Question

Write down the formula for the force in vector form, acting on a moving charged particle in a uniform electric and magnetic fields. Obtain the formula for the radius of the path of the particle entering perpendicular to the magnetic field only. What is the law for the direction of force acting on the particle?

Hint:

When dealing with the force on a charged particle in a magnetic field, always remember the right-hand rule for the direction of the force. The formula for the radius of the path is \( r = \frac{mv}{qB} \), which applies when the velocity is perpendicular to the magnetic field.

Answer 1

The force acting on a charged particle moving in the presence of both electric and magnetic fields is given by the Lorentz force law, which is: \[ \mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) \] Where: - \( \mathbf{F} \) is the total force on the particle.
- \( q \) is the charge of the particle.
- \( \mathbf{E} \) is the electric field.
- \( \mathbf{v} \) is the velocity of the particle.
- \( \mathbf{B} \) is the magnetic field.
- \( \mathbf{v} \times \mathbf{B} \) represents the cross product of the velocity and the magnetic field vectors, which gives the direction of the magnetic force. This formula accounts for both the electric force (\( q\mathbf{E} \)) and the magnetic force (\( q\mathbf{v} \times \mathbf{B} \)). Now, let’s focus on the case where the particle is moving perpendicular to the magnetic field, and the electric field is neglected (i.e., \( \mathbf{E} = 0 \)): \[ \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \] Radius of the Path:
When a charged particle moves perpendicular to a magnetic field, it experiences a magnetic force that causes it to move in a circular path. The force acts as a centripetal force, and the radius of the path can be derived using the equation for the centripetal force: \[ F_{\text{magnetic}} = F_{\text{centripetal}} \] The magnetic force is: \[ qvB = \frac{mv^2}{r} \] Where: - \( m \) is the mass of the particle,
- \( r \) is the radius of the circular path.
Solving for the radius \( r \), we get: \[ r = \frac{mv}{qB} \] Thus, the radius of the circular path of the particle is directly proportional to its mass and velocity, and inversely proportional to the charge of the particle and the magnetic field strength. Direction of the Force:
The direction of the magnetic force is given by the right-hand rule: - Point your fingers in the direction of the velocity \( \mathbf{v} \). - Curl your fingers towards the direction of the magnetic field \( \mathbf{B} \). - Your thumb will point in the direction of the force \( \mathbf{F} \) acting on a positive charge. For a negative charge, the force will be in the opposite direction.