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Chapter 20: Problem 2

An ion having charge \(+6 e\) is traveling horizontally to the left at 8.50 \(\mathrm{km} / \mathrm{s}\) when it enters a magnetic field that is perpendicular to its velocity and deflects it downward with an initial magnetic force of \(6.94 \times 10^{-15} \mathrm{N} .\) What are the direction and magnitude of this field? Illustrate your method of solving this problem with a diagram.

Short Answer

Expert verified
The magnitude of the magnetic field is 0.085 T, directed into the page.

Step by step solution

01

Identify given information

The charge of the ion is \(+6e\), where \(e\) is the elementary charge \(1.6 \times 10^{-19}\) C. The velocity \(v\) of the ion = 8.50 km/s = 8500 m/s. The magnetic force \(F\) acting on it is \(6.94 \times 10^{-15}\) N.
02

Use the magnetic force formula

The magnetic force on a charge moving in a magnetic field is given by \(F = qvB\sin\theta\). Here, \(\theta = 90^\circ\) since the velocity and magnetic field are perpendicular, making \(\sin\theta = 1\). Substitute the known values: \(F = (6 \times 1.6 \times 10^{-19} \text{ C}) \times 8500 \text{ m/s} \times B\).
03

Rearrange to solve for the magnetic field

Solve the equation \(6.94 \times 10^{-15} = (6 \times 1.6 \times 10^{-19}) \times 8500 \times B\) for \(B\).\[ B = \frac{6.94 \times 10^{-15}}{(6 \times 1.6 \times 10^{-19} \times 8500)} \]
04

Calculate the magnitude of the magnetic field

Calculate \(B\) using the formula: \[ B = \frac{6.94 \times 10^{-15}}{(6 \times 1.6 \times 10^{-19} \times 8500)} \approx 0.085 \text{ T} \].
05

Determine the direction of the magnetic field

Since the ion is deflected downward by the magnetic field and using the right-hand rule, if the charge is positive and moving to the left, then the magnetic field must be into the page to produce a downward force.

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Key Concepts

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

Ion Charge
An ion is a charged atom or molecule, which means it has an excess or deficit of electrons. The charge of an ion is critical in determining how it interacts with magnetic fields. In our context, the ion in question carries a charge of \(+6e\), meaning it has six times the elementary positive charge of an electron.

  • Elementary charge \(e\) is equal to \(1.6 \times 10^{-19}\) Coulombs.
  • The total charge of the ion in this exercise is \(+6 \times 1.6 \times 10^{-19}\) C.
Positive ion charge indicates a surplus of protons over electrons. This net charge determines how the ion behaves in a magnetic field, influencing both the force experienced and the direction the ion will move under that force.
Magnetic Field Direction
Understanding the direction of the magnetic field is key to solving problems that involve magnetic forces on charged particles. A magnetic field exerts a force on moving charges, and this force is maximized when the particle's velocity is perpendicular to the field direction.

In the provided exercise:
  • The ion is initially moving horizontally to the left.
  • It experiences an initial magnetic force downward.
  • According to the right-hand rule (explained in the next section), the magnetic field's direction results in the observed downward deflection.
Therefore, for a positively charged ion moving left and deflected downwards, the magnetic field direction is into the page (perpendicular to the screen or paper). This orthogonality is crucial as it maximizes the magnetic force that influences charged particles.
Right-Hand Rule
The right-hand rule is a simple yet effective tool to determine the direction of the magnetic force acting on a moving charged particle in a magnetic field.


Here's how you apply it:
  • Use your right hand.
  • Point your thumb in the direction of the velocity of the positive charge (in this case, to the left).
  • Extend your fingers in the direction of the magnetic field.
  • Your palm will naturally face in the direction of the magnetic force experienced by the charge (in this exercise, downward).
This rule is especially handy because it allows you to visualize the relationships between velocity, magnetic field orientation, and force direction without needing complex calculations. This intuitive approach reinforces your understanding of how magnetic forces operate.
Velocity of Charged Particles
The velocity of a charged particle plays a significant role in determining the magnitude and direction of the magnetic force it experiences. Velocity vector orientation and speed directly affect the force interaction with the magnetic field.

Consider the exercise scenario:
  • The ion has a velocity of 8.50 km/s (or 8500 m/s) to the left.
  • The magnetic force equation is \(F = qvB\sin\theta\), where \(\theta = 90^\circ\) because the velocity & magnetic field are perpendicular.
  • This relationship indicates that both the speed and direction of the ion, alongside the strength and orientation of the magnetic field, directly impact the magnitude of the force: \(F = qvB\).
A higher speed increases the magnitude of the magnetic force, while the direction influences how the force is directed (downwards in our scenario). Recognizing these aspects is crucial for understanding the full dynamics of the charged particle in a magnetic environment.

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Most popular questions from this chapter

A solenoid having 165 turns and a cross-sectional area of 6.75 \(\mathrm{cm}^{2}\) carries a current of 1.20 A. If it is placed in a uniform 1.12 T magnetic field, find the torque this field exerts on the solenoid if its axis is oriented (a) perpendicular to the field, (b) parallel to the field, (c) at 35. \(0^{\circ}\) with the field.

A straight vertical wire carries a current of 1.20 \(\mathrm{A}\) down- ward in a region between the poles of a large electromagnet where the field strength is 0.588 \(\mathrm{T}\) and is horizontal. What are the magnitude and direction of the magnetic force on a 1.00 \(\mathrm{cm}\) section of this wire if the magnetic-field direction is (a) toward the east, (b) toward the south, (c) \(30.0^{\circ}\) south of west?

Two long parallel transmission lines 40.0 \(\mathrm{cm}\) apart carry 25.0 \(\mathrm{A}\) and 75.0 A currents. Find all locations where the net magnetic field of the two wires is zero if these currents are in (a) the same direction, (b) opposite directions.

\(\bullet\) An electric bus operates by drawing current from two parallel overhead cables, at a potential difference of \(600 \mathrm{V},\) and spaced 55 \(\mathrm{cm}\) apart. When the power input to the bus's motor is at its maximum power of \(65 \mathrm{hp},(\) a) what current does it draw and (b) what is the attractive force per unit length between the cables?

\(\bullet\) A closely wound circular coil has a radius of 6.00 \(\mathrm{cm}\) and carries a current of 2.50 A. How many turns must it have if the magnetic field at its center is \(6.39 \times 10^{-4} \mathrm{T} ?\)
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