/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A particle of \(\operatorname{ma... [FREE SOLUTION] | 91Ó°ÊÓ

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A particle of \(\operatorname{mass} 6.0 \times 10^{-8} \mathrm{kg}\) and charge \(+7.2 \mu \mathrm{C}\) is traveling due east. It enters perpendicularly a magnetic field whose magnitude is 3.0 T. After entering the field, the particle completes one-half of a circle and exits the field traveling due west. How much time does the particle spend traveling in the magnetic field?

Short Answer

Expert verified
The particle spends approximately 0.0873 seconds in the magnetic field.

Step by step solution

01

Understanding the Problem

The problem involves a charged particle moving in a circular path due to a perpendicular magnetic field. The particle finishes half of this circle, and we need to determine how long it takes for this movement.
02

Identify Key Quantities

The given quantities are the mass of the particle, \(m = 6.0 \times 10^{-8} \text{ kg} \), the charge, \(q = +7.2 \times 10^{-6} \text{ C} \), and the magnetic field strength, \(B = 3.0 \text{ T}\). These will be used in calculations involving motion in magnetic fields.
03

Use the Formula for the Cyclotron Frequency

The time period for completing a full circle in a magnetic field is given by \( T = \frac{2 \pi m}{q B} \). Here, the cyclotron frequency \( f \) is the reciprocal of the time period, which is \( f = \frac{qB}{2\pi m} \). However, we only need half a cycle, so we will use half of the full period.
04

Calculate the Cyclotron Period

Substitute the given values into the formula for the period: \[ T = \frac{2 \pi \times 6.0 \times 10^{-8}}{7.2 \times 10^{-6} \times 3.0} \text{ seconds}. \]
05

Solve for the Half-Period

Calculate the full period using the above values, and then divide by 2 to get the time for half a circle:\[ T_{\text{half}} = \frac{\pi m}{q B} = \frac{\pi \times 6.0 \times 10^{-8}}{7.2 \times 10^{-6} \times 3.0} = \frac{\pi}{36} \text{ seconds}. \]
06

Final Calculation

Compute \( \frac{\pi}{36} \) to find the time spent in the magnetic field: \( T_{\text{half}} \approx 0.0873 \text{ seconds}. \)

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Key Concepts

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

Cyclotron Frequency
The cyclotron frequency is a concept that appears often in physics when dealing with charged particles in a magnetic field. Essentially, it describes the frequency of a charged particle undergoing circular motion due to the presence of a constant magnetic field. The particle moves like this because the magnetic field exerts a force perpendicular to the particle's velocity, causing it to follow a circular path rather than a straight line.

The formula for cyclotron frequency is derived from the balance of the magnetic force and the centripetal force necessary for circular movement. The cyclotron frequency, denoted by \( f \), is given by the equation:
  • \( f = \frac{qB}{2\pi m} \)
where \( q \) is the particle's charge, \( B \) is the strength of the magnetic field, and \( m \) is the particle's mass. This formula shows us that the frequency of rotation is independent of the particle's speed and depends solely on its charge-to-mass ratio and the magnetic field.

Understanding this frequency is crucial for applications like particle accelerators, where precise control over the paths of particles is necessary.
Circular Motion
When a charged particle enters a magnetic field perpendicularly, it experiences a force that continuously changes its direction but not its speed. As a result, the particle traces out a circular path. This is a beautiful demonstration of a fundamental concept in physics: circular motion in a magnetic field.

Circular motion occurs when a constant force acts perpendicular to an object's velocity, keeping it in a continuous curved path. For charged particles in a magnetic field, this perpendicular force is the Lorentz force given by:
  • \( F = qvB \)
where \( v \) is the speed of the particle, and the other variables are as previously defined. In this setup, the magnetic force provides the centripetal force necessary to keep the particle moving in a circle. The radius of the circle, called the "radius of curvature," depends on the particle's velocity and is given by:
  • \( r = \frac{mv}{qB} \)
This tells us that faster particles or those with smaller charge-to-mass ratios will follow wider arcs.
Charge-to-Mass Ratio
The charge-to-mass ratio is fundamental when studying a charged particle's motion in a magnetic field. This ratio, denoted by \( \frac{q}{m} \), influences the cyclotron frequency and the radius of the particle's path in the field.

Understanding the charge-to-mass ratio helps us predict and control how a particle will behave when exposed to a magnetic field. A high charge-to-mass ratio means the particle will experience stronger accelerations, resulting in tighter, more frequent circular paths.

For the given problem, calculating the time a particle spends in a magnetic field involves using formulas where this ratio features prominently. Specifically, in the formula for cyclotron frequency:
  • \( f = \frac{qB}{2\pi m} \)
And in the equation for the radius of curvature:
  • \( r = \frac{mv}{qB} \)
The charge-to-mass ratio allows us to bridge the properties of the particle, like mass and charge, with its motion in the field, giving comprehensive insight into the dynamics of charged particles.

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Most popular questions from this chapter

A wire carries a current of 0.66 A. This wire makes an angle of \(58^{\circ}\) with respect to a magnetic field of magnitude \(4.7 \times 10^{-5} \mathrm{T}\). The wire experiences a magnetic force of magnitude \(7.1 \times 10^{-5} \mathrm{N}\). What is the length of the wire?

Two pieces of the same wire have the same length. From one piece, a square coil containing a single loop is made. From the other, a circular coil containing a single loop is made. The coils carry different currents. When placed in the same magnetic field with the same orientation, they experience the same torque. What is the ratio \(I_{\text {square }} / I_{\text {circle }}\) of the current in the square coil to the current in the circular coil?

The maximum torque experienced by a coil in a 0.75-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 horizontal wire of length \(0.53 \mathrm{m}\), carrying a current of \(7.5 \mathrm{A}\), is placed in a uniform external magnetic field. When the wire is horizontal, it experiences no magnetic force. When the wire is tilted upward at an angle of \(19^{\circ},\) it experiences a magnetic force of \(4.4 \times 10^{-3} \mathrm{N}\). Determine the magnitude of the external magnetic field.

A charged particle with a charge-to-mass ratio \(|q| / m=5.7 \times 10^{8} \mathrm{C} / \mathrm{kg}\) travels on a circular path that is perpendicular to a magnetic field whose magnitude is 0.72 T. How much time does it take for the particle to complete one revolution?

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