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If the magnitude of the magnetic force on a proton is \(F\) when it is moving at \(15.0^{\circ}\) with respect to the field, what is the magnitude of the force (in terms of \(F\) ) when this charge is moving at \(30.0^{\circ}\) with respect to the field?

Short Answer

Expert verified
The new force is approximately 1.933 times the original force.

Step by step solution

01

Understanding Magnetic Force on a Moving Charge

The formula for the magnetic force on a charged particle moving in a magnetic field is given by \( F = qvB\sin(\theta) \), where \( q \) is the charge, \( v \) is the velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.
02

Expressing Initial Force

For the angle \( 15.0^{\circ} \), we have \( F = qvB\sin(15.0^{\circ}) \). This force is given to be \( F \). So, \( F = qvB\sin(15.0^{\circ}) \).
03

Determine New Magnetic Force at New Angle

Now consider when the proton is moving at an angle \( 30.0^{\circ} \). The magnetic force becomes \( F_{30} = qvB\sin(30.0^{\circ}) \).
04

Calculating the Force Ratio

The ratio of magnetic forces at \( 30.0^{\circ} \) and \( 15.0^{\circ} \) is \( \frac{F_{30}}{F} = \frac{qvB\sin(30.0^{\circ})}{qvB\sin(15.0^{\circ})} \). Since \( qvB \) is common, it cancels out. The resulting ratio is \( \frac{\sin(30.0^{\circ})}{\sin(15.0^{\circ})} \).
05

Calculate Sine Values

Calculate the sine values: \( \sin(30.0^{\circ}) = 0.5 \) and \( \sin(15.0^{\circ}) \approx 0.2588 \).
06

Final Calculation of Force Ratio

The force ratio is \( \frac{0.5}{0.2588} \approx 1.933 \). Hence, \( F_{30} = 1.933 \times F \).

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Key Concepts

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

Proton
A proton is a subatomic particle found within the nucleus of an atom. It carries a positive charge, denoted as \( +e \), where \( e = 1.6 \times 10^{-19} \) Coulombs, and has a mass of about \( 1.67 \times 10^{-27} \) kilograms. These positively charged particles are crucial in many physical interactions, such as electromagnetic forces.
When a proton moves through a magnetic field, it experiences a magnetic force. This force arises because of the proton's charge and velocity relative to the magnetic field.
  • The direction and magnitude of this force are entirely dependent on the charge, the speed at which the proton travels, and its angle to the field.
  • Protons are used in various scientific applications like MRI machines and particle accelerators due to their predictable interaction with magnetic fields.
Angle of Velocity to Magnetic Field
The angle between the velocity of a charged particle, like a proton, and the magnetic field is a critical factor in calculating the magnetic force it experiences. This angle, denoted as \( \theta \), influences how much of the velocity is effectively interacting with the magnetic field.
When a proton travels at different angles to a magnetic field:
  • The magnetic force is zero if the proton moves parallel (\( \theta = 0^{\circ} \)) or antiparallel (\( \theta = 180^{\circ} \)) to the field.
  • The force is maximal when the proton's path is perpendicular (\( \theta = 90^{\circ} \)) to the magnetic field.
  • At an angle like \( 15^{\circ} \) or \( 30^{\circ} \), the magnetic force varies based on the sine of the angle, as per the equation \( F = qvB\sin(\theta) \).
The sine function is used here because it describes the perpendicular component of the velocity that contributes to the magnetic force. Thus, the angle profoundly affects the magnitude of the force experienced.
Force Ratio
In situations where a proton changes its angle relative to a magnetic field, the force ratio becomes a valuable tool for comparison. This ratio tells us how the magnetic force changes with alterations in the angle of velocity.
Given two different angles, for instance, \( 15^{\circ} \) and \( 30^{\circ} \), the force experienced by the proton can vary significantly.
  • The force ratio is determined by taking the sine of both angles and forming a ratio of these values.
  • Thus, for angles of \( 15^{\circ} \) and \( 30^{\circ} \), the force ratio \( \frac{F_{30}}{F} \) is calculated as \( \frac{\sin(30^{\circ})}{\sin(15^{\circ})} \).
Since the sine values for \( 30^{\circ} \) and \( 15^{\circ} \) are known (\( 0.5 \) and approximately \( 0.2588 \), respectively), the force ratio comes out to roughly \( 1.933 \). This result indicates that the magnetic force when the proton moves at \( 30^{\circ} \) is about 1.933 times stronger than at \( 15^{\circ} \). Understanding this ratio is crucial for predicting and explaining the behavior of charged particles in magnetic fields.
Magnetic Field Strength
Magnetic field strength, denoted as \( B \), plays a pivotal role in determining the force exerted on charged particles like protons. It is measured in Tesla (T) and quantifies the field's intensity or power.
In practical terms:
  • A strong magnetic field results in a significant force exerted on a moving charged particle, thereby altering its path more effectively.
  • Conversely, a weaker field exerts less influence over the charged particle's trajectory.
As per the magnetic force equation \( F = qvB\sin(\theta) \), the force is directly proportional to the strength of the magnetic field. This means that any increase in \( B \) will linearly increase the magnitude of force experienced by the proton or any charged particle involved. Magnetic field strength is a key parameter in a wide array of scientific and engineering applications, from medical imaging to data storage technologies.

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Most popular questions from this chapter

\bullet A circular coil of 50 loops and diameter 20.0 \(\mathrm{cm}\) is lying flat on a tabletop, and carries a clockwise current of 2.50 A. A magnetic field of 0.450 \(\mathrm{T}\) , directed to the north and at an angle of \(45.0^{\circ}\) from the vertical down through the coil and into the tabletop is turned on. (a) What is the torque on the coil, and (b) which side of the coil (north or south) will tend to rise from the tabletop?

\(\bullet\) When a certain paramagnetic material is placed in an external magnetic field of 1.5000 \(\mathrm{T}\) , the field inside the material is measured to be 1.5023 \(\mathrm{T}\) . Find (a) the relative permeability and (b) the magnetic permeability of this material.

A deuteron particle (the nucleus of an isotope of hydrogen consisting of one proton and one neutron and having a mass of \(3.34 \times 10^{-27} \mathrm{kg}\) ) moving horizontally enters a uniform, vertical, 0.500 T magnetic field and follows a circular arc of radius 55.6 \(\mathrm{cm} .\) (a) How fast was this deuteron moving just before it entered the magnetic field and just after it came out of the field? (b) What would be the radius of the arc followed by a proton that entered the field with the same velocity as the deuteron?

\(\bullet\) A beam of protons is accelerated through a potential dif- ference of 0.745 \(\mathrm{kV}\) and then enters a uniform magnetic field traveling perpendicular to the field. (a) What magnitude of field is needed to bend these protons in a circular arc of diameter 1.75 \(\mathrm{m} ?\) (b) What magnetic field would be needed to produce a path with the same diameter if the particles were electrons having the same speed as the protons?

\(\bullet\) A beam of protons traveling at 1.20 \(\mathrm{km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field in a direction perpendicular to its original direction (Fig. \(20.60 ) .\) The beam travels a distance of 1.18 cm while in the field. What is the magnitude of the magnetic field?

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