/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Shows a snapshot of a proton mov... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Shows a snapshot of a proton moving at velocity \(\vec{v}=(-200 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) toward a long straight wire with current \(i=\) \(350 \mathrm{~m} \mathrm{~A}\). At the instant shown, the proton's distance from the wire is \(d=2.89 \mathrm{~cm} .\) In unit-vector nota- tion, what is the magnetic force on the proton due to the current?

Short Answer

Expert verified
The magnetic force is \( 7.74 \times 10^{-23} \hat{k} \, \text{N} \).

Step by step solution

01

Determine the Magnetic Field by the Wire

The magnetic field created by a straight current-carrying wire can be calculated using Ampère's law: \( B = \frac{\mu_0 I}{2 \pi d} \), where \( \mu_0 = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \), \( I = 350 \, \text{mA} = 0.350 \, \text{A} \), and \( d = 0.0289 \, \text{m} \). Thus, \( B = \frac{4\pi \times 10^{-7} \times 0.350}{2\pi \times 0.0289} \approx 2.42 \times 10^{-6} \, \text{T} \). The direction of \( B \) is circular around the wire.
02

Identify the Direction of the Magnetic Field

Using the right-hand rule, point the thumb of your right hand in the direction of current (upwards along +\(\hat{j}\)). Your fingers curl in the direction of the magnetic field, which is circling the wire, creating a field in the negative x-direction at the position of the proton.
03

Calculate the Magnetic Force on the Proton

The magnetic force on a moving charge is given by \( \vec{F} = q(\vec{v} \times \vec{B}) \). For a proton, \( q = 1.6 \times 10^{-19} \, \text{C} \). Given \( \vec{v} = (-200 \, \text{m/s}) \hat{j} \) and \( \vec{B} = 2.42 \times 10^{-6} \, \text{T} \, (-\hat{i}) \), the cross product \( \vec{v} \times \vec{B} \) results in: \( \vec{v} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -200 & 0 \ -2.42 \times 10^{-6} & 0 & 0 \end{vmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \hspace{0.4em} 0 & -200 & 0 \ \hspace{0.7em} \,-2.42 \times 10^{-6} & 0 & 0 \end{vmatrix} \).
04

Simplify the Cross Product

Evaluating the determinant gives: \[ \vec{v} \times \vec{B} = (-200 \cdot 0)\hat{k} - (0 \cdot 0)\hat{j} + (200 \cdot (-2.42 \times 10^{-6}))\hat{i} = 4.84 \times 10^{-4} \hat{k} \, \text{m/s}^2. \] Thus, the force becomes: \( \vec{F} = (1.6 \times 10^{-19} \, \text{C})(4.84 \times 10^{-4} \hat{k} \, \text{m/s}^2) \approx 7.74 \times 10^{-23} \hat{k} \, \text{N} \).
05

Interpret the Result

The magnetic force on the proton is \( \vec{F} \approx 7.74 \times 10^{-23} \hat{k} \, \text{N} \), which indicates the force is in the positive z-direction (along \( \hat{k} \)).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Magnetic Field
A magnetic field is a vector field that represents the magnetic influence of electric charges in relative motion and magnetic materials. It is denoted by the symbol \( \vec{B} \) and interacts with moving charges to exert a force.

In the context of the exercise, the magnetic field is generated by a long, straight current-carrying wire. The magnitude of this field at a distance \( d \) from the wire can be calculated using Ampère's law: \[ B = \frac{\mu_0 I}{2 \pi d} \] where \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \)), \( I \) is the current, and \( d \) is the distance from the wire. The direction of this field is circular around the wire.
  • The field circles around the wire forming closed loops.
  • Its strength decreases as you move farther from the wire.
Current-Carrying Wire
A current-carrying wire is a conductor through which electric current flows. The flow of electric charges within the wire creates a magnetic field in the surrounding space. This is due to the inherent nature of moving charges generating magnetic fields.
In this scenario, we have a long straight wire carrying a current of \( 350 \, \text{mA} \). This produces a magnetic field around the wire. It's important to visualize the field as circular loops forming around the wire.
  • The field's direction can be determined using the right-hand rule.
  • The magnitude is influenced by the current's strength and the distance from the wire.


