Problem 20 - College Physics

Question

\(\bullet\) A 3.25 g bullet picks up an electric charge of 1.65\(\mu C\) as it travels down the barrel of a rifle. It leaves the barrel at a speed of 425 \(\mathrm{m} / \mathrm{s}\) , traveling perpendicular to the earth's magnetic field, which has a magnitude of \(5.50 \times 10^{-4} \mathrm{T} .\) Calculate (a) the magnitude of the magnetic force on the bullet and (b) the magnitude of the bullet's acceleration due to the magnetic force at the instant it leaves the rifle barrel.

Step-by-Step Solution

Answer

The magnetic force is \(3.86 \times 10^{-7} \, N\) and the acceleration is \(0.119 \, \text{m/s}^2\).

Step 1: Identify Given Variables

We are given the following values: the charge on the bullet as \( q = 1.65 \, \mu C = 1.65 \times 10^{-6} \, C \), the velocity of the bullet as \( v = 425 \, \text{m/s} \), and the magnetic field as \( B = 5.50 \times 10^{-4} \, T \).

Step 2: Calculate the Magnetic Force

The formula for the magnetic force acting on a charged particle moving in a magnetic field is given by \( F = qvB \sin \theta \). Since the bullet's path is perpendicular to the magnetic field, \( \theta = 90^\circ \), thus \( \sin \theta = 1 \). So the force is \( F = qvB \).Substituting the given values, we get:\[F = (1.65 \times 10^{-6} \, C)(425 \, \text{m/s})(5.50 \times 10^{-4} \, T)\]Calculate the above expression to find \( F \).

Step 3: Perform the Force Calculation

Carry out the multiplication:\(F = 1.65 \times 10^{-6} \times 425 \times 5.50 \times 10^{-4}\)This yields:\(F = 3.861375 \times 10^{-7} \, N\)Thus, the magnetic force on the bullet is approximately \( 3.86 \times 10^{-7} \, N \).

Step 4: Calculate the Bullet's Acceleration

To find the acceleration, we use Newton's second law, \( F = ma \), where \( m \) is the mass of the bullet and \( a \) is its acceleration. Rearranging gives \( a = \frac{F}{m} \).Given the mass of the bullet as \( m = 3.25 \, g = 3.25 \times 10^{-3} \, kg \), substitute the values into the equation:\[a = \frac{3.861375 \times 10^{-7} \, N}{3.25 \times 10^{-3} \, kg}\]Calculate this to find \( a \).

Step 5: Perform the Acceleration Calculation

Divide the force by the mass:\(a = \frac{3.861375 \times 10^{-7}}{3.25 \times 10^{-3}}\)This yields:\(a = 0.1188 \, \text{m/s}^2\)Therefore, the bullet's acceleration due to the magnetic force is approximately \( 0.119 \, \text{m/s}^2 \).

Key Concepts

Electric Charge

When dealing with electromagnetism, the electric charge is a fundamental property. It characterizes how particles interact with electric and magnetic fields.
Electric charge can be positive or negative and comes in discrete quantities, typically given in units of Coulombs (C). For our bullet, it carries a charge expressed as microcoulombs, which is a millionth of a Coulomb.

Charged particles, like the bullet in this example, interact with magnetic fields if they are moving. This interaction is the basis of the magnetic force that affects the particle's motion. Therefore, electric charge is crucial in calculating such forces as it directly influences the magnitude of this force. Without a charge, the magnetic field cannot exert a force on the particle.

Magnetic Field

A magnetic field is a force field that surrounds magnetic materials and moving electric charges. It is usually represented by the symbol \( B \) and is measured in Tesla (T).

In this problem, the magnetic field is uniform, meaning it has the same magnitude and direction everywhere within a specified region.
The bullet moves perpendicular to this magnetic field—a key point—since the force exerted by the magnetic field is maximized when the motion of the charged particle is perpendicular to the field lines.

The interaction of a moving charged particle with a magnetic field gives rise to what we call the magnetic force. This force can be calculated using the formula \( F = qvB \sin \theta \), where \( \theta \) is the angle between the velocity (v) vector and the magnetic field (B) direction. With the bullet traveling perpendicular, \( \theta = 90^\circ \), simplifying our calculation of the magnetic force.

Acceleration Calculation

Acceleration is the rate of change of velocity of an object.
In this exercise, we calculate the acceleration due to the magnetic force using Newton's second law.
The law states that \( F = ma \), where \( F \) is the force exerted on an object, \( m \) is its mass, and \( a \) is the resulting acceleration.

To find acceleration, we rearrange the formula to \( a = \frac{F}{m} \).
This tells us how the force changes the velocity of our object.

The initial requirements are the force (calculated from the magnetic force formula, \( F = qvB \)) and the mass of the bullet.
Substituting these values helps us find how the bullet speeds up or slows down as it interacts with the magnetic field through the force applied.

Newton's Second Law

Newton's second law is a cornerstone of classical mechanics. It defines how the velocity of an object changes when it is subject to external forces, and is usually written as \( F = ma \).
According to this principle, an object's acceleration depends directly on the net force acting on it and inversely on its mass.

This law is invaluable when calculating motion-related phenomena.

• If the mass is greater, a larger force is needed to achieve the same acceleration.
• Lighter objects experience greater acceleration for the same amount of force.

Utilizing this in our scenario, we apply the calculated magnetic force on the small mass of the bullet to determine its acceleration.
It beautifully illustrates how force and mass interact in dynamics, explaining motion quantitatively and predictably.