Problem 2
Question
A particle of mass 0.195 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?
Step-by-Step Solution
Verified Answer
Magnitude: 0.47775 T; Direction: West.
1Step 1: Understanding the Forces
To keep the particle moving in a horizontal path under gravity's influence, the magnetic force must balance the gravitational force. This setup ensures no vertical motion. We have:- Gravitational force, \( F_g = m imes g \). - Magnetic force, \( F_m = q imes v imes B \), directed upward if it balances gravity.
2Step 2: Calculate the Gravitational Force
The gravitational force \( F_g \) acting on the particle is given by:\[ F_g = m imes g \]Given \( m = 0.195 \) g and \( g = 9.8 \, \text{m/s}^2 \), convert \( m \) to kg (\( 0.195 \, \text{g} = 0.000195 \, \text{kg} \)) and use:\[ F_g = 0.000195 \times 9.8 = 1.911 imes 10^{-3} \, \text{N} \]
3Step 3: Relate Magnetic and Gravitational Forces
For the particle to experience no vertical displacement, balance the gravitational force with the magnetic force:\[ q imes v imes B = m imes g \]Rearrange to find the magnetic field magnitude:\[ B = \frac{m imes g}{q imes v} \]
4Step 4: Compute the Magnetic Field Magnitude
Substitute the known values:- Mass \( m = 0.000195 \, \text{kg} \)- Gravitational acceleration \( g = 9.8 \, \text{m/s}^2 \)- Charge \( q = -2.50 imes 10^{-8} \, \text{C} \)- Velocity \( v = 4.00 imes 10^{4} \, \text{m/s} \)Plug these into the formula:\[ B = \frac{0.000195 \times 9.8}{-2.50 imes 10^{-8} \times 4.00 imes 10^4} = 1.911 imes 10^{-3} \div -1 \times 10^{-8} = -0.47775 \text{ T} \]
5Step 5: Determine the Direction of the Magnetic Field
Since the charge is negative, a northward velocity and upward force require using the left-hand rule for direction:
- Velocity is northward
- Magnetic force is upward
With the left-hand rule, the magnetic field points west. Therefore, the field's actual direction is westward to ensure the upward magnetic force opposes gravity.
Key Concepts
Magnetic ForceGravitational ForceParticle Motion in a Magnetic Field
Magnetic Force
Magnetic force is a fundamental concept in physics. It describes how moving charged particles interact with magnetic fields. For a charged particle moving in a magnetic field, the force acting on it can be determined using the formula:
One key point is the direction of the force. For positive charges, use the right-hand rule: your fingers follow the velocity direction, curl towards the magnetic field direction, and your thumb shows the force direction. Conversely, for negative charges, like in our exercise, use the left-hand rule to find that the magnetic force direction is opposite to what the right-hand rule suggests.
- Magnetic force, \( F_m = q \times v \times B \)
One key point is the direction of the force. For positive charges, use the right-hand rule: your fingers follow the velocity direction, curl towards the magnetic field direction, and your thumb shows the force direction. Conversely, for negative charges, like in our exercise, use the left-hand rule to find that the magnetic force direction is opposite to what the right-hand rule suggests.
Gravitational Force
Gravitational force is one of the forces that we experience daily. It is the force of attraction between two bodies with mass. For objects close to Earth's surface, this force can be simplified with the formula:
In the context of our exercise, the particle experiences a gravitational force pulling it downward. To keep it moving horizontally, we need a counteracting upward force, provided by the magnetic force. This balance is crucial to maintaining the particle's motion along the desired path.
- Gravitational force, \( F_g = m \times g \)
In the context of our exercise, the particle experiences a gravitational force pulling it downward. To keep it moving horizontally, we need a counteracting upward force, provided by the magnetic force. This balance is crucial to maintaining the particle's motion along the desired path.
Particle Motion in a Magnetic Field
Understanding the motion of particles in a magnetic field involves examining how forces work to change their path. A magnetic field can change a charged particle's direction, but not its speed. In our exercise, we balance magnetic and gravitational forces to keep the particle in horizontal motion.
When the gravitational force pulls downward, the magnetic force must counteract it to avoid any vertical movement. Thus, the magnetic field must be precisely calculated to ensure that the particle remains on its path. The direction of the magnetic field is critical and follows from the left-hand or right-hand rule, depending on the charge's polarity. In this case, a correctly oriented magnetic field is essential for the particle to continue traveling northward with no vertical deviation.
By aligning the forces, we ensure smooth, uninterrupted particle motion in a magnetic field, showcasing how electromagnetic and gravitational forces can work in harmony. This reveals the intricate dance of forces acting on particles in various fields.
When the gravitational force pulls downward, the magnetic force must counteract it to avoid any vertical movement. Thus, the magnetic field must be precisely calculated to ensure that the particle remains on its path. The direction of the magnetic field is critical and follows from the left-hand or right-hand rule, depending on the charge's polarity. In this case, a correctly oriented magnetic field is essential for the particle to continue traveling northward with no vertical deviation.
By aligning the forces, we ensure smooth, uninterrupted particle motion in a magnetic field, showcasing how electromagnetic and gravitational forces can work in harmony. This reveals the intricate dance of forces acting on particles in various fields.
Other exercises in this chapter
Problem 1
A particle with a charge of \(-1.24 \times 10^{-8} \mathrm{C}\) is moving with instantancous \(\quad\) velocity \(\quad \vec{v}=\left(4.19 \times 10^{4} \mathrm
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In a 1.25 - T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50\(\mu \mathrm{C}\) and initially moving northward at 4.75 \
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A particle with mass \(1.81 \times 10^{-3} \mathrm{kg}\) and a charge of \(1.22 \times\) \(10^{-8} \mathrm{C}\) has, at a given instant, a velocity \(\vec{v}=\l
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An electron experiences a magnetic force of magnitude \(4.60 \times 10^{-15} \mathrm{N}\) when moving at an angle of \(60.0^{\circ}\) with respect to a magnetic
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