/*! 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 59 You wish to hit a target from se... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You wish to hit a target from several meters away with a charged coin having a mass of 4.25\(\mathrm { g }\) and a charge of \(+ 2500 \mu \mathrm { C }\) . The coin is given an initial velocity of 12.8\(\mathrm { m } / \mathrm { s }\) , and a downward, uniform electric field with field strength 27.5\(\mathrm { N } / \mathrm { C }\) exists through-out the region. If you aim directly at the target and fire the coin horizontally, what magnitude and direction of uniform magnetic field are needed in the region for the coin to hit the target?

Short Answer

Expert verified
A magnetic field of 2.15 T directed into the page is needed.

Step by step solution

01

Understand the Problem

The coin is subject to both an electric and a magnetic field. We want the magnetic force to counteract the electric force so the coin hits the target.
02

Calculate the Electric Force

Use the formula for electric force: \( F_e = qE \), where \( q = 2500 \, \mu C = 2500 \times 10^{-6} \, C \) and \( E = 27.5 \, N/C \). \( F_e = (2500 \times 10^{-6} \, C)(27.5 \, N/C) = 0.06875 \, N \).
03

Determine the Required Magnetic Force

For the coin to travel straight and hit the target, the magnetic force must be equal and opposite to the electric force: \( F_m = F_e = 0.06875 \, N \).
04

Relate Magnetic Force to Magnetic Field

The magnetic force is given by \( F_m = qvB \sin \theta \), where \( \theta = 90^\circ \) because the velocity is horizontal. Thus, \( F_m = qvB \).
05

Solve for Magnetic Field Magnitude

Rearrange the equation to solve for \( B \): \( B = \frac{F_m}{qv} \). Substituting the known values \( F_m = 0.06875 \, N \), \( q = 2500 \times 10^{-6} \, C \), and \( v = 12.8 \, m/s \), \( B = \frac{0.06875}{(2500 \times 10^{-6} \times 12.8)} = 2.15 \, T \).
06

Determine the Magnetic Field Direction

By using the right-hand rule: thumb in the direction of velocity (horizontal), fingers in the direction of the electric field (downward), the palm facing the direction of force (opposite to gravity) requires the magnetic field to be directed into the page.

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 Force
Magnetic force plays a crucial role when charged particles move through a magnetic field. It can deflect the path of the particle depending on its charge, velocity, and the magnetic field's direction. This force is given by the equation \[ F_m = qvB \sin \theta \]where:
  • \( q \) is the charge of the particle
  • \( v \) is its velocity
  • \( B \) is the magnetic field strength
  • \( \theta \) is the angle between the direction of velocity and the magnetic field
In the case of our charged coin, the angle \( \theta \) is 90 degrees because the coin is moving perpendicular to the magnetic field. Therefore, \( \sin \theta = 1 \), simplifying the equation to \[ F_m = qvB \]. The right-hand rule helps determine the vector direction of the magnetic force, just like when holding a coin so that your thumb, fingers, and palm facilitate understanding of the path deflected by magnetic effects.
Electric Field
An electric field is a region around a charged particle that exerts force on other charged particles. In our exercise, the electric field is uniform, meaning it has a consistent strength and direction within the designated area. The force exerted by an electric field on a charge is expressed by \[ F_e = qE \]where:
  • \( q \) is the charge affected by the field
  • \( E \) is the electric field strength
The electric field in the scenario applies a downward force on the charged coin. Calculating this force is essential to balance it with a suitable magnetic force, ensuring the coin reaches its target without being dragged off its path by the electric field.
Lorentz Force
The Lorentz force is the total electromagnetic force acting on a particle moving in an electric and magnetic field. It combines both electric and magnetic forces. The equation representing the Lorentz force is:\[ F = F_e + F_m = qE + qvB \]This force dictates how charged objects move when subjected to electromagnetic fields. In our coin exercise, the desired equilibrium condition means we want the magnetic force to oppose the electric force precisely. This results in the Lorentz force being zero, allowing the coin to maintain a constant velocity straight towards the target. By accurately calculating and applying a magnetic field in the required direction, the coin's trajectory is unaffected by external electromagnetic influences.
Projectile Motion
Projectile motion describes the path of an object thrown into the air, subject to only gravity and no other forces. However, our situation involves a charged coin within both electric and magnetic fields. While you aim and fire the coin horizontally, these fields change its natural projectile path, necessitating precise corrections. The goal within this exercise is to make external forces like magnetic and electric fields cancel out. This way, the initial horizontal velocity of the coin remains constant, removing any arc usually seen in projectile motion. Thus, with the magnetic force exactly countering the electric force, comfort can be found knowing the coin will follow a straight path towards its target without deviation, and calculations will ensure this balance is achieved.

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

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section \(38.5 ) ,\) in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times\) \(10 ^ { 6 } \mathrm { m } / \mathrm { s }\) in a circular orbit of radius \(5.3 \times 10 ^ { - 11 } \mathrm { m } .\) (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I\) (c) What is the magnetic moment of the atom due to the motion of the electron?

A singly ionized (one electron removed) \(^{40} \mathrm{K}\) atom passes through a velocity selector consisting of uniform perpendicular electric and magnetic fields. The selector is adjusted to allow ions having a speed of 4.50 \(\mathrm{km} / \mathrm{s}\) to pass through undeflected when the magnetic field is 0.0250 T. The ions next enter a second uniform magnetic field \(\left(B^{\prime}\right)\) oriented at right angles to their velocity. 40 contains 19 protons and 21 neutrons and has a mass of \(6.64 \times 10^{-26} \mathrm{kg} .\) (a) What is the magnitude of the electric field in the velocity selector? (b) What must be the magnitude of \(B^{\prime}\) so that the ions will be bent into a semicircle of radius 12.5 \(\mathrm{cm} ?\)

A wire 25.0\(\mathrm { cm }\) long lies along the \(z\) -axis and carries a current of 7.40 A in the \(+ z\) -direction. The magnetic field is uniform and has components \(B _ { x } = - 0.242\) T, \(B _ { y } = - 0.985 \mathrm { T }\) , and \(B _ { z } = - 0.336 \mathrm { T }\) (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the magnetic force on the wire?

The magnetic poles of a small cyclotron produce a magnetic field with magnitude 0.85\(\mathrm { T }\) . The poles have a radius of \(0.40 \mathrm { m } ,\) which is the maximum radius of the orbits of the accelerated particles. (a) What is the maximum energy to which protons \(\left( q = 1.60 \times 10 ^ { - 19 } \mathrm { C } , m = 1.67 \times 10 ^ { - 27 } \mathrm { kg } \right)\) can be accelerated by this cyclotron? Give your answer in electron volts and in joules. (b) What is the time for one revolution of a proton orbiting at this maximum radius? (c) What would the magnetic-field magnitude have to be for the maximum energy to which a proton can be accelerated to be twice that calculated in part (a)? For \(B = 0.85 \mathrm { T } ,\) what is the maximum energy to which alpha particles \(\left( q = 3.20 \times 10 ^ { - 19 } \mathrm { C } , m = 6.65 \times 10 ^ { - 27 } \mathrm { kg } \right)\) can be accelerated by this cyclotron? How does this compare to the maximum energy for protons?

A straight, vertical wire carries a current of 1.20 A downward in a region between the poles of a large superconductingelectromagnet, where the magnetic field has magnitude \(B=\) 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 the wire that is in this uniform magnetic field, if the magnetic field direction is (a) east; (b) south; (c) \(30.0^{\circ}\) south of west?

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.