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A charged particle enters a uniform magnetic field and follows the circular path shown in the drawing. (a) Is the particle positively or negatively charged? Why? (b) The particle's speed is \(140 \mathrm{m} / \mathrm{s},\) the magnitude of the magnetic field is \(0.48 \mathrm{T}\), and the radius of the path is \(960 \mathrm{m}\). Determine the mass of the particle, given that its charge has a magnitude of \(8.2 \times 10^{-4} \mathrm{C}\).

Short Answer

Expert verified
(a) The particle is negatively charged. (b) The mass is approximately 0.0027 kg.

Step by step solution

01

Understand the direction of force and charge

When a charged particle enters a magnetic field, it experiences a force that is perpendicular to both the velocity of the particle and the magnetic field. This is given by the right-hand rule for positively charged particles. Since the particle follows a circular path in the given direction, observe how the particle curves to apply the right-hand rule: Point your fingers in the direction of velocity and curl them towards the magnetic field. Your thumb will point in the direction of the force for a positive charge. If the force direction agrees with the centripetal force towards the center of the circular path, the charge is positive; otherwise, it is negative.
02

Use the centripetal force formula

A charged particle moving in a magnetic field experiences a magnetic force that acts as a centripetal force. This force is given by the equation: \ \[ F = qvB = \frac{mv^2}{r} \]where \(F\) is the magnetic force, \(q\) is the charge, \(v\) is the speed of the particle, \(B\) is the magnetic field, \(m\) is the mass of the particle, and \(r\) is the radius of the path. Rearranging provides us a formula to solve for the mass, \(m\): \[ m = \frac{qBr}{v} \].
03

Substitute given values into the mass formula

With known values, substitute into the equation derived for mass: \ \[ m = \frac{(8.2 \times 10^{-4}) (0.48) (960)}{140} \]. \ Complete the calculation: \1. Multiply the values in the numerator: \(8.2 \times 10^{-4} \times 0.48 \times 960 = 0.377856\). \2. Divide this result by 140 to find \(m\).
04

Calculate the mass of the particle

Divide the result from Step 3 by the particle’s speed: \ \[ m = \frac{0.377856}{140} \approx 0.0027 \text{ kg}\]. \ Thus, the mass of the particle is approximately \(0.0027 \text{ kg}\).

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Key Concepts

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

Right-Hand Rule
The right-hand rule is a simple and effective technique used to determine the direction of the force experienced by a charged particle moving through a magnetic field. To apply it, place your right hand so that your fingers follow the direction of the particle's velocity. Then, curl your fingers towards the direction of the magnetic field lines. Your thumb will naturally point in the direction of the force that acts on the particle if it is positively charged. If the particle is negatively charged, the force direction will be opposite to that indicated by your thumb.

This rule arises from the vector nature of the magnetic force, which is defined by the cross-product in its equation: \[ \vec{F} = q( \vec{v} \times \vec{B} ) \]where \( \vec{F} \) is the magnetic force, \( q \) is the charge, \( \vec{v} \) is the velocity vector of the particle, and \( \vec{B} \) is the magnetic field vector. - To summarize: - Use your right hand for positively charged particles. - Opposite direction for negatively charged particles. - Ensures correct application of the force.
This method helps visualize and ensure you correctly comprehend the interaction of charged particles with magnetic fields.
Centripetal Force
Centripetal force is crucial when a particle moves in a circular path. It is the force that acts towards the center of the circle and keeps the particle on its curved path. For a charged particle in a magnetic field, the magnetic force doubles as the centripetal force. This is expressed by the equation:\[ F = \frac{mv^2}{r} \]where \( F \) is the centripetal force, \( m \) is the mass of the particle, \( v \) is the particle's speed, and \( r \) is the radius of the circular path.

Another critical formula links these concepts:\[ qvB = \frac{mv^2}{r} \]This equation equates the magnetic force \( qvB \) to the centripetal force \( \frac{mv^2}{r} \) demonstrating how the movement of the charged particle through the magnetic field keeps it in a circular trajectory.

- Key points to remember: - Magnetic force provides necessary centripetal force. - Allows calculation of mass using rearranged formulas. - Critical for particle motion in circular paths in magnetic fields.
Uniform Magnetic Field
A uniform magnetic field is characterized by magnetic field lines that are parallel and equidistant from each other. This means that the magnetic field strength is consistent and does not vary in magnitude or direction. It is crucial for analyses involving magnetic forces on charged particles, as it simplifies the calculations of their motions.

- Main characteristics: - Constant strength throughout the region. - Direction remains unchanged. - Field lines are parallel and evenly spaced.

In a uniform magnetic field, a charged particle entering the field experiences a steady magnetic force that remains constant in magnitude. This results in predictable and stable motion like the circular path seen in this exercise. The uniformity of the field ensures the effect of the magnetic force on the particle does not vary when the particle moves, allowing for straightforward calculations of motion dynamics like centripetal force and the radius of the path.

Understanding uniform magnetic fields is essential in applications ranging from MRI machines to particle accelerators, where controlling the path of charged particles is crucial.

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Most popular questions from this chapter

A long solenoid has a length of \(0.65 \mathrm{m}\) and contains 1400 turns of wire. There is a current of \(4.7 \mathrm{A}\) in the wire. What is the magnitude of the magnetic field within the solenoid?

At New York City, the earth's magnetic field has a vertical component of \(5.2 \times 10^{-5} \mathrm{T}\) that points downward (perpendicular to the ground) and a horizontal component of \(1.8 \times 10^{-5} \mathrm{T}\) that points toward geographic north (parallel to the ground). What are the magnitude and direction of the magnetic force on a \(6.0-\mathrm{m}\) long, straight wire that carries a current of \(28 \mathrm{A}\) perpendicularly into the ground?

The maximum torque experienced by a coil in a \(0.75-\mathrm{T}\) magnetic field is \(8.4 \times 10^{-4} \mathrm{N} \cdot \mathrm{m} .\) The coil is circular and consists of only one turn. The current in the coil is 3.7 A. What is the length of the wire from which the coil is made?

A wire has a length of \(7.00 \times 10^{-2} \mathrm{m}\) and is used to make a circular coil of one turn. There is a current of \(4.30 \mathrm{A}\) in the wire. In the presence of a \(2.50-\mathrm{T}\) magnetic field, what is the maximum torque that this coil can experience?

An ionized helium atom has a mass of \(6.6 \times 10^{-27} \mathrm{kg}\) and a speed of \(4.4 \times 10^{5} \mathrm{m} / \mathrm{s} .\) It moves perpendicular to a \(0.75-\mathrm{T}\) magnetic field on a circular path that has a 0.012-m radius. Determine whether the charge of the ionized atom is \(+e\) or \(+2 e\).

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