Problem 2
Question
A particle of mass 0.195 g carries a charge of -2.50 \(\times\) 10\(^{-8}\) C. The particle is given an initial horizontal velocity that is due north and has magnitude 4.00 \(\times\) 10\(^4\) m/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
The magnetic field has a magnitude of 0.191 T and is directed west.
1Step 1: Understand the Forces
The particle experiences a gravitational force downwards due to gravity and a magnetic force acting perpendicular to both its velocity and the magnetic field. For the particle to move horizontally, the magnetic force must counteract the gravitational force.
2Step 2: Express the Gravitational Force
The gravitational force can be expressed as \( F_g = m imes g \), where \( m = 0.195 \times 10^{-3} \) kg and \( g = 9.81 \) m/s extsuperscript{2}. Compute \( F_g \) as follows:\[ F_g = 0.195 \times 10^{-3} \times 9.81 = 1.91195 \times 10^{-3} \text{ N} \].
3Step 3: Determine the Necessary Magnetic Force
Since the magnetic force \( F_m \) must balance the gravitational force to maintain horizontal movement, we have \( F_m = F_g = 1.91195 \times 10^{-3} \) N.
4Step 4: Use the Lorentz Force Equation
The magnetic force on a charged particle moving in a magnetic field is given by \( F_m = qvB \sin\theta \). For a perpendicular magnetic field, \( \theta = 90^{\circ} \), hence \( \sin\theta = 1 \).We have \( q = -2.50 \times 10^{-8} \) C and \( v = 4.00 \times 10^4 \) m/s. Thus, \( F_m = |q|vB = 1.91195 \times 10^{-3} \) N.
5Step 5: Solve for the Magnetic Field Magnitude
Rearranging the equation \( |q|vB = 1.91195 \times 10^{-3} \), solve for \( B \):\[ B = \frac{F_m}{|q|v} = \frac{1.91195 \times 10^{-3}}{2.50 \times 10^{-8} \times 4.00 \times 10^4} \]Calculate \( B \):\[ B = 1.91195 \times 10^{-3} / (2.50 \times 10^{-8} \times 4.00 \times 10^4) = 0.191195 \text{ T} \].
6Step 6: Determine the Direction of the Magnetic Field
Using the right-hand rule, since the particle's velocity is northward and the force must act upward to counteract gravity, the magnetic field should point towards the west.
Key Concepts
Lorentz ForceGravitational ForceMagnetic Field DirectionCharged Particle Motion
Lorentz Force
The Lorentz force is the force exerted on a charged particle moving through both electric and magnetic fields. In many practical scenarios, like the one examined in our exercise, we consider only the magnetic component. This force can be calculated using the equation:
- \( F_m = qvB\sin\theta \)
Gravitational Force
The gravitational force is a fundamental force exerted by the Earth on objects, pulling them towards its center. It depends on the mass of the object and the acceleration due to gravity. The formula to calculate the gravitational force is:
- \( F_g = mg \)
Magnetic Field Direction
The direction of a magnetic field is significant when determining the net force acting on a charged particle. The exercise required maintaining a charged particle in horizontal motion while countering the downward gravitational pull.
To understand the direction of the magnetic force, we use the right-hand rule:
- Point your fingers in the direction of the particle's velocity (northward).
- Align your palm to "push" in the direction of the force required to balance gravity (upwards).
- Your thumb, extending perpendicular to both, indicates the magnetic field direction (west).
Charged Particle Motion
The motion of a charged particle in a magnetic field is distinctive because the magnetic force acts perpendicular to the motion of the particle. This force does not work on the particle (as work involves displacement in the direction of force), but it does change the direction of the velocity.
For a particle like ours with an initial horizontal velocity:
- The velocity vector points north.
- The magnetic force, exerting force perpendicular to its direction, keeps the particle moving in the plane by offsetting other forces, such as gravity.
- This keeps the particle in "circular" motion.
Other exercises in this chapter
Problem 1
A particle with a charge of -1.24 \(\times\) 10\(^{-8}\) C is moving with instantaneous velocity \(\vec{v} =\) 14.19 \(\times\) 10\(^4\) m/s)\(\hat{\imath}\) +
View solution Problem 5
An electron experiences a magnetic force of magnitude 4.60 \(\times\) 10\(^{-15}\) N when moving at an angle of 60.0\(^\circ\) with respect to a magnetic field
View solution Problem 6
An electron moves at 1.40 \(\times\) 10\(^6\) m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.40 \(\times\) 10\
View solution Problem 7
A particle with charge 7.80 \(\mu\)C is moving with velocity \(\vec{v} =\) - 13.80 \(\times\) 103m/s\(\hat{\jmath}\). The magnetic force on the particle is meas
View solution