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Chapter 27: Problem 2

A particle of mass \(0.195 \mathrm{~g}\) carries a charge of \(-2.50 \times 10^{-8} \mathrm{C}\). The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{~m} / \mathrm{s}\). What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

Short Answer

Expert verified
The magnitude of the magnetic field is determined by calculating the value of \( B \) using the formula \( B = \frac{mg}{qv} \) and substituting the given values. The direction of the magnetic field is eastward, determined using the right hand rule.

Step by step solution

01

Calculate the gravitational force

The gravitational force exerted by the Earth on the particle can be calculated using the formula \( F = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity. Here, \( m = 0.195 \times 10^{-3} \) kg (\( 0.195 \) g = \( 0.195 \times 10^{-3} \) kg) and \( g = 9.8 \) m/s\(^2\).
02

Equate the gravitational force to the magnetic force

The magnetic force exerted on a moving charged particle in a magnetic field can be calculated using the formula \( F = qvBsin(\theta) \), where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field, and (\theta) is the angle between the velocity and the magnetic field. Here, we want to find the magnetic field that will keep the particle moving in the same direction (therefore, \( \theta = 90 \) degrees) and the gravitational force and magnetic force to be the same. Thus, \( qvB = mg \).
03

Solve for the magnetic field

Rearranging the formula from Step 2 to solve for \( B \), we get \( B = \frac{mg}{qv} \). Substituting the given values of \( m = 0.195 \times 10^{-3} \) kg, \( q = -2.5 \times 10^{-8} \) C (a negative charge does not affect the magnitude of the force), \( g = 9.8 \) m/s\(^2\), and \( v = 4.00 \times 10^{4} \) m/s, we can calculate the value of \( B \).
04

Determine the direction of the magnetic field

The direction of the magnetic field can be found using the right hand rule. A negative charge moving north will experience a force pointing down (the direction of gravity) when the magnetic field is directed to the east. Therefore, the minimum magnetic field that will keep the particle moving in the same horizontal direction (northward) must be towards the east.

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental force of nature that acts between two masses. This force pulls objects towards one another. For objects on Earth, gravitational force pulls them towards the center of the Earth.
\[ F = mg \]
This formula calculates gravitational force. Here,
  • \( F \) is the gravitational force.
  • \( m \) is the mass of the object.
  • \( g \) is the acceleration due to gravity, approximately \(9.8 \text{ m/s}^2\) on Earth.
In our example, a particle with a mass of \(0.195 \text{ g}\) or \(0.195 \times 10^{-3} \text{ kg}\) experiences a gravitational pull. Although tiny, this force is crucial in maintaining balance in various physical processes including the motion of this particle within a magnetic field.
Magnetic Force
Magnetic force acts on a charged particle when it moves through a magnetic field. It is given by:
\[ F = qvB ext{sin}( heta) \]
Where:
  • \( F \) is the magnetic force.
  • \( q \) is the charge of the particle.
  • \( v \) is the velocity of the particle.
  • \( B \) is the magnetic field strength.
  • \( \theta \) is the angle between the velocity and magnetic field.
In the exercise, the angle \( \theta \) is \( 90 \degree \), making \( \text{sin}(\theta) = 1 \). The goal is to equate this magnetic force to the gravitational force to find the perfect balance. This balance ensures the particle continues moving in a straight line without deflecting upwards or downwards.
Particle Motion
The motion of a charged particle in a magnetic field is affected by the Lorentz force, which consists of both electric and magnetic components; however, in this case, only the magnetic component is considered. This force can change the direction of the particle's velocity but not its speed.
The particle in this scenario must maintain a steady northward trajectory, counteracting the downward gravitational pull with an eastward magnetic field. This requires:
  • The particle's velocity to be exactly perpendicular to the magnetic field.
  • Correct calculation of the magnetic field strength to just balance the gravitational pull.
The right-hand rule helps to understand the motion and direction. For a negatively charged particle moving north, point your left thumb north, your fingers east (direction of magnetic field), which leaves your palm facing down, aligning with the gravitational force. Thus, the magnetic field ensures a stable horizontal path, demonstrating the interaction between gravitational force, magnetic force, and particle motion.

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Most popular questions from this chapter

A particle with charge \(q\) and mass \(m\) is dropped at time \(t=0\) from rest at its origin in a region of constant magnetic field \(\overrightarrow{\boldsymbol{B}}\) that points horizontally. What happens? To answer, construct a Cartesian coordinate system with the \(y\) -axis pointing downward and the \(z\) -axis pointing in the direction of the magnetic field. At time \(t \geq 0\) the particle has velocity \(\overrightarrow{\boldsymbol{v}}=v_{x} \hat{\imath}+v_{y} \hat{\jmath} .\) The net force \(\overrightarrow{\boldsymbol{F}}=F_{x} \hat{\imath}+F_{y} \hat{\jmath}\) on the particle is the vector sum of its weight and the magnetic force. (a) Using Newton's second law, write equations for \(a_{x}\) and \(a_{y},\) where \(\vec{a}=a_{x} \hat{\imath}+a_{y} \hat{\jmath}\) is the acceleration of the particle. (b) Differentiate the second of these equations with respect to time. Then substitute your expression for \(a_{x}=d v_{x} / d t\) to determine an equation for \(d v_{y}^{2} / d t^{2}\) in terms of \(v_{y}\) (c) This result shows that \(v_{y}\) is a simple harmonic oscillator. Use the initial conditions to determine \(v_{y}(t) .\) Write your answer in terms of the angular frequency \(\omega=q B / m\). Note that \(\left(d v_{y} / d t\right)_{0}=g .\) (d) Substitute your result for \(v_{y}(t)\) into your equation for \(d v_{x} / d t .\) Integrate using the initial conditions to determine \(v_{x}(t) .\) (e) Integrate your expressions for \(v_{x}(t)\) and \(v_{y}(t)\) to determine \(x(t)\) and \(y(t) .\) (f) If \(m=1.00 \mathrm{mg}, q=19.6 \mu \mathrm{C}\) and \(B=10.0 \mathrm{~T}\), what maximum vertical distance does the particle drop before returning upward?

A straight, vertical wire carries a current of 2.60 A down- ward in a region between the poles of a large superconducting electromagnet, 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?

The plane of a \(5.0 \mathrm{~cm} \times 8.0 \mathrm{~cm}\) rectangular loop of wire is parallel to a 0.19 T magnetic field. The loop carries a current of 6.2 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 dc motor with its rotor and field coils connected in series has an internal resistance of \(3.2 \Omega .\) When the motor is running at full load on a \(120 \mathrm{~V}\) line, the emf in the rotor is \(105 \mathrm{~V}\). (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

A flat, square surface with side length \(3.40 \mathrm{~cm}\) is in the \(x y\) -plane at \(z=0 .\) Calculate the magnitude of the flux through this surface produced by a magnetic field \(\overrightarrow{\boldsymbol{B}}=(0.200 \mathrm{~T}) \hat{\imath}+(0.300 \mathrm{~T}) \hat{\jmath}-(0.500 \mathrm{~T}) \hat{\boldsymbol{k}}\).
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