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(II) A doubly charged helium atom whose mass is \(6.6 \times 10^{-27} \mathrm{kg}\) is accelerated by a voltage of 2700 \(\mathrm{V}\) . (a) What will be its radius of curvature if it moves in a plane perpendicular to a uniform \(0.340-\mathrm{T}\) field? (b) What is its period of revolution?

Short Answer

Expert verified
(a) The radius is approximately 0.09 m. (b) The period is about \( 1.3 \times 10^{-7} \) s.

Step by step solution

01

Calculate Kinetic Energy

The kinetic energy (KE) gained by the helium ion when it is accelerated through a potential difference \( V \) is given by the formula \( KE = qV \), where \( q \) is the charge of the ion. For a doubly charged helium ion, \( q = 2e \), where \( e = 1.6 \times 10^{-19} \text{ C} \). Thus, \( KE = 2 \times 1.6 \times 10^{-19} \times 2700 \). Calculate this to find the kinetic energy in joules.
02

Determine Speed from Kinetic Energy

Use the kinetic energy formula \( KE = \frac{1}{2} mv^2 \) to find the speed \( v \) of the ion. Rearrange the formula to solve for \( v \): \( v = \sqrt{\frac{2 \times KE}{m}} \). Substitute \( KE \) calculated in the previous step and \( m = 6.6 \times 10^{-27} \) kg to find \( v \).
03

Calculate Radius of Curvature

The radius of curvature in a magnetic field \( B \) is given by \( r = \frac{mv}{qB} \). Use the previously calculated \( v \), \( q = 3.2 \times 10^{-19} \text{ C} \), and \( B = 0.340 \text{ T} \) to compute the radius \( r \).
04

Determine Cyclotron Frequency

The cyclotron frequency \( f \) is given by \( f = \frac{qB}{2\pi m} \). Substitute \( q = 3.2 \times 10^{-19} \text{ C} \), \( B = 0.340 \text{ T} \), and \( m = 6.6 \times 10^{-27} \) kg to find the cyclotron frequency.
05

Calculate Period of Revolution

The period of revolution \( T \) is the reciprocal of the frequency, given by \( T = \frac{1}{f} \). Use the frequency calculated in the previous step to determine the period \( T \).

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Key Concepts

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

Kinetic Energy
Every moving object carries kinetic energy, which is essentially the energy it possesses due to its motion. When we talk about a charged particle like a helium ion, this kinetic energy can be gained from being accelerated by an electric field. The relationship is described by the formula: \[ KE = qV \]Here, \( KE \) represents the kinetic energy, \( q \) is the charge of the particle, and \( V \) is the voltage through which it is accelerated.
In the case of a doubly charged helium ion, the charge \( q \) equals twice the charge of an electron because the helium ion has lost two electrons. The elementary charge \( e \) is \( 1.6 \times 10^{-19} \text{ C} \). Hence, \( q = 2e = 3.2 \times 10^{-19} \text{ C} \). Accordingly, as this ion moves through a potential difference, it acquires energy, which can be computed in units of joules.
Understanding kinetic energy in this context helps in determining how fast the particle is moving once it has been accelerated. The rooted formula \( KE = \frac{1}{2} mv^2 \) then allows us to find the speed \( v \) when kinetic energy and mass \( m \) (in this instance, \( 6.6 \times 10^{-27} \text{ kg} \)) are known.
Magnetic Field
A magnetic field is a region in which a magnetic force acts on moving charges, like a helium ion in this example.
The nature of a magnetic field is described by a vector quantity, denoted as \( B \), and it is measured in teslas (T). The behavior of charged particles in a magnetic field is quite fascinating. When such particles enter a magnetic field perpendicularly, they experience a force that acts in a direction perpendicular to their velocity and to the magnetic field lines.
  • This force does not change the speed of the particle but may change its direction, causing it to move in a circular path, a phenomenon known as circular or cyclotronic motion.
  • The extent of this curvature or radius of the motion is determined by the formula: \[ r = \frac{mv}{qB} \]
  • In this case, \( v \) is the speed of the ion, \( m \) its mass, and \( B \) the strength of the magnetic field.
Using these parameters, one can ascertain the radius of curvature once the particle enters the magnetic field. This depicts how well the path is curved and provides insights into the nature of particle movement in magnetic fields.
Cyclotron Frequency
The cyclotron frequency represents how fast a charged particle, such as a helium ion, revolves while moving through a magnetic field.
It is a critical concept for understanding particle dynamics and is calculated using the formula: \[ f = \frac{qB}{2\pi m} \]
  • In this expression, \( f \) is the frequency of the oscillations or revolutions, \( q \) is the particle charge, \( B \) is the magnetic field strength, and \( m \) is the particle's mass.
  • The cyclotron frequency is inversely related to the mass, meaning lighter particles revolve quicker at certain magnetic field strengths.
  • This frequency does not depend on the velocity of the particle, illustrating that in a uniform magnetic field, all particles with the same charge-to-mass ratio revolve at the same frequency.
Understanding cyclotron frequency is also useful because it allows you to determine the period of revolution \( T \), which is simply the time it takes for one complete oscillation and is given by the inverse of the frequency: \[ T = \frac{1}{f} \]By understanding these concepts, we can predict and comprehend the movement of charged particles in magnetic environments, a principle that serves as a foundation for applications such as cyclotrons and magnetic resonance imaging.

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Most popular questions from this chapter

The power cable for an electric trolley (Fig. 56\()\) carries a horizontal current of 330 \(\mathrm{A}\) toward the east. The Earth's magnetic field has a strength \(5.0 \times 10^{-5} \mathrm{T}\) and makes an angle of dip of \(22^{\circ}\) at this location. Calculate the magnitude and direction of the magnetic force on a \(5.0-\) m length of this cable.

What is the velocity of a beam of electrons that goes undeflected when passing through perpendicular electric and magnetic fields of magnitude \(8.8 \times 10^{3} \mathrm{~V} / \mathrm{m}\) and \(7.5 \times 10^{-3} \mathrm{~T},\) respectively? What is the radius of the electron orbit if the electric field is turned off?

(1I) In a probe that uses the Hall effect to measure magnetic fields, a 12.0 -A current passes through a 1.50 -cm-wide 1.30 -mm-thick strip of sodium metal. If the Hall emf is 1.86\(\mu V\) , what is the magnitude of the magnetic field (take it perpendic- ular to the flat face of the strip)? Assume one free electron per atom of Na, and take its specific gravity to be \(0.971 .\)

(II) The Hall effect can be used to measure blood flow rate because the blood contains ions that constitute an electric current. \((a)\) Does the sign of the ions influence the emf? (b) Determine the flow velocity in an artery 3.3 \(\mathrm{mm}\) in diameter if the measured emf is 0.13 \(\mathrm{mV}\) and \(B\) is 0.070 \(\mathrm{T}\) . (In actual practice, an alternating magnetic field is used.)

An oil drop whose mass is determined to be \(3.3 \times 10^{-15} \mathrm{~kg}\) is held at rest between two large plates separated by \(1.0 \mathrm{~cm}\) as in Fig. \(27-31 .\) If the potential difference between the plates is \(340 \mathrm{~V}\), how many excess electrons does this drop have?

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