Problem 4
Question
\(\cdot\) A particle having a mass of 0.195 g carries a charge of \(-2.50 \times 10^{-8} \mathrm{C} .\) The particle is given an initial horizontal northward velocity of \(4.00 \times 10^{4} \mathrm{m} / \mathrm{s} .\) What are the magnitude and direction of the minimum magnetic field that will balance the earth's gravitational pull on the particle?
Step-by-Step Solution
Verified Answer
The minimum magnetic field has a magnitude of 0.019 T and is directed eastward.
1Step 1: Understand the Gravitational Force on the Particle
The gravitational force acting on the particle is given by the formula \( F_g = m imes g \), where \( m \) is the mass of the particle and \( g \) is the acceleration due to gravity (approximately \( 9.81 \text{ m/s}^2 \)). Convert the mass from grams to kilograms, so \( m = 0.195 \text{ g} = 0.195 \times 10^{-3} \text{ kg} \). Thus, \( F_g = 0.195 \times 10^{-3} \times 9.81 \approx 1.91 \times 10^{-3} \text{ N} \).
2Step 2: Determine the Magnetic Force Required to Balance the Gravitational Force
The magnetic force \( F_m \) must be equal in magnitude and opposite in direction to the gravitational force to balance it. Therefore, \( F_m = F_g = 1.91 \times 10^{-3} \text{ N} \).
3Step 3: Use the Magnetic Force Formula to Determine the Magnetic Field
The force on a particle moving in a magnetic field is given by \( F_m = q imes v imes B \), where \( q \) is the charge of the particle, \( v \) is the velocity, and \( B \) is the magnetic field. Rearrange to find \( B = \frac{F_m}{q imes v} \). Substitute \( F_m = 1.91 \times 10^{-3} \text{ N} \), \( q = -2.50 \times 10^{-8} \text{ C} \), and \( v = 4.00 \times 10^{4} \text{ m/s} \), which yields \( B = \frac{1.91 \times 10^{-3}}{-2.50 \times 10^{-8} \times 4.00 \times 10^{4}} \approx 0.019 \text{ T} \).
4Step 4: Determine the Direction of the Magnetic Field
To balance the gravitational force acting downward, the magnetic force (which is upward in this situation) must be generated by the particle moving in the chosen field. Use the right-hand rule to determine the direction: point your thumb northward (velocity), the fingers in the direction of the force (upward), and the palm faces the direction of the magnetic field. The magnetic field should be directed eastward. Since charge is negative, the actual magnetic force direction is opposite to that indicated by the right-hand rule, confirming the eastward field.
Key Concepts
Gravitational ForceMagnetic ForceRight-Hand RulePhysics Problem Solving
Gravitational Force
Gravitational force is what keeps us grounded on Earth. It is the force by which a planet or other celestial body draws objects toward its center. For any object near the Earth's surface, this force can be calculated using the formula \( F_g = m \times g \). Here, \( m \) is the object's mass and \( g \) is the gravitational acceleration, approximately \( 9.81 \, \text{m/s}^2 \).
For our tiny particle of mass \( 0.195 \, \text{g} \), which we convert to kilograms \((0.195 \times 10^{-3} \, \text{kg})\), the gravitational force comes out to be \( 1.91 \times 10^{-3} \, \text{N} \).
This calculation reveals the force the Earth exerts on the particle, urging it downward.
For our tiny particle of mass \( 0.195 \, \text{g} \), which we convert to kilograms \((0.195 \times 10^{-3} \, \text{kg})\), the gravitational force comes out to be \( 1.91 \times 10^{-3} \, \text{N} \).
This calculation reveals the force the Earth exerts on the particle, urging it downward.
Magnetic Force
Magnetic force acts on charged particles as they move through a magnetic field. The formula for this force is \( F_m = q \times v \times B \), where:
Using the formula for magnetic force, we rearrange to find the magnetic field \( B \). With the given values, the calculated magnetic field strength is approximately \( 0.019 \, \text{T} \).
- \( q \) is the charge on the particle,
- \( v \) is the velocity,
- \( B \) is the magnetic field.
Using the formula for magnetic force, we rearrange to find the magnetic field \( B \). With the given values, the calculated magnetic field strength is approximately \( 0.019 \, \text{T} \).
Right-Hand Rule
The right-hand rule is a clever mnemonic used to find the direction of the magnetic force or field when a charged particle moves through a magnetic field. Here's how you can use it:
- Point your thumb in the direction of the particle's velocity (in this problem, towards the north).
- Extend your fingers in the direction of the magnetic force (upward to balance gravity here).
- Your palm indicates the direction of the magnetic field.
Physics Problem Solving
Physics problem solving involves breaking down a problem into manageable steps. In this example, we:
- Identified the gravitational force first using object properties and known constants.
- Equated this force with the magnetic force required to counteract it.
- Applied the formula for magnetic force to find the necessary magnetic field.
- Used the right-hand rule to determine the correct direction for the magnetic field.
Other exercises in this chapter
Problem 2
An ion having charge \(+6 e\) is traveling horizontally to the left at 8.50 \(\mathrm{km} / \mathrm{s}\) when it enters a magnetic field that is perpendicular t
View solution Problem 3
\(\bullet\) A proton traveling at 3.60 \(\mathrm{km} / \mathrm{s}\) suddenly enters a uniform magnetic field of 0.750 \(\mathrm{T}\) , traveling at an angle of
View solution Problem 5
At a given instant, a particle with a mass of \(5.00 \times 10^{-3} \mathrm{kg}\) and a charge of \(3.50 \times 10^{-8} \mathrm{C}\) has a velocity with a magni
View solution Problem 6
If the magnitude of the magnetic force on a proton is \(F\) when it is moving at \(15.0^{\circ}\) with respect to the field, what is the magnitude of the force
View solution