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Chapter 29: Problem 19

Positron A positron with kinetic energy \(2.0 \mathrm{keV}\) is projected into a uniform magnetic field \(\vec{B}\) of magnitude \(0.10 \mathrm{~T}\), with its velocity vector making an angle of \(89^{\circ}\) with \(\vec{B}\). Find (a) the period, (b) the pitch \(p\), and (c) the radius \(r\) of its helical path.

Short Answer

Expert verified
Period: \(3.58 \times 10^{-10} \, \text{s}\); Pitch: \(5.26 \times 10^{-5} \, \text{m}\); Radius: \(4.78 \times 10^{-3} \, \text{m}\).

Step by step solution

01

- Identify given values

Kinetic energy, \(K = 2.0 \, \text{keV}\)Magnetic field, \(B = 0.10 \, \text{T}\)Velocity angle, \(\theta = 89^{\circ}\)Charge of positron, \(e = 1.6 \times 10^{-19} \, \text{C}\)Mass of positron, \(m = 9.11 \times 10^{-31} \, \text{kg}\)
02

- Calculate the velocity of the positron

Convert the kinetic energy from keV to Joules: \(2.0 \, \text{keV} = 2.0 \times 10^{3} \times 1.6 \times 10^{-19} \, \text{J} = 3.2 \times 10^{-16} \, \text{J}\)Use the kinetic energy equation, \(K = \frac{1}{2}mv^2\), to find the velocity:\[v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2 \times 3.2 \times 10^{-16}}{9.11 \times 10^{-31}}} \approx 8.4 \times 10^7 \, \text{m/s}\]
03

- Find the components of velocity

Calculate the velocity components parallel and perpendicular to the magnetic field:\[v_{\parallel} = v \cos(\theta)\]\[v_{\perp} = v \sin(\theta)\]\[v_{\parallel} = 8.4 \times 10^7 \cos(89^{\circ}) \approx 1.47 \times 10^5 \, \text{m/s}\]\[v_{\perp} = 8.4 \times 10^7 \sin(89^{\circ}) \approx 8.4 \times 10^7 \, \text{m/s}\]
04

- Calculate the period (T)

The period of the helical motion is given by:\[T = \frac{2\pi m}{eB}\]\[T = \frac{2\pi \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 0.10} \approx 3.58 \times 10^{-10} \, \text{s}\]
05

- Compute the pitch (p)

Pitch of the helical path is given by:\[p = v_{\parallel} T\]\[p = 1.47 \times 10^5 \times 3.58 \times 10^{-10} \approx 5.26 \times 10^{-5} \, \text{m}\]
06

- Determine the radius (r)

The radius of the helical path is:\[r = \frac{m v_{\perp}}{eB}\]\[r = \frac{9.11 \times 10^{-31} \times 8.4 \times 10^7}{1.6 \times 10^{-19} \times 0.10} \approx 4.78 \times 10^{-3} \, \text{m}\]