These properties make current-carrying wires useful in generating magnetic fields for a wide array of applications including electromagnets and inductors.
Right-Hand Rule
The right-hand rule is a helpful mnemonic for determining the direction of the magnetic field relative to a current or the force relative to a magnetic field and current. When dealing with a current-carrying wire, the rule is applied as follows:

  • Point your thumb in the direction of the electric current.
  • Your fingers will naturally curl in the direction of the magnetic field created by the current.


In our exercise, the current in the wire is upward (in the positive \( \hat{j} \) direction). Using the right-hand rule, you will find that the magnetic field forms concentric circles around the wire. At the position of the proton, the field direction is towards the negative \( \hat{i} \) direction (or the negative x-axis). This understanding is crucial in solving magnetic force problems.
Proton Charge
Protons, fundamental particles of the atomic nucleus, carry a positive charge. The charge of a proton is exactly \( 1.6 \times 10^{-19} \) Coulombs. Understanding the charge of a proton is essential when calculating the magnetic force it experiences.

The magnetic force \( \vec{F} \) exerted on a proton moving with velocity \( \vec{v} \) in a magnetic field \( \vec{B} \) is given by: \[ \vec{F} = q(\vec{v} \times \vec{B}) \] where \( q \) represents the charge of the proton. Knowing this allows us to compute the resulting force from magnetic interactions.
  • The proton has a positive charge.
  • This charge interacts with magnetic fields to experience forces.
Cross Product
The cross product is a mathematical operation that finds a vector perpendicular to two given vectors. In physics, it is used to calculate the magnetic force on a charged particle moving in a magnetic field. The cross product \( \vec{v} \times \vec{B} \) where \( \vec{v} \) is velocity and \( \vec{B} \) is magnetic field, informs us about the direction and magnitude of the force:

  • It utilizes a determinant format for calculation.
  • The result is perpendicular both to velocity and the magnetic field vectors.

In the exercise, the proton's velocity is along the negative y-axis and the magnetic field is along the negative x-axis. By solving the determinant, you determine the direction of the force along the positive z-axis.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A 200 -turn solenoid having a length of \(25 \mathrm{~cm}\) and a diameter of \(10 \mathrm{~cm}\) carries a current of \(0.29\) A. Calculate the magnitude of the magnetic field \(\vec{B}\) inside the solenoid.

A long wire is known to have a radius greater than \(4.0 \mathrm{~mm}\) and to carry a current that is uniformly distributed over its cross section. The magnitude of the magnetic field due to that current is \(0.28 \mathrm{mT}\) at a point \(4.0 \mathrm{~mm}\) from the axis of the wire, and \(0.20 \mathrm{mT}\) at a point 10 \(\mathrm{mm}\) from the axis of the wire. What is the radius of the wire?

A current \(i=10 \mathrm{~A}\) is set up in a long hairpin conductor formed by bending a wire into a semicircle of radius \(R=5.0 \mathrm{~mm}\). Point \(b\) is midway between the straight sections and so distant from the semicircle that each straight section can be approximated as being an infinite wire. What are the (a) magnitude and (b) direction (into or out of the page) of \(\vec{B}\) at \(a\) and the (c) magnitude and (d) direction of \(\vec{B}\) at \(b\) ?

Three long wires are parallel to a \(z\) axis, and each carries a current of \(10 \mathrm{~A}\) in the positive \(z\) direction. Thein points of intersection with the \(x y\) plane form an equilateral triangle with sides of \(50 \mathrm{~cm}\), as shown in Fig. \(29-77\). A fourth wire (wire \(b\) ) passes through the midpoint of the base of the triangle and is parallel to the other three wires. If the net magnetic force on wire \(a\) is zero, what are the (a) size and (b) direction \((+z\) or \(-z)\) of the current in wire \(b ?\)

A conductor carries \(6.0 \mathrm{~A}\) along the closed path abcdefgha running along 8 of the 12 edges of a cube of edge length 10 \(\mathrm{cm}\). (a) Taking the path to be a combination of three square current loops \((b c f g b, a b g h a\), and \(c d e f c)\), find the net magnetic moment of the path in unit- vector notation. (b) What is the magnitude of the net magnetic field at the \(x y z\) coordinates of \((0,5.0 \mathrm{~m}, 0)\) ?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.