Problem 24 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

The magnitude of the magnetic field is approximately 1.67 T.

Step 1: 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.

Step 2: 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 \(d = 1.18 \; \text{cm}\).

Step 3: Calculate the Radius of the Circular Path

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

Step 4: 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: \[ qvB = \frac{mv^2}{R} \].

Step 5: 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 \( B \approx 1.67 \; \text{T} \).

Key Concepts

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’s 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.