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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

Suppose the electric field between the plates in Fig. 27.22 is \(1.88 \times 10^{4} \mathrm{~V} / \mathrm{m}\) and the magnetic field in both regions is \(0.682 \mathrm{~T}\). If the source contains the three isotopes of krypton, \({ }^{82} \mathrm{Kr},{ }^{84} \mathrm{Kr},\) and \({ }^{86} \mathrm{Kr},\) and the ions are singly charged, find the distance between the lines formed by the three isotopes on the particle detector. Assume the atomic masses of the isotopes (in atomic mass units) are equal to their mass numbers, \(82,84,\) and \(86 .\) (One atomic mass unit \(=1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{~kg} .\) )

A coil with magnetic moment \(1.45 \mathrm{~A} \cdot \mathrm{m}^{2}\) is oriented initially with its magnetic moment antiparallel to a uniform 0.835 T magnetic field. What is the change in potential energy of the coil when it is rotated \(180^{\circ}\) so that its magnetic moment is parallel to the field?

A particle of mass \(0.195 \mathrm{~g}\) carries a charge of \(-2.50 \times 10^{-8} \mathrm{C}\). The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\). What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

A small particle with positive charge \(q=+3.75 \times 10^{-4} \mathrm{C}\) and mass \(m=5.00 \times 10^{-5} \mathrm{~kg}\) is moving in a region of uniform electric and magnetic fields. The magnetic field is \(B=4.00 \mathrm{~T}\) in the \(+z\) -direction. The electric field is also in the \(+z\) -direction and has magnitude \(E=60.0 \mathrm{~N} / \mathrm{C}\). At time \(t=0\) the particle is on the \(y\) -axis at \(y=+1.00 \mathrm{~m}\) and has velocity \(v=30.0 \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction. Neglect gravity. (a) What are the \(x\) -, \(y\) and \(z\) -coordinates of the particle at \(t=0.0200 \mathrm{~s} ?\) (b) What is the speed of the particle at \(t=0.0200 \mathrm{~s} ?\)

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is \(155 \mathrm{~V} / \mathrm{m}\) and the magnetic field is \(0.0315 \mathrm{~T}\). The ions next enter a uniform magnetic field of magnitude \(0.0175 \mathrm{~T}\) that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is \(17.5 \mathrm{~cm},\) what is their mass?
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