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Key Concepts

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

Kinetic Energy of the Positron
We start by examining the kinetic energy of the positron. Kinetic energy represents the energy a particle has due to its motion. In this exercise, the kinetic energy of the positron is given as 2.0 keV (kilo-electronvolts). To work with this in our calculations, we convert keV to Joules. By knowing that 1eV = 1.6 x 10^-19 Joules, the kinetic energy becomes 3.2 x 10^-16 Joules. The kinetic energy is crucial because it allows us to determine the velocity of the positron using the formula: \[ K = \frac{1}{2} mv^2 \]. This highlights that kinetic energy is proportional to the square of the velocity.
Understanding the Magnetic Field
The magnetic field ( B ) affects the motion of charged particles like positrons. The strength of the magnetic field is given in Teslas (T), and in this problem, the magnetic field is 0.10 T.The main role of the magnetic field is to exert a force perpendicular to the positron's motion, causing it to move in a helical path. This force is called the Lorentz force, calculated using: \( F = q( v \times B) \), where q is the charge of the positron.
Exploring Positron Dynamics
The positron, a positively charged particle, behaves in a unique manner in a magnetic field. Positrons move in a helical path because their velocity has two components: one parallel and one perpendicular to the field. The parallel component allows the positron to move forward, while the perpendicular component makes it follow a circular path. The interaction between the magnetic field and the positron results in a helical motion, combining both linear and circular movement.
Velocity Components
Velocity components are critical in understanding helical motion. The total velocity ( v ) of the positron can be split into perpendicular ( v_\perp ) and parallel ( v_\parallel ) to the magnetic field direction. This is done using trigonometric functions:\( v_\parallel = v cos( \theta ) \)and\( v_\perp = v sin( \theta ) \). Theta is the angle between the velocity vector and the magnetic field, which is 89掳. The nearly perpendicular orientation results in a small v_\parallel and a large v_\perp.
Period of Motion
The period ( T ) refers to the time it takes for the positron to complete one full cycle of its circular path. It is independent of the velocity and is calculated using the formula: \[ T = \frac{2 pi m}{eB} \].inserting the values for mass (m), charge (e), and magnetic field (B), the period comes out to be 3.58 x 10^-10 seconds. This value represents the time interval for one circular revolution.
Calculating the Pitch
The pitch ( p ) of the helical path measures the distance along the magnetic field direction (longitudinal displacement) during one complete cycle of circular motion. It is calculated by multiplying the parallel component of the velocity by the period:\( p = v_\parallel T \). With v_\parallel around 1.47 x 10^5 m/s and T approximately 3.58 x 10^-10 sec, the pitch is approximately 5.26 x 10^-5 meters. This small pitch reflects that the positron moves forward very slowly compared to its circular motion.
Radius of Helical Path
The radius ( r ) of the helical path determines the size of the circles the positron makes due to the magnetic field. This radius is given by: \( r = \frac{m v_\perp}{eB} \). Using the given value of the magnetic field and derived values of mass and perpendicular velocity, the radius calculates to approximately 4.78 x 10^-3 meters (4.78 mm). This gives us a clear picture of how tightly coiled the helical path is.

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Most popular questions from this chapter

At One Instant A proton travels through uniform magnetic and electric fields. The magnetic field is \(\vec{B}=(-2.5 \mathrm{mT}) \hat{\mathrm{i}}\). At one instant the velocity of the proton is \(\vec{v}=(2000 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) \((4.0 \mathrm{~V} / \mathrm{m}) \hat{\mathrm{k}}\) and \((\mathrm{b})(4.0 \mathrm{~V} / \mathrm{m}) \mathrm{i}\) ?

Three Force Fields We have studied three long-range forces: gravity, electricity, and magnetism. Compare and contrast these three forces giving at least one feature that all three forces have in common, and at least one feature that distinguishes each force.

Horizontal Motion An electron with kinetic energy \(2.5 \mathrm{keV}\) moves horizontally into a region of space in which there is a downward-directed uniform electric field of magnitude \(10 \mathrm{kV} / \mathrm{m}\). (a) What are the magnitude and direction of the (smallest) uniform magnetic field that will cause the electron to continue to move horizontally? Ignore the gravitational force, which is small. (b) Is it possible for a proton to pass through the combination of fields undeflected? If so, under what circumstances?

Zero Total Force A closed wire loop with current \(i\) is in a uni. form magnetic field \(\vec{B}\), with the plane of the loop at angle \(\theta\) to the direction of \(\vec{B}\). Show that the total magnetic force on the loop is zero. Does your proof also hold for a nonuniform magnetic field?

Circular Path An alpha particle \((q=+2 e, m=4.00 \mathrm{u})\) travels in a circular path of radius \(4.50 \mathrm{~cm}\) in a uniform magnetic field with magnitude \(B=1.20 \mathrm{~T}\). Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy in electron-volts, and (d) the potential difference through which it would have to be accelerated to achieve this energy.
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