Problem 55 - University Physics with Modern Physics

Question

Two identical circular, wire loops 40.0 \(\mathrm{cm}\) in diameter each carry a current of 1.50 \(\mathrm{A}\) in the same direction. These loops are parallel to each other and are 25.0 \(\mathrm{cm}\) apart. Line \(a b\) is normal to the plane of the loops and passes through their centers. A proton is fired at 2400 \(\mathrm{km} / \mathrm{sperpendicular}\) to line \(a b\) from a point midway between the centers of the loops. Find the magnitude and direction of the magnetic force these loops exert on the proton just after it is fired.

Step-by-Step Solution

Answer

The magnitude of the force is approximately \( 3.8 \times 10^{-16} \) N, directed radially outward.

Step 1: Calculate the Magnetic Field from One Loop

To find the magnetic field at the center, we use the formula for the magnetic field along the axis of a current loop: \[ B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \] where \( R = 0.2 \) meters (radius of the loop), \( x = 0.125 \) meters (half the distance between the two loops), \( I = 1.5 \) A, and \( \mu_0 = 4\pi \times 10^{-7} \) T·m/A is the permeability of free space. Substitute these values to find \( B \).

Step 2: Calculate Total Magnetic Field at the Midpoint

Since both loops contribute to the magnetic field and carry current in the same direction, we sum the magnetic fields from each loop at the midpoint. So, the total magnetic field is twice the field due to one loop: \[ B_{\text{total}} = 2 \times B \] Use the value from Step 1 for \( B \).

Step 3: Find the Force on the Proton

The magnetic force on a moving charge in a magnetic field is given by the Lorentz force equation: \[ F = qvB_{\text{total}}\sin\theta \] where \( q = 1.6 \times 10^{-19} \) C (charge of a proton), \( v = 2400 \times 10^3 \) m/s (velocity of proton), and \( \theta = 90^\circ \) since the velocity is perpendicular to \( B \). Note \( \sin 90^\circ = 1 \). Substitute these values to find \( F \).

Step 4: Determine Force Direction Using Right-Hand Rule

The direction of the magnetic force can be determined using the right-hand rule for forces: Point your fingers in the direction of velocity (perpendicular to \( ab \)), then curl them towards the direction of the magnetic field. Your thumb will point in the force's direction. Since both loops' magnetic fields point in the same direction, calculate the proton's force direction accordingly.

Key Concepts

Magnetic Field

Magnetic fields are fundamental concepts in physics, representing areas where magnetic forces can be felt. They are generated by moving electric charges, such as current in a wire loop. In this exercise, we have two identical circular wire loops that produce a magnetic field because they each carry a current of 1.50 A. The formula used to calculate the magnetic field at the center of a loop along its axis is: \[ B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}} \] where:

• \( B \) is the magnetic field.
• \( \mu_0 \) is the permeability of free space.
• \( I \) is the electric current.
• The radius \( R \) and distance \( x \) each correspond to the setup.

In this instance, the radius \( R \) of each loop is 0.2 meters, and they are placed 0.25 meters apart. The knowledge of magnetic fields and how they are calculated is crucial to understanding the effects that moving charges experience in such fields.

Lorentz Force

The Lorentz force refers to the force experienced by a charged particle when it moves through an electric and magnetic field. It's an essential concept when determining how forces act on particles like protons. For our exercise, we focus on the magnetic component of the Lorentz force, given by: \[ F = qvB_{\text{total}}\sin\theta \] where:

• \( F \) is the magnetic force experienced by the proton.
• \( q \) is the charge of the proton, approximately \( 1.6 \times 10^{-19} \) C.
• \( v \) is the velocity of the proton.
• \( B_{\text{total}} \) is the total magnetic field acting on the proton.
• \( \theta \) is the angle between the velocity and the magnetic field.

For this problem, the proton moves perpendicular to the axis connecting the loops, implying \( \theta = 90^\circ \) and \( \sin 90^\circ = 1 \). Thus, the force can be simplified to \( F = qvB_{\text{total}} \), where we need the magnitude of the total magnetic field and the velocity to find \( F \). Knowing the Lorentz force allows us to understand the magnitude and direction of the motion of charged particles in magnetic fields.

Right-Hand Rule

The right-hand rule is a straightforward, easy-to-use tool for finding the direction of the magnetic force on a moving charge. Given a vector cross-product situation, it remains essential in electromagnetism. Follow these steps to apply the right-hand rule for the exercise: 1. Position your right hand so your fingers point in the direction of the proton's velocity. In this setup, since the proton is fired perpendicular to the line connecting the centers of the loops, your fingers should point away from the center.2. Rotate your hand so you can curl your fingers towards the direction of the magnetic field \( B \), which would be in the same plane as the loops since the currents are parallel.3. Extend your thumb; it will point in the direction of the Lorentz force acting on the proton.By using the right-hand rule, you can easily determine the direction of any forces involved, making it a powerful tool in the study of electromagnetic forces.

Current Loop

A current loop is simply a loop of wire or conductor through which an electric current flows. The circular loop in our exercise generates a magnetic field, a common element in electromagnetics. The total magnetic field in the setup is due to two identical loops, both carrying a current of 1.50 A. These loops have a significant impact on the dynamics of charged particles like the proton in this problem. The current in each loop contributes to a combined magnetic field at the center of the two loops. Key features of a current loop include:

• Circular Shape: Influences the symmetry of the resulting magnetic field.
• Current Direction: Influences the direction of the magnetic field.
• Magnetic Dipole Moment: Arises due to the loop, aligning with the field produced.

In our problem, an understanding of how current loops generate magnetic fields helps us ascertain how the external magnetic field influences the motion of particles, hence affecting the proton's trajectory.