the magnetic field strength, measured in Teslas (T)
  • \( \theta \) is the angle between the velocity and the magnetic field direction
In the given exercise, the angle is 90 degrees, which means \( \sin \theta = 1 \), simplifying the equation to \( F = qvB \). This highlights how the Lorentz force depends on both the velocity of the particle and the magnetic field's characteristics.
Newton's Second Law
Newton's second law of motion is pivotal when analyzing the motion of particles under the influence of forces. It states that the acceleration of a particle is directly proportional to the net force acting on it and inversely proportional to its mass. This relationship is mathematically given by:
  • \( F = ma \)
Where:
  • \( F \) is the net force applied, measured in Newtons (N)
  • \( m \) is the mass of the particle, measured in kilograms (kg)
  • \( a \) is the acceleration of the particle, measured in meters per second squared (m/s虏)
In the provided exercise, once the force is calculated using the Lorentz equation, you can solve for acceleration by rearranging the formula: \( a = \frac{F}{m} \). This permits a clear understanding of how external forces impact the motion characteristics of particles.
Right-Hand Rule
The right-hand rule is a mnemonic tool used to determine the direction of vectors in three-dimensional space, especially in electromagnetism. When dealing with forces involving charged particles, magnetic fields, and velocities, it provides a simple way to predict the direction of the resulting force:
  • Point your thumb in the direction of the velocity of the particle.
  • Point your fingers in the direction of the magnetic field.
  • The palm then faces the direction of the force (or acceleration).
In the problem at hand, since the beryllium-9 nucleus is moving upward, and the magnetic field is directed from west to east, the right-hand rule shows that the force鈥攁nd the resultant acceleration鈥攚ould direct towards the north, giving a comprehensive understanding of spatial relationships in vector calculation.
Beryllium-9 Nucleus
The beryllium-9 nucleus is an atomic nucleus consisting of four protons and five neutrons. In problems involving magnetic fields, attention is usually given to the charge due to protons since neutrons are neutral. In terms of fundamental particles:
  • Protons are positively charged particles with a charge of \( 1.60 \times 10^{-19} \) C each.
  • Assuming an isolated nucleus, the charge is the sum of the charges of its protons.
  • For beryllium-9, this total charge is \( 4 \times 1.60 \times 10^{-19} \) C which equals \( 6.40 \times 10^{-19} \) C.
Understanding the charge of a nucleus allows us to effectively apply the Lorentz force formula, especially since only the moving charged particles interact with magnetic fields, thereby influencing their motion.
Acceleration Calculation
Calculating acceleration involves combining the concepts of force and mass. Once the force exerted on a particle, like the beryllium-9 nucleus, is known, determining its acceleration is straightforward using Newton's second law. For this:
  • Recognize the force obtained from the Lorentz equation: \( F = 9.63 \times 10^{-16} \) N.
  • Plug the values into the rearranged formula \( a = \frac{F}{m} \) for acceleration.
  • With \( m = 1.50 \times 10^{-26} \) kg, compute \( a = 6.42 \times 10^{10} \) m/s虏.
This calculation reflects how a magnetic field can significantly affect a charged particle, altering its path and speed. It also emphasizes the practical use of theoretical physics in determining measurable outcomes like acceleration in dynamic systems.

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Most popular questions from this chapter

A circular coil of wire 8.6 \(\mathrm{cm}\) in diameter has 15 turns and carries a current of 2.7 \(\mathrm{A}\) . The coil is in a region where the magnetic field is 0.56 \(\mathrm{T}\) (a) What orientation of the coil gives the maximum torque on the coil, and what is this maximum torque? (b) For what orientation of the coil is the magnitude of the torque 71\(\%\) of the maximum found in part (a)?

. You have 25 \(\mathrm{m}\) of wire, which you want to use to construct a 44 \(\mathrm{cm}\) diameter coil whose magnetic field at its center will exactly cancel the earth's field of 0.55 gauss. What current will your coil require?

A closely wound circular coil with a diameter of 4.00 \(\mathrm{cm}\) has 600 turns and carries a current of 0.500 A. What is the magnetic field at the center of the coil?

\(\bullet\) Two circular concentric loops of wire lie on a tabletop, one inside the other. The inner loop has a diameter of 20.0 \(\mathrm{cm}\) and carries a clockwise current of 12.0 \(\mathrm{A}\) , as viewed from above, and the outer wire has a diameter of 30.0 \(\mathrm{cm} .\) What must be the magnitude and direction (as viewed from above) of the current in the outer loop so that the net magnetic field due to this combination of loops is zero at the common center of the loops?

A circular metal loop is 22 \(\mathrm{cm}\) in diameter. (a) How large a current must flow through this metal so that the magnetic field at its center is equal to the earth's magnetic field of \(0.50 \times\) \(10^{-4} \mathrm{T}\) (b) Show how the loop should be oriented so that it can cancel the earth's magnetic field at its center.
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