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Magazine Chapter 21: Problem 16
A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.50 T. The ficld is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron?
Short Answer
Expert verified
The new magnetic field magnitude is approximately 0.00273 T.
Step by step solution
01
Understand the Problem
We need to calculate the new magnetic field strength that allows an electron, with the same speed as the proton, to travel in a circular path with the same radius. The initial magnetic field for the proton is given as 0.50 T.
02
Recall Relevant Formula
The radius of the circular path for a charged particle moving in a magnetic field is given by the equation \( r = \frac{mv}{qB} \), where \( r \) is the radius, \( m \) is the mass of the particle, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field strength.
03
Set Up Equations for Proton and Electron
For the proton, the equation becomes \( r_{p} = \frac{m_{p}v}{q_{p}B_{p}} \). For the electron, it becomes \( r_{e} = \frac{m_{e}v}{q_{e}B_{e}} \). Since \( r_{p} = r_{e} \) and \( v \) is the same for both, we equate the expressions.
04
Equate and Simplify
Equate the two radius formulas: \( \frac{m_{p}v}{q_{p}B_{p}} = \frac{m_{e}v}{q_{e}B_{e}} \). Simplifying gives \( \frac{m_{p}}{q_{p}B_{p}} = \frac{m_{e}}{q_{e}B_{e}} \).
05
Isolate the Magnetic Field for Electron
Rearrange the equation to solve for \( B_{e} \): \( B_{e} = \frac{m_{e}q_{p}B_{p}}{m_{p}q_{e}} \).
06
Use Known Values
Substitute known values: \( m_{p} = 1.67 \times 10^{-27} \) kg, \( m_{e} = 9.11 \times 10^{-31} \) kg, \( q_{p} = 1.6 \times 10^{-19} \) C, \( q_{e} = -1.6 \times 10^{-19} \) C, \( B_{p} = 0.50 \) T. The negative of the electron's charge cancels as we consider magnitudes: \( B_{e} = \frac{9.11 \times 10^{-31} \times 1.6 \times 10^{-19} \times 0.50}{1.67 \times 10^{-27} \times 1.6 \times 10^{-19}} \).
07
Calculate the Solution
Calculate the expression: \( B_{e} = \frac{9.11 \times 0.50}{1.67} \approx 0.00273 \) T.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Charged Particle Motion
When a charged particle moves through a magnetic field, it experiences a magnetic force. This force acts perpendicular to the direction of both the magnetic field and the particle's velocity. As a result, the charged particle tends to follow a curved path. This principle is fundamental to the behavior of charged particles under magnetic influence.
The motion of charged particles is described by several factors:
The motion of charged particles is described by several factors:
- Direction: The right-hand rule helps determine the direction of force on a positive charge. For electrons, which are negatively charged, the shift is in the opposite direction.
- Velocity: A constant velocity results in a circular motion, while changing velocities can result in varying paths.
- Mass and charge: Heavier particles or those with greater charge require a stronger magnetic field to follow the same path as lighter or less charged counterparts.
Radius of Circular Path
The radius of the path a charged particle takes in a magnetic field depends on several factors. An essential formula used to determine this is:
This formula helps solve problems where similarly charged particles like protons and electrons are compared in various fields.
- \[ r = \frac{mv}{qB} \]
- Where:
- \( r \) is the radius of the circular path
- \( m \) is the mass of the particle
- \( v \) is the velocity of the particle
- \( q \) is the charge of the particle
- \( B \) is the magnetic field strength
This formula helps solve problems where similarly charged particles like protons and electrons are compared in various fields.
Proton
Protons are positively charged particles found in the nucleus of an atom. They have a relatively larger mass compared to electrons. In the context of magnetic fields:
- Charge: The charge of a proton is \( 1.6 \times 10^{-19} \) Coulombs.
- Mass: A proton's mass is approximately \( 1.67 \times 10^{-27} \) kilograms.
- Motion in Magnetic Fields: When a proton enters a magnetic field perpendicularly, it follows a circular path. The radius of this path can be calculated using the formula discussed. As protons are heavier than electrons, for a given velocity and magnetic field, they require a stronger field to achieve the same path radius as electrons.
Electron
Electrons are negatively charged particles that orbit the nucleus of an atom. They are much lighter than protons, which significantly impacts their behavior in a magnetic field.
- Charge: Electrons have a charge of \(-1.6 \times 10^{-19}\) Coulombs. In calculations, we use the absolute value.
- Mass: An electron's mass is about \( 9.11 \times 10^{-31} \) kilograms, making them significantly lighter than protons.
- Motion in Magnetic Fields: When projected into a magnetic field, electrons will follow a circular path much smaller in radius compared to that of protons, given the same field strength and velocity. This smaller mass and same charge density result in a distinct path characteristic that must be carefully calculated when solving problems.
