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Chapter 22: Problem 27

A proton with a kinetic energy of \(4.9 \times 10^{-16}\) J moves perpendicular to a magnetic field of 0.26 T. What is the radius of its circular path?

Short Answer

Expert verified
The radius of the proton's circular path is approximately 0.122 m.

Step by step solution

01

Convert Kinetic Energy to Velocity

The kinetic energy (\(KE\)) of the proton is given by \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the proton (\(1.67 \times 10^{-27} \text{ kg}\)) and \(v\) is its velocity. Rearrange the formula to solve for \(v\): \(v = \sqrt{\frac{2KE}{m}}\). Substitute the known values: \(v = \sqrt{\frac{2 \times 4.9 \times 10^{-16}}{1.67 \times 10^{-27}}}\). Calculate \(v\).
02

Determine Radius of Circular Path

The radius \(r\) of the circular path of a charged particle in a magnetic field is given by \(r = \frac{mv}{qB}\), where \(m\) is the mass, \(v\) is the velocity, \(q\) is the charge of the proton (\(1.6 \times 10^{-19} \text{ C}\)), and \(B\) is the magnetic field. Substitute for \(v\) from Step 1 and the given values: \(r = \frac{(1.67 \times 10^{-27}) \times v}{1.6 \times 10^{-19} \times 0.26}\). Calculate \(r\).
03

Calculate Velocity

For the velocity, substitute previous step's equation: \(v = \sqrt{\frac{2 \times 4.9 \times 10^{-16}}{1.67 \times 10^{-27}}}\). On calculation, the velocity \(v\) is approximately \(3.09 \times 10^7 \text{ m/s}\).
04

Calculate Radius

Substitute the calculated velocity into the equation for the radius: \(r = \frac{(1.67 \times 10^{-27}) \times 3.09 \times 10^7}{1.6 \times 10^{-19} \times 0.26}\). Solving this gives \(r \approx 0.122 \text{ m}\).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It can be calculated using the formula:
  • \[ KE = \frac{1}{2}mv^2 \]
where \( m \) is the mass of the object and \( v \) is its velocity.
In our context, the kinetic energy is used to find the velocity of a proton, given by \( 4.9 \times 10^{-16} \) joules.
To do this, we rearrange the kinetic energy formula to solve for velocity:
  • \[ v = \sqrt{\frac{2KE}{m}} \]
By substituting the proton's known mass \( 1.67 \times 10^{-27} \text{ kg} \), we can compute the velocity.
Understanding kinetic energy is crucial for determining how fast the proton is moving as it enters the magnetic field.
Magnetic Field
A magnetic field is a region where a magnetic force is exerted on moving charge particles, such as protons. The strength of the magnetic field is denoted by the symbol \( B \) and measured in Teslas (T).
For this exercise, the proton experiences a magnetic field strength of 0.26 T, which is perpendicular to its motion.
When a charged particle like a proton enters a magnetic field at right angles, it experiences a force that changes the direction of its motion, leading to circular motion.
  • The force exerted on a charged particle by a magnetic field is calculated using the Lorentz force formula: \[ F = qvB \]where \( q \) is the charge of the proton and \( v \) is its velocity.
This force is responsible for the circular path the proton follows and depends linearly on the magnetic field strength \( B \).
Thus, understanding the magnetic field helps us predict and calculate the proton's motion in the magnetic environment.
Circular Motion
When a charged particle, like a proton, travels through a magnetic field, it can undergo circular motion. This occurs because the magnetic force acts perpendicular to its velocity, causing it to change direction rather than speed.
The radius \( r \) of this circular path can be calculated with the formula:
  • \[ r = \frac{mv}{qB} \]
Here, \( m \) is the mass of the proton, \( v \) is its velocity, \( q \) is its charge, and \( B \) is the magnetic field's strength.
In this given problem, substituting the known values allows us to find the radius of about 0.122 m.
The concept of circular motion is vital to understanding how magnetic fields influence charged particles, ensuring their motion is not linear but curved.
Circular motion, therefore, provides a way to measure the effects of a magnetic field on moving charges.

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Most popular questions from this chapter

In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius \(5.29 \times 10^{-11} \mathrm{m}\) about the nucleus. Given that the charge on the electron is \(-1.60 \times 10^{-19} \mathrm{C}\), and that its speed is \(2.2 \times 10^{6} \mathrm{m} / \mathrm{s},\) find the magnitude of the magnetic field the electron produces at the nucleus of the atom.

A long, straight wire carries a current of 7.2 A. How far from this wire is the magnetic field it produces equal to the Earth's magnetic field, which is approximately \(5.0 \times 10^{-5}\) T?

Two long, straight wires are separated by a distance of \(9.25 \mathrm{cm} .\) One wire carries a current of \(2.75 \mathrm{A},\) the other carries a current of 4.33 A. (a) Find the force per meter exerted on the 2.75-A wire. (b) Is the force per meter exerted on the 4.33-A wire greater than, less than, or the same as the force per meter exerted on the \(2.75-\mathrm{A}\) wire? Explain.

Two charged particles with different speeds move one at a time through a region of uniform magnetic field. The particles move in the same direction and experience equal magnetic forces. (a) If particle 1 has four times the charge of particle 2 , which particle has the greater speed? Explain. (b) Find the ratio of the speeds, \(v_{1} / v_{2}\).

Medical X-rays An electron in a medical X-ray machine is accelerated from rest through a voltage of \(10.0 \mathrm{kV}\). (a) Find the maximum force a magnetic field of \(0.957 \mathrm{T}\) can exert on this electron. (b) If the voltage of the \(X\) -ray machine is increased, does the maximum force found in part (a) increase, decrease, or stay the same? Explain. (c) Repeat part (a) for an electron accelerated through a potential of \(25.0 \mathrm{kV}\).
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