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Chapter 5: Problem 9

A proton moves at \(7.50 \times 10^{7} \mathrm{~m} / \mathrm{s}\) perpendicular to a magnetic field. The field causes the proton to travel in a circular path of radius \(0.800 \mathrm{~m}\). What is the field strength?

Short Answer

Expert verified
The magnetic field strength is approximately \( 0.783 \text{ T} \).

Step by step solution

01

Identify Given Information

We know the velocity of the proton is \( v = 7.50 \times 10^{7} \text{ m/s} \) and the radius \( r = 0.800 \text{ m} \). We need to find the magnetic field strength \( B \).
02

Use the Formula for Magnetic Force

The magnetic force acting on a charged particle in a magnetic field is given by \( F = qvB \sin \theta \). Here, \( \theta = 90^\circ \) since the velocity is perpendicular to the field, so \( \sin \theta = 1 \). Thus, \( F = qvB \).
03

Relate Magnetic Force to Centripetal Force

For circular motion, the magnetic force is the centripetal force keeping the proton in a circular path. Therefore, \( qvB = \frac{mv^2}{r} \), where \( m \) is the mass of the proton and \( v \) is its velocity.
04

Solve for Magnetic Field Strength

Rearrange the formula to solve for the magnetic field strength \( B \): \( B = \frac{mv}{qr} \).
05

Substitute Known Values

The charge of a proton is \( q = 1.602 \times 10^{-19} \text{ C} \) and the mass of a proton is \( m = 1.673 \times 10^{-27} \text{ kg} \). Substituting these values into the equation, we get:\[ B = \frac{(1.673 \times 10^{-27} \text{ kg})(7.50 \times 10^{7} \text{ m/s})}{(1.602 \times 10^{-19} \text{ C})(0.800 \text{ m})} \]
06

Calculate the Result

By calculating the expression, we find \( B \approx 0.783 \text{ T} \) (Tesla).

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Key Concepts

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

Magnetic Force
Magnetic force is created when charged particles, like protons, move through a magnetic field. This force is important in many applications, from electric motors to MRI machines. The magnetic force experienced by a charged particle depends on three key things: the particle's charge, its velocity, and the strength of the magnetic field it moves through.
This relationship is mathematically expressed by the equation:
  • \( F = qvB \sin \theta \)
Here, \(F\) stands for the magnetic force, \(q\) is the particle鈥檚 charge, \(v\) its velocity, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the velocity and the magnetic field direction.
In cases where the velocity is perpendicular to the magnetic field, the angle \(\theta\) is 90 degrees, meaning that \(\sin\theta = 1\). This simplifies the formula to:
  • \( F = qvB \)
This formula helps us understand how a magnetic field can influence a moving charge, dictating its path and direction.
Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path. It acts inward, towards the center of the circle. For objects moving in a circle with constant speed, this force is essential to maintaining circular motion. Without it, an object would move off in a straight line due to inertia.
For a proton moving in a magnetic field, the magnetic force acts as the centripetal force. The relationship between centripetal force \(F_c\), mass \(m\), velocity \(v\), and radius \(r\) is given by:
  • \( F_c = \frac{mv^2}{r} \)
When a proton moves in a circle under the influence of a magnetic field, its centripetal force is provided by the magnetic force. Therefore, the equation becomes:
  • \( qvB = \frac{mv^2}{r} \)
Understanding this equivalence is crucial for analyzing the motion of charged particles in magnetic fields, as it shows how magnetic forces result in stable circular paths.
Proton Motion
Proton motion in a magnetic field is a fascinating example of physics in action. When in a magnetic field, a proton's path can bend and circle due to the magnetic force. This is crucial in fields like particle physics, where we study the properties and behaviors of elementary particles.
In this exercise, we have a proton traveling perpendicularly to a magnetic field. This causes the proton to move in a circular path, a scenario where we often think about rotational motion and magnetic effects. The radius of this path can tell us a lot about the strength of the magnetic field and the proton's interactions with it.
To solve for the magnetic field strength \(B\), we leverage its path characterized by velocity \(v\) and radius \(r\). By using the formula:
  • \( B = \frac{mv}{qr} \)
We substitute known values for the proton's charge \(q\) and mass \(m\) to find the field strength. This shows how the magnetic field strength directly influences the motion and path of subatomic particles like protons.

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Most popular questions from this chapter

High-velocity charged particles can damage biological cells and are a component of radiation exposure in a variety of locations ranging from research facilities to natural background. Describe how you could use a magnetic field to shield yourself.

(a) Find the charge on a baseball, thrown at \(35.0 \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's \(5.00 \times 10^{-5} \mathrm{~T}\) field, that experiences a 1.00-N magnetic force. (b) What is unreasonable about this result? (c) Which assumption or premise is responsible?

A cosmic ray electron moves at \(7.50 \times 10^{6} \mathrm{~m} / \mathrm{s}\) perpendicular to the Earth's magnetic field at an altitude where field strength is \(1.00 \times 10^{-5} \mathrm{~T}\). What is the radius of the circular path the electron follows?

(a) A 0.750-m-long section of cable carrying current to a car starter motor makes an angle of \(60^{\circ}\) with the Earth's \(5.50 \times 10^{-5} \mathrm{~T}\) field. What is the current when the wire experiences a force of \(7.00 \times 10^{-3} \mathrm{~N} ?(\mathrm{~b})\) If you run the wire between the poles of a strong horseshoe magnet, subjecting \(5.00 \mathrm{~cm}\) of it to a \(1.75-\mathrm{T}\) field, what force is exerted on this segment of wire?

An electron moving at \(4.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\) in a 1.25-T magnetic field experiences a magnetic force of \(1.40 \times 10^{-16} \mathrm{~N}\). What angle does the velocity of the electron make with the magnetic field? There are two answers. (a) A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less than \(1.00 \times 10^{-12} \mathrm{~N}\). What is the greatest the charge can be if it moves at a maximum speed of \(30.0 \mathrm{~m} / \mathrm{s}\) in the Earth's field? (b) Discuss whether it would be difficult to limit the charge to less than the value found in (a) by comparing it with typical static electricity and noting that static is often absent.
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