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Chapter 27: Problem 24

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, leaving the field in a direction perpendicular to its original direction (Fig. E27.24). The beam travels a distance of 1.18 \(\mathrm{cm}\) while in the field. What is the magnitude of the magnetic field?

Short Answer

Expert verified
The magnitude of the magnetic field is approximately 1.67 T.

Step by step solution

01

Understand the Circular Path

When charged particles like protons enter a magnetic field perpendicularly, they move in a circular path due to the Lorenz force acting as the centripetal force. In this scenario, the protons exit perpendicular to their initial direction, implying they have completed a quarter of a circular path.
02

Relate Distance to the Radius

The distance traveled by the protons while in the magnetic field corresponds to one-quarter of the circumference of the circle. So, the relation is given by: \(d = \frac{1}{4} (2\pi R)\), where \(d = 1.18 \; \text{cm}\).
03

Calculate the Radius of the Circular Path

By rearranging the relation from Step 2, we find the radius \(R\): \[ R = \frac{d}{\frac{1}{2}\pi} = \frac{1.18 \times 10^{-2}}{\frac{\pi}{2}} \approx 7.51 \times 10^{-3} \; \text{m} \].
04

Use Centripetal Force Formula

The centripetal force is provided by the magnetic force, which is \( F = qvB \). This force equals the centripetal force \( F = \frac{mv^2}{R} \). Set these equal: \[ qvB = \frac{mv^2}{R} \].
05

Solve for Magnetic Field B

Rearrange the equation from Step 4 to solve for \(B\): \[ B = \frac{mv}{qR} \]. For the proton, \(m = 1.67 \times 10^{-27} \; \text{kg}\), \(v = 1.20 \times 10^3 \; \text{m/s}\), \(q = 1.60 \times 10^{-19} \; \text{C}\), and \(R = 7.51 \times 10^{-3} \; \text{m}\). Substitute these values to find \( B \approx 1.67 \; \text{T} \).

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Key Concepts

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

Lorentz Force
When protons, or any charged particles, enter a magnetic field, they're subjected to the Lorentz Force. This force causes them to move in a specific way. It's important to remember that this force acts perpendicular to both the velocity of the particles and the direction of the magnetic field. This perpendicularity triggers the curved paths we observe.

The size of this force is represented by the formula: \( F = qvB \). Here, \( q \) is the charge of the particle, \( v \) is its velocity, and \( B \) is the magnetic field. Let's break it down further:
  • In our case, the protons have a positive charge, making the direction of force predictable using the right-hand rule.
  • The force doesn't change the speed of the particle, only its direction, bending it into a circular path.
This phenomenon is crucial for many technologies, like cyclotrons used in particle physics.
Centripetal Force
The circular motion of charged particles like protons in a magnetic field relies on centripetal force. This force keeps the particles from flying off tangentially by constantly pulling them towards the circle鈥檚 center. In our exercise, it's the Lorentz Force fulfilling the role of centripetal force.

We use the formula \( F_c = \frac{mv^2}{R} \) for centripetal force:
  • \( m \) is the mass of the particle.
  • \( v \) is the velocity, and \( R \) is the radius of the circle.
With the Lorentz force acting as this centripetal force, we set: \( qvB = \frac{mv^2}{R} \). This equation helps connect the magnetic field's properties to the particle's path.
Circular Motion
When the protons enter the magnetic field perpendicularly, they start moving in circular motion. Imagine the protons following the circumference of a circle. The path they take is ultimately determined by the Lorentz force acting as the centripetal force.

In the context of our exercise:
  • The distance traveled within the field equals one-quarter of the circle's circumference. This is essential for calculating the radius of the circular path.
  • To compute the radius \( R \), use \( R = \frac{d}{\frac{1}{2}\pi} \), where \( d \) is the known distance the protons traveled (1.18 cm).
  • Knowing the radius and velocity allows calculating the magnetic field strength using the rearranged centripetal equation: \( B = \frac{mv}{qR} \). This results in \( B \approx 1.67 \; \text{T} \).
Understanding this circular motion is pivotal to many areas of physics and technology, including designing instruments that control charged particles.

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Most popular questions from this chapter

A circular loop of wire with area \(A\) lies in the \(x y\) -plane. As viewed along the \(z\) -axis looking in the \(- z\) -direction toward the origin, a current \(I\) is circulating clockwise around the loop. The torque produced by an external magnetic field \(\vec { B }\) is given by \(\vec { \tau } = D ( 4 \hat { \imath } - 3 \hat { \jmath } ) ,\) where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U = - \vec { \mu } \cdot B\) is negative. The magnitude of the magnetic field is \(B _ { 0 } = 13 D / I A\) . (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B _ { x } , B _ { y }\) and \(B _ { z }\) of \(\vec { B } .\)

A deuteron (the nucleus of an isotope of hydrogen) has a mass of \(3.34 \times 10^{-27} \mathrm{kg}\) and a charge of \(+e .\) The deuteron travels in a circular path with a radius of 6.96 \(\mathrm{mm}\) in a magnetic field with magnitude 2.50 \(\mathrm{T}\) (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconducting electromagnet of the cyclotron produces a 2.9 -T magnetic field perpendicular to the proton orbits. (a) When the protons have achieved a kinetic energy of 2.7\(\mathrm { MeV }\) what is the radius of their circular orbit and what is their angular speed? (b) Repeat part (a) when the protons have achieved their final kinetic energy of 5.4\(\mathrm { MeV }\) .

A straight, \(2.5-\mathrm{m}\) wire carries a typical household current of 1.5 \(\mathrm{A}\) (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects undernormal household conditions?

A particle with charge \(-5.60 \mathrm{nC}\) is moving in a uniform magnetic field \(\vec{\boldsymbol{B}}=-(1.25 \mathrm{T}) \hat{\boldsymbol{k}}\) . The magnetic force on the particle is measured to be \(\vec{\boldsymbol{F}}=-\left(3.40 \times 10^{-7} \mathrm{N}\right) \hat{\imath}+(7.40 \times\) \(10^{-7} \mathrm{N} ) \hat{J}\) . (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{v} \cdot\) What is the angle between \(\vec{v}\) and \(\vec{F} ?\)
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