Problem 25 - University Physics with Modern Physics

Question

An electron in the beam of a TV picture tube is accelerated by a potential difference of 2.00 \(\mathrm{kV}\) . Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius 0.180 \(\mathrm{m}\) . What is the magnitude of the field?

Step-by-Step Solution

Answer

The magnetic field magnitude is \(5.23 \times 10^{-4} \; \mathrm{T}\).

Step 1: Convert Voltage to Energy

The potential difference that accelerates the electron is 2.00 \( \mathrm{kV} \), which is equal to 2000 volts. The energy gained by the electron can be calculated using the formula \( E = qV \), where \( q \) is the charge of the electron \( (1.60 \times 10^{-19} \; \mathrm{C}) \) and \( V \) is the potential difference. Thus, the energy is \( E = 1.60 \times 10^{-19} \; \mathrm{C} \times 2000 \; \mathrm{V} = 3.20 \times 10^{-16} \; \mathrm{J} \).

Step 2: Determine Electron Velocity

The kinetic energy of the electron, which equals the energy gained, can be expressed as \( \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron \( (9.11 \times 10^{-31} \; \mathrm{kg}) \) and \( v \) is the velocity. Solving for \( v \), we have \[ v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 3.20 \times 10^{-16} \; \mathrm{J}}{9.11 \times 10^{-31} \; \mathrm{kg}}} = 2.65 \times 10^{7} \; \mathrm{m/s} \].

Step 3: Use Circular Motion in Magnetic Field

In a magnetic field, the electron follows a circular path due to the Lorentz force, and the centripetal force required is \( F = \frac{mv^2}{r} \). The magnetic force can be calculated with \( F = qvB \), where \( B \) is the magnetic field we need to find. Equating both expressions gives: \[ qvB = \frac{mv^2}{r} \].

Step 4: Solve for Magnetic Field Strength

Rearrange the expressions to solve for \( B \): \[ B = \frac{mv}{qr} = \frac{9.11 \times 10^{-31} \; \mathrm{kg} \times 2.65 \times 10^{7} \; \mathrm{m/s}}{1.60 \times 10^{-19} \; \mathrm{C} \times 0.180 \; \mathrm{m}} = 5.23 \times 10^{-4} \; \mathrm{T} \]. This is the strength of the magnetic field.

Key Concepts

Magnetic Field Strength

Magnetic field strength is the measure of the force that a magnetic field exerts on moving charges, like electrons. In our exercise, an electron moves through a magnetic field after being accelerated by a voltage. The effect of this field on the electron is observed as circular motion.
When we want to determine the strength of the magnetic field, we analyze how much force the field applies on the electron to keep it moving in a circle. This requires a balance between magnetic force and the centripetal force that keeps the electron on its path. The magnetic force is calculated with the formula: \( F = qvB \), where:

• \( F \) is the force on the electron,
• \( q \) is the charge of the electron,
• \( v \) is its velocity,
• and \( B \) is the magnetic field strength.

To find \( B \), the strength of the magnetic field, this force must equal the centripetal force \( F = \frac{mv^2}{r}\) that keeps the electron in circular motion.

Electron Acceleration

Electron acceleration is the process of increasing the speed of an electron, typically done by passing it through a potential difference, like in a TV picture tube. In the given problem, the electron is accelerated by a potential difference of 2.00 kV, converting electrical energy into kinetic energy.
The formula used to determine the energy gained by the electron is \( E = qV \), where:

• \( E \) is the energy,
• \( q \) is the electron charge, and
• \( V \) is the voltage.

This energy then appears as kinetic energy, calculated by \( \frac{1}{2}mv^2 \). By setting the kinetic energy equal to the energy gained from the potential difference, we can solve for the electron's velocity \( v \). This velocity is crucial for determining the magnetic field strength as it affects how much force the field exerts on the electron.

Circular Motion in Magnetic Fields

When an electron moves through a magnetic field, it often travels in a circular path. This phenomenon is due to the Lorentz force, which causes the electron to experience a constant force perpendicular to its velocity, resulting in circular motion.
For circular motion in a magnetic field, we equate the magnetic force to the necessary centripetal force. The centripetal force needed to keep an object moving in a circle is \( F = \frac{mv^2}{r} \), while the magnetic force it experiences is \( F = qvB \).
With these formulas, equate to solve for the magnetic field strength \( B \) by rearranging to \( B = \frac{mv}{qr} \). Here, \( m \) is the mass of the electron, \( v \) its velocity, \( q \) the charge, and \( r \) the radius of the circular path. This formula provides insights into how fast an electron moves under the influence of magnetic fields.