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Chapter 21: Problem 76

Multiple-Concept Example 7 discusses how problems like this one can be solved. A \(+6.00 \mu C\) charge is moving with a speed of \(7.50 \times 10^{4} \mathrm{m} / \mathrm{s}\) parallel to a very long, straight wire. The wire is \(5.00 \mathrm{cm}\) from the charge and carries a current of \(67.0 \mathrm{A}\) in a direction opposite to that of the moving charge. Find the magnitude and direction of the force on the charge.

Short Answer

Expert verified
The force's magnitude is approximately \( 6.03 \times 10^{-5} \text{ N} \) to the left.

Step by step solution

01

Determine the Magnetic Field due to the Wire

Use Amp猫re's Law to determine the magnetic field, \( B \), at the position of the charge due to the infinitely long wire. The magnetic field from a long straight wire is given by the formula \[B = \frac{\mu_0 I}{2\pi r}\]where \( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \) is the permeability of free space, \( I = 67.0 \text{ A} \) is the current in the wire, and \( r = 0.050 \text{ m} \) is the distance from the wire to the charge.
02

Calculate the Magnetic Field

Substitute the values into the formula to find the magnetic field:\[B = \frac{4\pi \times 10^{-7} \times 67.0}{2\pi \times 0.050}\]Simplify the equation to find \( B \).
03

Apply the Right-Hand Rule

Determine the direction of the magnetic field around the wire using the right-hand rule. With the current flowing upwards (opposite to the moving charge's direction), the magnetic field will circle the wire. At the location of the charge, the field would be directed into the plane of the charge's path if the wire is on the left side.
04

Calculate the Magnetic Force

Use the magnetic force formula on a moving charge:\[F = qvB\sin(\theta)\]where \( q = 6.00 \times 10^{-6} \text{ C} \), \( v = 7.50 \times 10^4 \text{ m/s} \), and \( \theta = 90^\circ \) because the motion of the charge is perpendicular to the wire. Hence, \( \sin(90^\circ) = 1 \).
05

Substitute and Solve for Force

Substitute the known values into the equation:\[F = 6.00 \times 10^{-6} \times 7.50 \times 10^4 \times B\]Calculate \( F \) using the magnetic field \( B \) obtained from Step 2.
06

Determine the Direction of the Force

The direction of the force on the positive charge can be determined using the right-hand rule for force direction: Point your thumb in the direction of the charge's velocity (opposite the current direction) and your fingers in the direction of the magnetic field (into the plane), and the palm points toward the force direction. This indicates the force is directed to the left.

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Key Concepts

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

Amp猫re's Law
Amp猫re's Law is a key principle in electromagnetism that helps us calculate the magnetic field generated by electric currents. It states that the magnetic field in space around an electric current is proportional to the total current that passes through any closed loop.
This can be mathematically expressed by the integral form of Amp猫re's Law, where the line integral of the magnetic field \(B\) around a closed path is equal to the permeability of free space \(\mu_0\) times the current \(I\) enclosed by the path.To apply this to a long straight wire carrying a current, the formula simplifies to:
  • \(B = \frac{\mu_0 I}{2\pi r}\)
Here, \(\mu_0\) is a constant, \(I\) is the current, and \(r\) is the distance from the wire.
This formula is extremely useful for calculating the magnetic field when the wire is much longer than the distance to the point where the field is being calculated.
Right-Hand Rule
Understanding the right-hand rule is crucial for determining the direction of the magnetic fields and forces in electromagnetism. The right-hand rule is a mnemonic device used to predict the direction of the magnetic field lines around a current-carrying wire or the direction of force on a moving charge.
For a straight wire carrying a current, you can use the right-hand rule by doing the following:
  • Point your right thumb in the direction of the current flow.
  • Curl your fingers around the wire.
Your fingers will point in the direction of the magnetic field loops created around the wire.
When applying the right-hand rule to force on a charged particle, point your right thumb in the direction of the velocity of the positive charge, your fingers in the direction of the magnetic field, and the palm will face in the direction of the magnetic force.
Magnetic Field Calculation
Calculating the magnetic field involves using Amp猫re's Law to understand the influence of a current on the surrounding space. In the example problem, the magnetic field at the charge's location due to the infinitely long wire is calculated using:
  • \(B = \frac{\mu_0 I}{2\pi r}\)
Where \(\mu_0\) is the permeability of free space, \(I\) is the current, and \(r\) is the distance between the wire and the charge.In this case:
  • \(\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}\)
  • \(I = 67.0 \text{ A}\)
  • \(r = 0.050 \text{ m}\)
By substituting these values, you find the precise magnetic field acting at the point where the charge is located, an important step for understanding how forces will act on a moving charge.
Force Direction
In electromagnetism, determining the direction of the magnetic force on a moving charge is vital to understanding how charges behave in magnetic fields. The force on a charge moving in a magnetic field is given by the formula:
  • \(F = qvB\sin(\theta)\)
Here, \(q\) is the charge, \(v\) is the velocity, \(B\) is the magnetic field, and \(\theta\) is the angle between the direction of the velocity and the magnetic field.
For our exercise, since the motion of the charge is perpendicular to the wire, \(\theta = 90^\circ\), making \(\sin(90^\circ) = 1\). This simplifies the force calculation.To find the direction of this force, use the right-hand rule for force direction:
  • Point your thumb in the direction of the velocity (opposite the current direction).
  • Fingers point in the direction of the magnetic field.
  • The palm direction shows the force, which in this case, faces to the left.
This provides insight into the trajectory and behavior of charged particles in magnetic fields.

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Most popular questions from this chapter

A solenoid is formed by winding \(25.0 \mathrm{m}\) of insulated silver wire around a hollow cylinder. The turns are wound as closely as possible without overlapping, and the insulating coat on the wire is negligibly thin. When the solenoid is connected to an ideal (no internal resistance) 3.00-V battery, the magnitude of the magnetic field inside the solenoid is found to be \(6.48 \times 10^{-3} \mathrm{T}\). Determine the radius of the wire. (Hint: Because the solenoid is closely coiled, the number of turns per unit length depends on the radius of the wire.)

A piece of copper wire has a resistance per unit length of \(5.90 \times 10^{-3} \Omega / \mathrm{m} .\) The wire is wound into a thin, flat coil of many turns that has a radius of \(0.140 \mathrm{m}\). The ends of the wire are connected to a \(12.0-\mathrm{V}\) battery. Find the magnetic field strength at the center of the coil.

At a certain location, the horizontal component of the earth's magnetic field is \(2.5 \times 10^{-5} \mathrm{T}\), due north. A proton moves eastward with just the right speed for the magnetic force on it to balance its weight. Find the speed of the proton.

A small compass is held horizontally, the center of its needle a distance of \(0.280 \mathrm{m}\) directly north of a long wire that is perpendicular to the earth's surface. When there is no current in the wire, the compass needle points due north, which is the direction of the horizontal component of the earth's magnetic field at that location. This component is parallel to the earth's surface. When the current in the wire is \(25.0 \mathrm{A}\), the needle points \(23.0^{\circ}\) east of north. (a) Does the current in the wire flow toward or away from the earth's surface? (b) What is the magnitude of the horizontal component of the earth's magnetic field at the location of the compass?

A particle that has an \(8.2 \mu \mathrm{C}\) charge moves with a velocity of magnitude \(5.0 \times 10^{5} \mathrm{m} / \mathrm{s}\) along the \(+x\) axis. It experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge could experience has a magnitude of \(0.48 \mathrm{N}\). Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field.
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