Problem 4 - Physics

Question

When a charged particle moves at an angle of \(25^{\circ}\) with respect to a magnetic field, it experiences a magnetic force of magnitude \(F\). At what angle (less than \(90^{\circ}\) ) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude \(2 F ?\)

Step-by-Step Solution

Answer

The angle is approximately \(57^{\circ}\).

Step 1: Understanding the Magnetic Force

The force exerted on a charged particle moving through a magnetic field is given by the equation \( F = qvB \sin(\theta) \), where \( q \) is the charge of the particle, \( v \) is its velocity, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.

Step 2: Relating Force to the Given Angle

The initial force experienced is \( F = qvB \sin(25^{\circ}) \). We need to find another angle \( \theta \) such that the force becomes \( 2F = qvB \sin(\theta) \).

Step 3: Set up the Equation for New Force

Since we want the new force to be double the original force, set the equation \( qvB \sin(\theta) = 2 \times qvB \sin(25^{\circ}) \). This simplifies to \( \sin(\theta) = 2 \sin(25^{\circ}) \).

Step 4: Solving for the New Angle

To find \( \theta \), we need to take the inverse sine (arcsin) of both sides. Calculate \( \sin(25^{\circ}) \), which is approximately \( 0.4226 \). Thus, \( \sin(\theta) = 2 \times 0.4226 = 0.8452 \). Use a calculator to find \( \theta = \arcsin(0.8452) \), resulting in \( \theta \approx 57.0^{\circ} \).

Key Concepts

Angle with Magnetic Field

When a charged particle moves in a magnetic field, the angle between its velocity and the magnetic field lines greatly affects the experienced magnetic force. This angle, denoted as \( \theta \), determines how much of the particle's motion actually interacts with the field.
Since magnetism acts perpendicular to motion, not all the movement contributes to the force. Only the component of velocity that is perpendicular to the magnetic field line interacts.
In the provided exercise, the particle initially moves at an angle of \( 25^{\circ} \) to the magnetic field. This means part of its velocity is contributing to the magnetic force, resulting in a force \( F \).
If the angle changes, the effect on force changes, which is why understanding this relationship is critical.

Trigonometric Functions in Physics

Trigonometric functions, especially the sine function, play a key role in physics to describe how different components of vectors influence outcomes.
The magnetic force on a particle is calculated using the sine of the angle \( \theta \) between its velocity and the magnetic field. The force formula is \( F = qvB\sin(\theta) \), where each symbol has a specific meaning:

• \( F \) is the magnetic force.
• \( q \) is the charge of the particle.
• \( v \) is the velocity of the particle.
• \( B \) is the magnetic field strength.
• \( \theta \) is the angle with the magnetic field.

In our exercise, we started with \( \sin(25^{\circ}) \) and needed to find an angle \( \theta \) where \( \sin(\theta) = 2\sin(25^{\circ}) \). Using a calculator, we saw that \( \sin(25^{\circ}) \approx 0.4226 \), making \( \sin(\theta) \approx 0.8452 \). Thus, \( \theta \approx 57.0^{\circ} \).
This calculation demonstrates how trigonometry aids in translating angles into meaningful physical phenomena.

Magnitude of Magnetic Force

Understanding the magnitude of the magnetic force experienced by a particle requires an insight into both the physical properties involved and the geometric positioning, particularly the angle.
The magnetic force magnitude is directly proportional to the product of the charge \( q \), the particle’s velocity \( v \), the magnetic field strength \( B \), and the sine of the angle \( \theta \).
The equation \( F = qvB\sin(\theta) \) tells us that non-zero force only occurs when the charge is moving at an angle other than parallel to the magnetic field. At \( \theta = 0^{\circ} \), no force acts on the particle, as \( \sin(0^{\circ}) = 0 \).
With our aim of doubling the original force to \( 2F \), we used the angle such that the sine of that angle doubled, which directly increased the force magnitude. This emphasizes the power of understanding not just the quantities, but the trigonometric relationships that govern physical interactions.