Question

In a region of a uniform electric field \( \mathbf{E} \), a negatively charged particle is moving with a constant velocity \( \mathbf{v} = -v_0 \hat{i} \) near a long straight conductor coinciding with XX' axis and carrying current \( I \) towards -X axis. The particle remains at a distance \( d \) from the conductor.

Answer 1

Forces Acting on a Negatively Charged Particle in Electric and Magnetic Fields

Given Conditions:

• The particle possesses a negative charge and moves with a constant velocity defined as \( \mathbf{v} = -v_0 \hat{i} \).
• The particle is situated in proximity to a long, straight conductor aligned with the XX' axis, through which current \( I \) flows in the -X direction.
• The particle maintains a fixed distance \( d \) from the conductor.

1. Calculation of Magnetic Field from the Conductor:

The magnetic field strength \( B \) at a distance \( d \) from an infinitely long straight conductor carrying current \( I \) is determined by the formula: \[ B = \frac{\mu_0 I}{2 \pi d} \] where: - \( B \) represents the magnetic field intensity, - \( \mu_0 \) is the magnetic constant (permeability of free space), with a value of \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \), - \( I \) is the magnitude of the current, and - \( d \) is the perpendicular distance from the conductor to the point of measurement.

2. Magnetic Force Exerted on the Particle:

The magnetic force on a charged particle in motion is governed by the Lorentz force equation: \[ \mathbf{F}_B = q \mathbf{v} \times \mathbf{B} \] where: - \( q \) denotes the charge of the particle (in this case, negative), - \( \mathbf{v} \) is the particle's velocity vector, and - \( \mathbf{B} \) is the magnetic field vector. Given that the magnetic field is directed into the page and the particle's velocity is along the -X axis, the right-hand rule applied to \( \mathbf{v} \times \mathbf{B} \) indicates that the magnetic force is directed upwards, along the positive \( Y \)-axis. The magnitude of this magnetic force is calculated as: \[ F_B = |q| v_0 \frac{\mu_0 I}{2 \pi d} \]

3. Electric Force on the Particle:

The particle is subjected to an electric force as a consequence of a uniform electric field \( \mathbf{E} \). This force is quantified by: \[ \mathbf{F}_E = q \mathbf{E} \] Due to the particle's negative charge, the electric force \( \mathbf{F}_E \) acts in the opposite direction to the electric field \( \mathbf{E} \). Specifically, if \( \mathbf{E} \) is oriented along the positive \( Y \)-axis, then \( \mathbf{F}_E \) will be directed along the negative \( Y \)-axis. The magnitude of the electric force is given by: \[ F_E = |q| E \]

4. Criterion for Maintaining Constant Velocity:

For the particle to maintain a constant velocity, the resultant force acting upon it must be zero. This necessitates that the magnetic and electric forces be equal in magnitude and opposite in direction, thus balancing each other out: \[ F_B = F_E \] Substituting the derived expressions for these forces: \[ |q| v_0 \frac{\mu_0 I}{2 \pi d} = |q| E \] Upon canceling the charge magnitude \( |q| \) from both sides of the equation, we obtain: \[ v_0 \frac{\mu_0 I}{2 \pi d} = E \]

Summary:

The equilibrium condition for the particle to maintain constant velocity is established by the relationship between its speed \( v_0 \), the current \( I \) in the conductor, the distance \( d \) from the conductor, the electric field strength \( E \), and the particle's charge magnitude \( |q| \). This relationship can be expressed as: \[ v_0 = \frac{2 \pi d E}{\mu_0 I} \] This equation demonstrates that constant velocity is achieved when the electric force precisely counteracts the magnetic force.