Problem 21 - College Physics

Question

An electron in the beam of a TV picture tube is accelerated through a potential difference of 2.00 \(\mathrm{kV}\) . It then passes into a magnetic field perpendicular to its path, where it moves in a circular arc of diameter 0.360 \(\mathrm{m} .\) What is the magnitude of this field?

Step-by-Step Solution

Answer

The magnitude of the magnetic field is approximately 0.019 T.

Step 1: Determine the Kinetic Energy of the Electron

An electron accelerated through a potential difference of 2.00 kV gains kinetic energy. The kinetic energy given to the electron is equal to the electric potential energy, given by \( KE = qV \), where \( q \) is the charge of the electron (\( 1.6 \times 10^{-19} \) C) and \( V \) is the potential difference (2000 V). The kinetic energy is thus \( KE = 1.6 \times 10^{-19} \times 2000 \) J.

Step 2: Relate Kinetic Energy to Electron's Speed

The kinetic energy can also be expressed as \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \) kg) and \( v \) is its speed. Set \( qV = \frac{1}{2}mv^2 \) and solve for \( v \):\[v = \sqrt{\frac{2qV}{m}}\]Substitute the given values to find \( v \).

Step 3: Understand Motion in a Magnetic Field

Within the magnetic field, the force acting on the electron is the magnetic force, which provides the centripetal force to keep the electron in a circular path. This force is \( F = qvB = \frac{mv^2}{r} \), where \( B \) is the magnetic field strength, and \( r \) is the radius of the circular path.

Step 4: Calculate the Radius of the Circle

Determine the radius \( r \) of the path as half the diameter: \( r = \frac{0.360}{2} = 0.180 \) m.

Step 5: Solve for the Magnetic Field Strength

Rearrange the formula \( qvB = \frac{mv^2}{r} \) to solve for \( B \):\[B = \frac{mv}{qr}\]Substitute the values of \( m \), \( v \), \( q \), and \( r \) to calculate \( B \).

Step 6: Final Step: Calculate and Provide the Answer

Using the values from previous calculations, compute \( B B = \frac{(9.11 \times 10^{-31}) v}{(1.6 \times 10^{-19})(0.180)}\). Insert the previously calculated value of \( v \). Calculate \( B \) to find the magnitude of the magnetic field.

Key Concepts

Electron Acceleration

Understanding electron acceleration involves looking at how electrons gain energy and speed when they move through an electric field. In this case, a potential difference, measured in volts, provides the means to accelerate the electron. Here, the electron is exposed to a potential difference of 2.00 kV, which stands for kilovolts, equivalent to 2000 volts.
This potential difference causes the electron to gain kinetic energy. The energy associated with the change in potential (potential energy) is transferred into kinetic energy as the electron moves, according to the equation:

• \( KE = qV \)

where \( KE \) is the kinetic energy, \( q \) is the charge of the electron \((1.6 \times 10^{-19} \text{ C})\), and \( V \) is the potential difference. The result is an accelerated electron with increased speed as it leaves the electric field.

Magnetic Field Strength

The magnetic field strength, denoted as \( B \), describes how intense the magnetic field is in the region surrounding the electron. When an electron enters a perpendicular magnetic field, its path changes to a circular motion due to the Lorentz force, which acts perpendicular to the velocity of the electron and the magnetic field.
In this exercise, you're required to determine the magnitude of the magnetic field such that the electron travels in a circular path. The relationship involves the force acting on the moving charge (electron), described by the formula:

• \( F = qvB \)

This force must equal the centripetal force required to keep the electron moving in a circular path:

• \( \frac{mv^2}{r} \)

By setting these two equal, and rearranging to solve for \( B \):

• \( B = \frac{mv}{qr} \)

This provides the magnetic field strength, an essential part of understanding how magnetic influences affect the motion of charged particles.

Kinetic Energy

Kinetic energy is the energy an object possesses due to its motion. For an electron moving through space, this energy is acquired as it is accelerated by a potential difference. The kinetic energy must first be calculated in order to understand how the electron behaves once it's impacted by the magnetic field.
The kinetic energy is given by the formula:

• \( KE = qV = \frac{1}{2}mv^2 \)

This is where the potential energy from the electric field is transferred into motion, increasing the electron’s speed. Rearranging this formula allows for solving the speed \( v \) of the electron:

• \( v = \sqrt{\frac{2qV}{m}} \)

This speed is vital in determining subsequent forces on the electron in the magnetic field.

Centripetal Force

When a charged particle like an electron moves in a circular path within a magnetic field, it experiences a centripetal force. This is the force that acts towards the center of the circular path, maintaining the electron's curved trajectory.
The central idea behind centripetal force is to understand that for circular motion to occur, a force must be acting continuously towards the center of the circle. In the context of an electron moving within a magnetic field, the centripetal force is provided by the magnetic force, expressed by:

• \( F = \frac{mv^2}{r} \)

This equation helps determine the forces at play, allowing calculation of quantities like radius and speed, which are pivotal for solving and understanding the electron's motion. The characteristics of centripetal force, combined with knowledge of magnetic force, give insights into the dynamics of charged particles in magnetic fields.