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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 is negative. (b) The magnetic force is approximately 5.05 x 10^-2 N.

Step by step solution

01

Analyze the Direction of Force Using Right-Hand Rule

To determine the sign of the charge, use the right-hand rule for magnetic force, which applies to positive charges. According to the rule, point your thumb in the direction of the particle's velocity (north), your fingers in the direction of the magnetic field (upward), and your palm will face the direction of the force. Here, the particle is deflected east, which would align with your palm if the charge were positive. However, since it's deflected in the opposite direction dictated by the right-hand rule for a positive charge, it must be negative.
02

Calculate the Magnetic Force on the Particle

The magnetic force on a charged particle moving in a magnetic field is given by the formula \[ F = qvB \sin(\theta) \]where \( F \) is the magnetic force, \( q = 8.50 \times 10^{-6} \, \mathrm{C} \) is the charge, \( v = 4.75 \times 10^3 \, \mathrm{m/s} \) is the velocity, \( B = 1.25 \, \mathrm{T} \) is the magnetic field strength, and \( \theta = 90^\circ \) is the angle between velocity and magnetic field direction (since they are perpendicular). Substituting the values:\[ F = (8.50 \times 10^{-6} \, \mathrm{C})(4.75 \times 10^3 \, \mathrm{m/s})(1.25 \, \mathrm{T}) \sin(90^\circ) = 5.046875 \times 10^{-2} \, \mathrm{N} \]Thus, the magnetic force on the particle is approximately \( 5.05 \times 10^{-2} \, \mathrm{N} \).

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Key Concepts

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

Right-Hand Rule
The Right-Hand Rule is a handy method to determine the direction of force exerted on a charged particle moving through a magnetic field. It's a straightforward technique you can use to predict how positive charges behave. Imagine you have a positive charge moving northward in a magnetic field that points upward.
Here鈥檚 what you do:
  • Point your right thumb in the direction of the particle鈥檚 velocity 鈥� towards the north.
  • Let your fingers follow the direction of the magnetic field, which in this case is upwards.
  • Your palm will naturally face the direction of the magnetic force.
If the force is towards the east, as your palm would show here, the charge is indeed positive. However, if the particle deflects in the opposite direction (west here), the charge is negative. This rule works consistently for resulting force directions, letting you intuitively understand charge signs.
Charged Particle Motion
Understanding the motion of a charged particle in a magnetic field is crucial in physics. Here's the basic idea: when charged particles move through a magnetic field, they experience a magnetic force that influences their trajectory.
Certain aspects affecting this motion include:
  • The particle's speed and charge magnitude determines how much force it will experience.
  • The angle between the velocity of the particle and the magnetic field, typically represented as \( \theta \).
  • The magnetic field strength, or how intense the magnetic field is.
When these forces act, they change the motion path of the particle. A perpendicular arrangement, as found in the exercise, maximizes the effect of the magnetic force. This results in a circular motion if not balanced by other forces. It's interesting to note that the force doesn't do work on the particle, but changes the direction of its velocity.
Magnetic Field
A magnetic field is an invisible force field that originates from magnets, electric currents, or changing electric fields. It exerts a force on particles affected by it, specifically those which are charged. How does this field work? By influencing the motion of charged particles without touching them.
Several points to understand magnetic fields:
  • They have both a magnitude (strength) and a direction.
  • The unit measuring magnetic field strength is the Tesla (T).
  • They can be visualized as a series of lines emanating from the north pole of a magnet to the south pole in a closed-loop formation.
Magnetic fields play a crucial role in the functioning of a wide array of electronic devices and are fundamental to electromagnetism. By interacting with charged particles, these fields help us understand phenomena from compasses pointing to Earth's magnetic poles to complex operations in particle accelerators.

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Most popular questions from this chapter

The plane of a \(5.0 \mathrm{cm} \times 8.0 \mathrm{cm}\) rectangular loop of wire is parallel to a \(0.19-\mathrm{T}\) magnetic field. The loop carries a current of 6.2 \(\mathrm{A}\) . (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?

A straight piece of conducting wire with mass \(M\) and length \(L\) is placed on a friction- less incline tilted at an angle \(\theta\) from the horizontal (Fig. 27.61 ) There is a uniform, vertical magnetic field \(\overrightarrow{\boldsymbol{B}}\) at all points (produced by an arrangement of magnets not shown in the figure). To keep the wire from shiding down the incline, a voltage source is attached to the ends of the wire. When just the right amount of current flows through the wire, the wire remains at rest. Determine the magnitude and direction of the current in the wire that will cause the wire to remain at rest. Copy the figure and draw the direction of the current on your copy. In addition, show in a free-body diagram all the forces that act on the wire.

A particle with negative charge \(q\) and mass \(m=2.58 \times\) 10 \(^{-15} \mathrm{kg}\) is traveling through a region containing a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=-(0.120 \mathrm{T}) \hat{\boldsymbol{k}} .\) At a particular instant of time the velocity of the particle is \(\overrightarrow{\boldsymbol{v}}=\left(1.05 \times 10^{6} \mathrm{m} / \mathrm{s}\right)(-3 \hat{\imath}+4 \hat{\jmath}+\) 12\(\hat{k} )\) and the force \(\vec{F}\) on the particle has a magnitude of 1.25 \(\mathrm{N}\) . (a) Determine the charge \(q\) . (b) Determine the acceleration \(\overrightarrow{\boldsymbol{d}}\) of the particle. (c) Explain why the path of the particle is a helix, and determine the radius of curvature \(R\) of the circular component of the helical path. (d) Determine the cyclotron frequency of the particle. (e) Although helical motion is not periodic in the full sense of the word, the \(x\) - and \(y\) -coordinates do vary in a periodic way. If the coordinates of the particle at \(t=0\) are \((x, y, z)=(R, 0,0),\) determine its coordinates at a time \(t=2 T,\) where \(T\) is the period of the motion in the \(x y\) -plane.

A particle with charge \(-5.60 \mathrm{nC}\) is moving in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=-(1.25 \mathrm{T}) \hat{\boldsymbol{k}} .\) The magnetic force on the particle is measured to be \(\overrightarrow{\boldsymbol{F}}=-\left(3.40 \times 10^{-7} \mathrm{N}\right) \hat{\boldsymbol{i}}+\left(7.40 \times 10^{-7} \mathrm{N}\right) \hat{\boldsymbol{j}}\) (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\overrightarrow{\boldsymbol{F}} \cdot \overrightarrow{\boldsymbol{F}}\) . What is the angle between \(\overrightarrow{\boldsymbol{v}}\) and \(\overrightarrow{\boldsymbol{F}} ?\)

An Electromagnetic Rail Gun. 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 \(\overrightarrow{\boldsymbol{B}}\) fills the region between the rails (Fig. 27.63\()\) . (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 \(B=0.50 \mathrm{T}, \quad I=2.0 \times 10^{3} \mathrm{A}\) \(m=25 \mathrm{kg},\) 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.
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