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 on a Charged Particle in Electric and Magnetic Fields

Given:

• The particle is negatively charged and moving with a constant velocity \( \mathbf{v} = -v_0 \hat{i} \).
• The particle is near a long straight conductor along the XX' axis carrying current \( I \) towards the -X axis.
• The particle remains at a constant distance \( d \) from the conductor.

1. Magnetic Field due to the Current-Carrying Conductor:

The magnetic field at a distance \( d \) from a long straight conductor carrying current \( I \) is given by: \[ B = \frac{\mu_0 I}{2 \pi d} \] where: - \( B \) is the magnetic field, - \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)), - \( I \) is the current, - \( d \) is the distance from the conductor to the point where the magnetic field is calculated.

2. Magnetic Force on the Particle:

The magnetic force on a moving charge is given by the Lorentz force law: \[ \mathbf{F}_B = q \mathbf{v} \times \mathbf{B} \] where: - \( q \) is the charge of the particle (negative charge), - \( \mathbf{v} \) is the velocity of the particle, and - \( \mathbf{B} \) is the magnetic field. Since the magnetic field is directed into the page and the particle is moving along the -X axis, applying the right-hand rule for \( \mathbf{v} \times \mathbf{B} \), we find that the magnetic force will be directed **upwards**, along the positive \( Y \)-axis. The magnitude of the magnetic force is: \[ F_B = |q| v_0 \frac{\mu_0 I}{2 \pi d} \]

3. Electric Force on the Particle:

The particle experiences an electric force due to the uniform electric field \( \mathbf{E} \). The electric force is given by: \[ \mathbf{F}_E = q \mathbf{E} \] Since the particle is negatively charged, the electric force will be in the **opposite direction** to the electric field. If the electric field is directed along the positive \( Y \)-axis, the electric force will be directed along the **negative \( Y \)-axis**. The magnitude of the electric force is: \[ F_E = |q| E \]

4. Condition for Constant Velocity:

For the particle to move with a constant velocity, the net force on the particle must be zero. Therefore, the magnetic and electric forces must balance each other. Thus: \[ F_B = F_E \] Substituting the expressions for \( F_B \) and \( F_E \): \[ |q| v_0 \frac{\mu_0 I}{2 \pi d} = |q| E \] Canceling \( |q| \) from both sides: \[ v_0 \frac{\mu_0 I}{2 \pi d} = E \]

Conclusion:

The relationship between the velocity \( v_0 \), the current \( I \), the distance \( d \), the electric field \( E \), and the charge \( q \) is given by: \[ v_0 = \frac{2 \pi d E}{\mu_0 I} \] This equation shows that the particle will continue to move with a constant velocity as long as the electric force balances the magnetic force.