Problem 59 - University Physics with Modern Physics

Question

You wish to hit a target from several meters away with a charged coin having a mass of 4.25\(\mathrm { g }\) and a charge of \(+ 2500 \mu \mathrm { C }\) . The coin is given an initial velocity of 12.8\(\mathrm { m } / \mathrm { s }\) , and a downward, uniform electric field with field strength 27.5\(\mathrm { N } / \mathrm { C }\) exists through-out the region. If you aim directly at the target and fire the coin horizontally, what magnitude and direction of uniform magnetic field are needed in the region for the coin to hit the target?

Step-by-Step Solution

Answer

A magnetic field of 2.15 T directed into the page is needed.

Step 1: Understand the Problem

The coin is subject to both an electric and a magnetic field. We want the magnetic force to counteract the electric force so the coin hits the target.

Step 2: Calculate the Electric Force

Use the formula for electric force: \( F_e = qE \), where \( q = 2500 \, \mu C = 2500 \times 10^{-6} \, C \) and \( E = 27.5 \, N/C \). \( F_e = (2500 \times 10^{-6} \, C)(27.5 \, N/C) = 0.06875 \, N \).

Step 3: Determine the Required Magnetic Force

For the coin to travel straight and hit the target, the magnetic force must be equal and opposite to the electric force: \( F_m = F_e = 0.06875 \, N \).

Step 4: Relate Magnetic Force to Magnetic Field

The magnetic force is given by \( F_m = qvB \sin \theta \), where \( \theta = 90^\circ \) because the velocity is horizontal. Thus, \( F_m = qvB \).

Step 5: Solve for Magnetic Field Magnitude

Rearrange the equation to solve for \( B \): \( B = \frac{F_m}{qv} \). Substituting the known values \( F_m = 0.06875 \, N \), \( q = 2500 \times 10^{-6} \, C \), and \( v = 12.8 \, m/s \), \( B = \frac{0.06875}{(2500 \times 10^{-6} \times 12.8)} = 2.15 \, T \).

Step 6: Determine the Magnetic Field Direction

By using the right-hand rule: thumb in the direction of velocity (horizontal), fingers in the direction of the electric field (downward), the palm facing the direction of force (opposite to gravity) requires the magnetic field to be directed into the page.

Key Concepts

Magnetic Force

Magnetic force plays a crucial role when charged particles move through a magnetic field. It can deflect the path of the particle depending on its charge, velocity, and the magnetic field's direction. This force is given by the equation \[ F_m = qvB \sin \theta \]where:

• \( q \) is the charge of the particle
• \( v \) is its velocity
• \( B \) is the magnetic field strength
• \( \theta \) is the angle between the direction of velocity and the magnetic field

In the case of our charged coin, the angle \( \theta \) is 90 degrees because the coin is moving perpendicular to the magnetic field. Therefore, \( \sin \theta = 1 \), simplifying the equation to \[ F_m = qvB \]. The right-hand rule helps determine the vector direction of the magnetic force, just like when holding a coin so that your thumb, fingers, and palm facilitate understanding of the path deflected by magnetic effects.

Electric Field

An electric field is a region around a charged particle that exerts force on other charged particles. In our exercise, the electric field is uniform, meaning it has a consistent strength and direction within the designated area. The force exerted by an electric field on a charge is expressed by \[ F_e = qE \]where:

• \( q \) is the charge affected by the field
• \( E \) is the electric field strength

The electric field in the scenario applies a downward force on the charged coin. Calculating this force is essential to balance it with a suitable magnetic force, ensuring the coin reaches its target without being dragged off its path by the electric field.

Lorentz Force

The Lorentz force is the total electromagnetic force acting on a particle moving in an electric and magnetic field. It combines both electric and magnetic forces. The equation representing the Lorentz force is:\[ F = F_e + F_m = qE + qvB \]This force dictates how charged objects move when subjected to electromagnetic fields. In our coin exercise, the desired equilibrium condition means we want the magnetic force to oppose the electric force precisely. This results in the Lorentz force being zero, allowing the coin to maintain a constant velocity straight towards the target. By accurately calculating and applying a magnetic field in the required direction, the coin's trajectory is unaffected by external electromagnetic influences.

Projectile Motion

Projectile motion describes the path of an object thrown into the air, subject to only gravity and no other forces. However, our situation involves a charged coin within both electric and magnetic fields. While you aim and fire the coin horizontally, these fields change its natural projectile path, necessitating precise corrections. The goal within this exercise is to make external forces like magnetic and electric fields cancel out. This way, the initial horizontal velocity of the coin remains constant, removing any arc usually seen in projectile motion. Thus, with the magnetic force exactly countering the electric force, comfort can be found knowing the coin will follow a straight path towards its target without deviation, and calculations will ensure this balance is achieved.