An electron in a cathode-ray beam passes between 2.5-cm-long parallel-plate electrodes that are 5.0 mm apart. A 2.0 mT, 2.5-cm-wide magnetic field is perpendicular to the electric field between the plates. The electron passes through the electrodes without being deflected if the potential difference between the plates is 600 V. If the potential difference between the plates is set to zero, what is the electron's radius of curvature in the magnetic field?

Verified step by step guidance

1. Step 1: Understand the problem. The electron is initially passing through parallel-plate electrodes without deflection due to the balance between the electric and magnetic forces. When the electric field is removed, the electron will experience only the magnetic force, causing it to move in a circular path. The goal is to find the radius of curvature of this path.
2. Step 2: Recall the formula for the radius of curvature of a charged particle moving in a magnetic field. The radius of curvature (r) is given by: $r=\frac{\mathrm{mv}}{\mathrm{qB}}$, where m is the mass of the electron, v is its velocity, q is the charge of the electron, and B is the magnetic field strength.
3. Step 3: Determine the velocity of the electron. When the electric field was present, the electron's velocity was determined by the balance between the electric force and the magnetic force. Use the relationship: $v=\frac{E}{B}$, where E is the electric field strength and B is the magnetic field strength. The electric field strength can be calculated using $E=\frac{V}{d}$, where V is the potential difference and d is the separation between the plates.
4. Step 4: Substitute the given values to calculate the velocity. Use V = 600 V, d = 5.0 mm (converted to meters), and B = 2.0 mT (converted to teslas). Calculate E first, then use $v=\frac{E}{B}$ to find the velocity of the electron.
5. Step 5: Use the calculated velocity and the given values for the mass of the electron ($m=9.11\times {10}^{-}$ kg), charge of the electron ($q=1.6\times {10}^{-}$ C), and magnetic field strength (B = 2.0 mT) in the formula $r=\frac{\mathrm{mv}}{\mathrm{qB}}$ to calculate the radius of curvature.

Key Concepts

Lorentz Force

The Lorentz force is the force experienced by a charged particle moving through electric and magnetic fields. It is given by the equation F = q(E + v × B), where F is the force, q is the charge, E is the electric field, v is the velocity of the particle, and B is the magnetic field. This concept is crucial for understanding how the electron behaves in the presence of both electric and magnetic fields.

Radius of Curvature

The radius of curvature of a charged particle's path in a magnetic field is determined by the balance between the magnetic force and the centripetal force required to keep the particle in circular motion. The radius can be calculated using the formula r = mv/(qB), where m is the mass of the particle, v is its velocity, q is its charge, and B is the magnetic field strength. This concept is essential for determining how the electron will curve in the magnetic field when the electric field is turned off.

Kinematics of Charged Particles

Understanding the kinematics of charged particles involves analyzing their motion under the influence of forces. In this scenario, the electron's velocity can be derived from the potential difference between the plates, which accelerates the electron. This concept is important for calculating the electron's speed as it enters the magnetic field, which directly affects its radius of curvature.