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

Short Answer

Expert verified
The time for one revolution is approximately \(1.55 \times 10^{-8}\) seconds.

Step by step solution

01

Identify Key Variables

First, understand the given variables from the problem. The charge-to-mass ratio \( \frac{|q|}{m} = 5.7 \times 10^8 \mathrm{C/kg} \) and the magnetic field \( B = 0.72 \mathrm{T} \). We need to find the time period \( T \) for one complete revolution of the particle.
02

Use Relevant Formula

Know the formula related to the motion of a charged particle in a magnetic field. The formula linking the period of revolution \( T \) to the charge-to-mass ratio and magnetic field is \( T = \frac{2\pi m}{|q|B} \).
03

Substitute the Charge-to-Mass Ratio

Substitute \( \frac{1}{T} = \frac{|q|B}{2\pi m} \) with the given charge-to-mass ratio. Given \( \frac{|q|}{m} = 5.7 \times 10^8 \mathrm{C/kg} \), we can use it directly to modify the formula to \( T = \frac{2\pi}{|q|/m \times B} \).
04

Calculate the Time Period

Substitute the known values into the modified equation to find \( T \). Compute: \( T = \frac{2\pi}{5.7 \times 10^8 \times 0.72} \approx 1.55 \times 10^{-8} \text{ seconds} \).

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Key Concepts

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

charge-to-mass ratio
In physics, the charge-to-mass ratio is a critical concept when dealing with the motion of charged particles. This ratio \(\frac{|q|}{m}\) represents the amount of charge a particle carries relative to its mass. A higher ratio implies that a particle will respond more vigorously to electromagnetic fields. This is particularly important in scenarios involving magnetic or electric fields. The charge-to-mass ratio provides an insight into how easily a particle can be accelerated by these forces.

In problems involving circular motion in a magnetic field, knowing the charge-to-mass ratio allows us to predict the particle's motion characteristics, such as its speed and how tight its circular path will be. For instance, a particle in a strong magnetic field with a high charge-to-mass ratio will have a smaller radius path compared to a particle with a lower ratio.

  • A large charge-to-mass ratio means high sensitivity to fields.
  • Essential in determining orbital paths in magnetic fields.
  • Significantly impacts calculations involving magnetic motion.
magnetic field
Magnetic fields are invisible fields that exert force on particles that possess charge and are in motion. They are described by the symbol \(B\) and are measured in teslas (T), which indicate the field's strength. In the context of charged particle motion like the one in our problem, the magnetic field is responsible for altering the trajectory of the charged particle, causing it to move in a circular path.

The fundamental rule guiding this behavior is known as the Lorentz force, which describes how a charge in a magnetic field experiences a force perpendicular to both its velocity and the magnetic field. This results in circular motion for a particle moving with a velocity at a right angle to the field.

Key aspects of magnetic fields include:

  • Direction and strength determine the particle's path.
  • The force felt is perpendicular to both velocity and magnetic field.
  • Directly influencing the period of revolution.
period of revolution
The period of revolution is the time taken for a particle to complete one loop around its circular path in a magnetic field. For charged particles moving perpendicular to a magnetic field, this period is constant, depending primarily on the charge-to-mass ratio and the strength of the magnetic field.

The formula for the period of revolution \(T\) for such systems is derived from equating the centripetal force required for circular motion to the Lorentz force exerted by the magnetic field:
\[ T = \frac{2\pi \ m}{|q|B} \]

This formula helps us calculate the time it takes for a particle to complete one circular path, given the involved magnetic field and charge-to-mass ratio.

  • Determined by charge-to-mass and magnetic field strength.
  • Inversely proportional to the strength of the magnetic field.
  • Direct application provides the time for one full orbit.
circular motion
Circular motion refers to the movement of a particle along a circular path. In our context, it stems from a charged particle traveling perpendicularly through a uniform magnetic field. The magnetic force acts as a centripetal force that continuously redirects the particle towards the center of its path, maintaining a circular trajectory.

This type of motion is characterized by a few key features:

  • Uniform repetition around the path, provided the magnetic field and conditions remain constant.
  • Constant speed of the particle, although its velocity is continuously changing direction.
  • Radius of path determined by balancing the magnetic force and the particle's inertia. \(r = \frac{m v}{|q| B} \) where \(v\) is the speed of the particle.


Understanding circular motion helps us predict aspects such as the revolution period and the energy dynamics involved in the system.
time calculation
Calculating the time a charged particle takes to complete one revolution in a magnetic field involves using a formula that connects time to known physical constants and measured quantities. Using the formula from the period of revolution:

\[ T = \frac{2\pi}{|q|/m \cdot B} \]

We can directly substitute the charge-to-mass ratio and magnetic field values provided in the problem to find the time \(T\).

In our scenario:
\[ T = \frac{2\pi}{5.7 \times 10^8 \times 0.72} \approx 1.55 \times 10^{-8} \text{ seconds} \]

This calculation simplifies to efficiently provide the time, emphasizing how mathematical expressions relate physical phenomena to measurable outcomes in a straightforward manner. This approach can be applied to similar problems dealing with periodic motion in magnetic fields.

Some notes on time calculation:
  • It is essential to get the correct ratio and field strength.
  • Involves basic algebra once formula is substituted.
  • Provide clear, logical steps to verify outcome.

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Most popular questions from this chapter

Two of the isotopes of cartoon, carbon-12 and carbon-13, have masses of \(19.93 \times 10^{-27} \mathrm{kg}\) and \(21.59 \times 10^{-27} \mathrm{kg}\) , respectively. These two isotopes are singly ionized \((+e),\) each given a speed of \(6.667 \times 10^{5} \mathrm{m} / \mathrm{s}\) . The ions then enter the bending region of a mass spectrometer where the magnetic field is 0.8500 \(\mathrm{T}\) . Determine the spatial separation between the two isotopes after they have traveled through a half-circle.

A positively charged particle of mass \(7.2 \times 10^{-8} \mathrm{kg}\) is traveling due east with a speed of 85 \(\mathrm{m} / \mathrm{s}\) and enters a \(0.31-\mathrm{T}\) uniform magnetic field. The particle moves through one- quarter of a circle in a time of \(2.2 \times 10^{-3} \mathrm{s}\) , at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.

Particle 1 and particle 2 have masses of \(m_{1}=2.3 \times 10^{-8} \mathrm{kg}\) and \(m_{2}=5.9 \times 10^{-8} \mathrm{kg},\) but they carry the same charge \(q .\) The two particles accelerate from rest through the same electric potential difference \(V\) and enter the same magnetic field, which has a magnitude \(B\). The particles travel perpendicular to the magnetic field on circular paths. The radius of the circular path for particle 1 is \(r_{1}=12 \mathrm{cm} .\) What is the radius (in cm) of the circular path for particle 2\(?\)

Two circular coils of current-carrying wire have the same magnetic moment. The first coil has a radius of \(0.088 \mathrm{m},\) has 140 turns, and carries a current of 4.2 \(\mathrm{A}\) . The second coil has 170 turns and carries a current of 9.5 \(\mathrm{A} .\) What is the radius of the second coil?

ssm The electrons in the beam of a television tube have a kinetic energy of \(2.40 \times 10^{-15} \mathrm{J}\) . Initially, the clectrons move horizontally from west to east. The vertical component of the earth's magnetic field points down, toward the surface of the earth, and has a magnitude of \(2.00 \times 10^{-5} \mathrm{T} .\) (a) In what direction are the electrons deflected by this field component? (b) What is the acceleration of an electron in part (a)?

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