91Ó°ÊÓ

Understanding the electric force is crucial when analyzing the interaction of charged particles with electric fields. The electric force, often simply described as the force exerted by an electric field on a charged object, can be calculated using the formula \( \mathbf{F}_E = q \mathbf{E} \). Here, \( q \) represents the charge of the particle, while \( \mathbf{E} \) denotes the electric field.This force acts on the particle in the direction of the field if the charge is positive, or opposite if the charge is negative.
In our problem, the particle has a charge of \( 1.0 \mathrm{C} \) and is in an electric field of \( (1.0 \mathrm{V/m}) \hat{\mathbf{y}} \). Therefore, the electric force is:

• Calculated specifically as \( \mathbf{F}_E = 1.0 \mathrm{C} \times (1.0 \mathrm{V/m}) \hat{\mathbf{y}} \)
• Resulting in a force of \( (1.0 \mathrm{N}) \hat{\mathbf{y}} \)

This calculation tells us that the force is directed along the \( \hat{\mathbf{y}} \) axis, parallel to the field and the strength is proportional to the charge value.

Magnetic force plays a different role compared to electric force. It depends on the motion of the charged particle through a magnetic field. The expression for magnetic force is \( \mathbf{F}_B = q (\mathbf{v} \times \mathbf{B}) \), where \( \mathbf{v} \) is the velocity and \( \mathbf{B} \) is the magnetic field.This force acts perpendicular to both the velocity of the charge and the magnetic field direction, resulting in a force perpendicular to both due to the cross product nature.
For our charged particle, which moves at a velocity \( (1.0 \mathrm{m/s}) \hat{\mathbf{x}} \) through a magnetic field \( (2.0 \mathrm{T}) \hat{\mathbf{y}} \), the magnetic force is calculated through:

• First finding the cross product \( \mathbf{v} \times \mathbf{B} \)
• This cross product is \( (1.0 \, \mathrm{m/s}) \hat{\mathbf{x}} \times (2.0 \mathrm{T}) \hat{\mathbf{y}} = (2.0 \, \mathrm{T\cdot m/s}) \hat{\mathbf{z}} \)
• The magnetic force is then \( \mathbf{F}_B = 1.0 \, \mathrm{C} \times (2.0 \, \mathrm{T\cdot m/s}) \hat{\mathbf{z}} = (2.0 \, \mathrm{N}) \hat{\mathbf{z}} \)

This force ensures that the charged particle experiences a force in the \( \hat{\mathbf{z}} \) direction, demonstrating the perpendicular attributes of magnetic effects.

The cross product is a vector operation that is especially relevant in physics, particularly when determining magnetic forces. The cross product formula for two vectors \( \mathbf{A} \) and \( \mathbf{B} \) is \( \mathbf{A} \times \mathbf{B} \), and it produces a vector perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \).This product is critical in calculating the magnetic force because it aligns with the principle of perpendicular force application. When working with the Lorentz force law involving magnetic field interactions, the cross product between velocity and the magnetic field vector represents the magnetic force's directional characteristics.
For our scenario:

• The velocity \( \mathbf{v} = (1.0 \, \mathrm{m/s}) \hat{\mathbf{x}} \)
• The magnetic field \( \mathbf{B} = (2.0 \, \mathrm{T}) \hat{\mathbf{y}} \)
• The cross product \( \mathbf{v} \times \mathbf{B} \) results in \( (2.0 \, \mathrm{T\cdot m/s}) \hat{\mathbf{z}} \), which is perpendicular to both
• This cross product gives the direction and magnitude of the resulting magnetic force

By utilizing the right-hand rule, one can determine the direction of the resultant vector, ensuring a complete understanding of how charged particles behave under magnetic influences.