91Ó°ÊÓ

You measure the charge-to-mass ratio \(q / m\) for a particle with positive charge in the following way: The particle starts from rest, is accelerated through a potential difference \(\Delta V,\) and attains a velocity with magnitude \(v .\) It then enters a region of uniform magnetic field \(B=0.200 \mathrm{~T}\) that is directed perpendicular to the velocity; the particle moves in a path that is an arc of a circle of radius \(R .\) You measure \(R\) as a function of \(\Delta V\). You plot your data as \(R^{2}\) (in units of \(\mathrm{m}^{2}\) ) versus \(\Delta V\) (in \(\mathrm{V}\) ) and find that the values lie close to a straight line that has slope \(1.04 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{V} .\) What is the value of \(q / m\) for this particle?

Step by step solution

01

Accelerating the particle

First, the particle is accelerated from rest through a potential difference \(\Delta V\). The energy gained by the particle is equal to the work done on it, which is \(q \Delta V\). This energy is all converted into kinetic energy as non-conservative forces are not doing work, thus \(q \Delta V = 1/2 m v^2\). Solving for \(v\), we get \(v = \sqrt{2q \Delta V/m}\).

02

Travelling through magnetic field

Next, the particle enters a region of uniform magnetic field \(B\), where it moves along a circular path of radius \(R\). The force acting on the particle due to the magnetic field is what keeps it moving in a circle. The magnetic force is given by \(qvB = m v^2 / R\), from which we can solve for \(R\) and get \(R = m v / (q B)\).

03

Relating linear distance and potential difference

Now, data is plotted as \(R^2\) versus \(\Delta V\) and it is given that the plot is approximately a straight line with slope \(1.04 \times 10^{-6}\). This means \(R^2 = k \Delta V\) where \(k\) is the slope equals to \(1.04 \times 10^{-6}\). Substituting \(R\) from the previous step and squaring, we get \(m^2 v^2 /(q^2 B^2) = k \Delta V\).

04

Calculating q/m

Substituting \(v\) from step 1 into the equation from step 3, we get \(m^2 2q \Delta V / m^2 q^2 B^2 = k \Delta V\). Simplifying this, we get \(2 / (qB) = k\). Solving for \(q / m\), we get \(q / m = 2 / (kB) = 2 / (1.04 \times 10^{-6} \times 0.2) = 9569.23 \, C/kg\).

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Key Concepts

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

Potential Difference

In physical terms, potential difference, often referred to as voltage, is the amount of work needed to move a charge from one point to another in an electric field.
It dictates how much energy a charge will gain or lose as it moves within the field. In our exercise, the potential difference (\( \Delta V \)) is what accelerates the particle from rest.
When a particle with charge \( q \) is accelerated through this potential difference, it gains an amount of energy equal to \( q \Delta V \).
This gained energy is then converted into kinetic energy, the energy associated with motion. This relationship gives rise to the equation \( q \Delta V = \frac{1}{2} mv^2 \), linking potential difference to velocity, mass, and charge of the particle.

Magnetic Field

A magnetic field is an invisible field that exerts force on electric charges that are in motion, like our particle.
In this exercise, the particle, after being accelerated, moves into a region where a uniform magnetic field (\( B \)) of magnitude 0.200 T (Tesla) is present. This magnetic field interacts with the moving particle, causing it to travel in a circular path.
The force exerted by the magnetic field is known as the Lorentz force and it is calculated as \( qvB \), where \( q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field strength.
This force is always perpendicular to the direction of velocity, creating centripetal motion that keeps the particle moving in a circle.

Kinetic Energy

Kinetic energy (KE) is the energy that an object possesses due to its motion.
For a particle starting from rest and being accelerated, as in our exercise, the kinetic energy is given by \( KE = \frac{1}{2} mv^2 \).
This expression denotes that the kinetic energy is directly proportional to the mass \( m \) and the square of the velocity \( v \) of the particle.
After the particle is accelerated by the potential difference, all the energy it gains is converted into kinetic energy.
Therefore, the relationship \( q \Delta V = \frac{1}{2} mv^2 \) can be used to determine the velocity, and thus kinetic energy, gained by the particle.
This essential concept explains how energy conservation guides particle acceleration.

Circular Motion

When a charged particle moves into a perpendicular magnetic field, it experiences circular motion.
This is due to the continuous force acting at right angles, the result of a magnetic field on a moving charge.
For circular motion to occur, the force exerted on the particle must be centripetal, pointing towards the center of the circular path.
In our problem, this is achieved by the magnetic force \( qvB \), which equates to the required centripetal force \( \frac{mv^2}{R} \).
Solving for the radius \( R \) gives \( R = \frac{mv}{qB} \).
This important relationship allows us to relate the path radius and magnetic field strength to the velocity and charge of the particle, playing a crucial role in the determination of the charge-to-mass ratio.