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Chapter 27: Problem 3

In a 1.25 T magnetic field directed vertically upward, a particle having a charge of magnitude \(8.50 \mu \mathrm{C}\) and initially moving northward at \(4.75 \mathrm{~km} / \mathrm{s}\) is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.

Short Answer

Expert verified
A) The charge of the particle is positive. B) The magnetic force on the particle is 0.05 N.

Step by step solution

01

Determine the sign of the charge

The right-hand rule can be used to determine the direction of the force, and hence, the sign of the charge. When the fingers are pointed in the direction of the velocity (northward) and then bent in the direction of the magnetic field (upward), the thumb points to the direction of the force on a positive charge (east). Therefore, the charge is positive.
02

Find the magnetic force on the particle

The magnetic force can be calculated using the formula \(F = qvBsin(胃)\). Convert the velocity from km/s to m/s by multiplying by 1000, so \(v = 4.75 km/s = 4750 m/s\). Given that the magnetic field strength is \(B = 1.25 T\), the particle's charge is \(q = 8.5 \mu C = 8.5 * 10^{-6} C\) and the angle is \(胃 = 90掳\), the magnetic force is: \[F = (8.5 * 10^{-6} C) * (4750 m/s) * (1.25 T) * sin(90掳) = 0.05 N.\]

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Key Concepts

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

Understanding the Right-Hand Rule
The right-hand rule is a simple yet powerful tool to determine the direction of the magnetic force on a charged particle. Hold your right hand with the fingers extended.
  • Point your fingers in the direction of the particle's velocity.
  • Bend your fingers in the direction of the magnetic field.
  • Your thumb will point in the direction of the force experienced by a positive charge.
This technique helps visualize interactions between velocity, magnetic field, and force.
In the exercise, with velocity northward and magnetic field upward, the force is directed east if the charge is positive. This process allows for a clear understanding of how forces act within magnetic fields.
Exploring Particle Charge
Particle charge refers to the electrical property of particles, noted as positive, negative, or neutral. Charged particles interact with magnetic fields in unique ways.
In this problem, the particle's charge is given, but determining its effect is essential. Using the right-hand rule, we concluded the charge was positive because the force directed east aligns with the positive force direction.
Charge magnitude, given as \(8.50 \mu C\), indicates how strong the particle's interaction will be with the magnetic field.
This highlights the charge's significance in determining both force direction and magnitude.
Understanding Magnetic Fields
Magnetic fields are invisible forces that affect charged particles. They are measured in Tesla (T).
Fields have both direction and magnitude, guiding the motion of particles within them. When a charged particle enters a magnetic field, it experiences a force at right angles to both its velocity and the field direction.
  • In this scenario, the magnetic field is 1.25 T, directed vertically upward.
  • This causes the northward-moving particle to deflect eastward.
Magnetic field interactions are fundamental in technologies like motors and generators.
Velocity Conversion Essentials
Understanding how to convert velocity is crucial in solving problems involving magnetic forces. In this exercise, the velocity was initially given as \(4.75 \text{ km/s}\).
To use this in calculations, it must be converted to meters per second (m/s), since standard units are essential in physics.
  • Multiply by 1000: \(4.75 \text{ km/s} = 4750 \text{ m/s}\).
Using consistent units, especially SI units, simplifies the mathematical process.
This ensures accuracy in determining forces and other physical quantities.

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Most popular questions from this chapter

A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\vec{B}\) fills the region between the rails (Fig. \(\mathbf{P 2 7 . 5 9}\) ). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m,\) find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\). (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{~km} / \mathrm{s}) .\) Let \(B=0.80 \mathrm{~T}, I=2.0 \times 10^{3} \mathrm{~A}, m=25 \mathrm{~kg}\) and \(L=50 \mathrm{~cm} .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

A dc motor with its rotor and field coils connected in series has an internal resistance of \(3.2 \Omega .\) When the motor is running at full load on a \(120 \mathrm{~V}\) line, the emf in the rotor is \(105 \mathrm{~V}\). (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section 39.3 ), in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{~m}\). (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I ?\) (c) What is the magnetic moment of the atom due to the motion of the electron?

A particle with initial velocity \(\overrightarrow{\boldsymbol{v}}_{0}=\left(5.85 \times 10^{3} \mathrm{~m} / \mathrm{s}\right) \hat{\jmath}\) enters a region of uniform electric and magnetic fields. The magnetic field in the region is \(\overrightarrow{\boldsymbol{B}}=-(1.35 \mathrm{~T}) \hat{\boldsymbol{k}} .\) Calculate the magnitude and direction of the electric field in the region if the particle is to pass through undeflected, for a particle of charge (a) \(+0.640 \mathrm{nC}\) and (b) \(-0.320 \mathrm{nC}\). You can ignore the weight of the particle.

An electron moves at \(1.40 \times 10^{6} \mathrm{~m} / \mathrm{s}\) through a region in which there is a magnetic field of unspecified direction and magnitude \(7.40 \times 10^{-2} \mathrm{~T}\). (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?
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