Problem 2 - Fundamentals of Physics

Question

A particle of mass \(10 \mathrm{~g}\) and charge \(80 \mu \mathrm{C}\) moves through a uniform magnetic field, in a region where the free-fall acceleration is \(-9.8 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}^{2} .\) The velocity of the particle is a constant \(20 \mathrm{i} \mathrm{km} / \mathrm{s},\) which is perpendicular to the magnetic field. What, then, is the magnetic field?

Step-by-Step Solution

Answer

The magnetic field is approximately 61.25 T.

Step 1: Understanding the Given Information

We are given a mass of the particle as \( 10 \, \text{g} \) or \( 0.01 \, \text{kg} \), charge \( 80 \, \mu C \) or \( 80 \times 10^{-6} \, \text{C} \), and velocity \( \mathbf{v} = 20 \hat{i} \, \text{km/s} \) or \( 20,000 \hat{i} \, \text{m/s} \). The free-fall acceleration is \( -9.8 \hat{j} \, \text{m/s}^2 \).

Step 2: Analyzing Forces

The particle is moving in a constant velocity, implying a net force of zero on the particle. Therefore, the gravitational force \( \mathbf{F}_g = m\mathbf{g} = (0.01) \times (-9.8 \hat{j}) \) is balanced by the magnetic force \( \mathbf{F}_B \).

Step 3: Calculating Magnetic Force

The formula for the magnetic force on a charged particle is \( \mathbf{F}_B = q \mathbf{v} \times \mathbf{B} \). Given the net force is zero, \( \mathbf{F}_B = -\mathbf{F}_g \).Thus, \( 80 \times 10^{-6} \times 20,000 \times B \hat{k} = 0.01 \times 9.8 \hat{j} \).

Step 4: Solving for Magnetic Field

From the equality \( 1.6 \times 10^{-3} B = 0.098 \), solve for \( B \), i.e. \[ B = \frac{0.098}{1.6 \times 10^{-3}} \approx 61.25 \, \text{T} \].

Key Concepts

Magnetic Force

The magnetic force is a critical concept when considering how charged particles interact with magnetic fields. It is given by the equation \( \mathbf{F}_B = q(\mathbf{v} \times \mathbf{B}) \), where \( q \) is the charge of the particle, \( \mathbf{v} \) is its velocity, and \( \mathbf{B} \) is the magnetic field vector.
The magnetic force acts perpendicular to both the velocity of the charged particle and the magnetic field. This particular arrangement is due to the nature of the cross product in the equation:

• **Direction**: The force direction follows the right-hand rule. If you point your fingers in the direction of velocity (\( \mathbf{v} \)), and curl them towards the magnetic field (\( \mathbf{B} \)), your thumb points in the direction of the magnetic force (\( \mathbf{F}_B \)).
• **Magnitude**: The magnitude is proportional to the sine of the angle between \( \mathbf{v} \) and \( \mathbf{B} \). Maximum force occurs when the angle is 90 degrees, meaning \( \mathbf{v} \) is perpendicular to \( \mathbf{B} \).

This concept is essential for solving problems where charged particles move through magnetic fields, like determining the motion or the magnetic field present.

Charged Particle Motion

When a charged particle moves through a magnetic field, its motion is influenced by the magnetic force exerting a perpendicular effect.
This typically results in circular or spiral trajectories because:

• The force is always perpendicular to its velocity, never changing the speed, only the direction.
• The motion forms a centripetal force situation, causing circular motion if the particle moves perfectly perpendicular to a uniform field.

Mathematically, the radius of this circular path can be determined by balancing the magnetic force against the centripetal force needed to keep an object in circular motion: \[ qvB = \frac{mv^2}{r} \] solving for \( r \) gives \[ r = \frac{mv}{qB} \]
This expression shows that the radius depends on mass \( m \), velocity \( v \), charge \( q \), and magnetic field \( B \). The understanding of charged particle motion is vital for applying magnetic fields in technology and sciences, such as in particle accelerators.

Uniform Magnetic Field

A uniform magnetic field is one where the magnetic field lines are parallel and equally spaced, meaning the magnetic field strength (\( \mathbf{B} \)) is constant throughout.
This kind of field is an idealized model often used for simplifying problems involving magnetic forces and particle trajectories.

• **Characteristics**: In a uniform magnetic field, the forces exerted on a moving charged particle will have both magnitude and direction that remain constant at given points, producing a predictable motion path.
• **Applications**: It is commonly applied in scenarios like the field inside a solenoid (a coil of wire), or in magnetic resonance imaging (MRI) where uniform magnetic fields help produce clear imaging.

In exercises analyzing motions of charged particles, assuming a uniform magnetic field helps use simple mathematical expressions, as consistent magnetic forces and predictable circular paths simplify calculations. This was exactly the case in understanding how the magnetic force balanced with gravitational force in the given problem scenario.