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Chapter 28: Problem 24

An electron is accelerated from rest by a potential difference of \(350 \mathrm{~V}\). It then enters a uniform magnetic field of magnitude 200 \(\mathrm{mT}\) with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

Short Answer

Expert verified
(a) Speed: \(1.1 \times 10^7\,\text{m/s}\); (b) Radius: \(3.13\,\text{mm}\).

Step by step solution

01

Electron's Kinetic Energy

The first step is to calculate the kinetic energy gained by the electron when it is accelerated. The potential difference accelerates the electron, providing it energy equal to the electric potential energy change. This is given by \[ KE = e \cdot V \]where \( e \) is the electron charge \( 1.6 \times 10^{-19} \, \text{Coulombs} \) and \( V \) is the potential difference \( 350 \, \text{Volts} \). Thus,\[ KE = 1.6 \times 10^{-19} \, \text{C} \times 350 \, \text{V} = 5.6 \times 10^{-17} \, \text{Joules} \].
02

Electron's Speed

Next, use the kinetic energy to find the speed of the electron. The kinetic energy is related to speed by the formula,\[ KE = \frac{1}{2} m v^2 \]where \( m \) is the mass of the electron \( 9.11 \times 10^{-31} \, \text{kg} \). Solving for \( v \),\[ v = \sqrt{\frac{2 \cdot KE}{m}} = \sqrt{\frac{2 \cdot 5.6 \times 10^{-17}}{9.11 \times 10^{-31}}} \approx 1.1 \times 10^7 \, \text{m/s} \].
03

Lorentz Force and Circular Motion

The Lorentz force will cause the electron to move in a circular path. For an electron moving perpendicular to a magnetic field, the magnetic force provides the necessary centripetal force\[ F_{mag} = F_{centripetal} \]\[ e v B = \frac{m v^2}{r} \]where \( B \) is the magnetic field \( 200 \, \text{mT} = 0.2 \, \text{T} \), and \( r \) is the radius of the path. Simplifying for \( r \),\[ r = \frac{m v}{e B} \].
04

Radius of Electron's Path

Plug the known values into the radius equation calculated in Step 3:\[ r = \frac{9.11 \times 10^{-31} \times 1.1 \times 10^7}{1.6 \times 10^{-19} \times 0.2} \approx 3.13 \times 10^{-3} \, \text{m} = 3.13 \, \text{mm} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Acceleration
When an electron is accelerated, it gains energy from the electric potential difference it passes through. This potential difference, measured in volts, provides energy to the electron that initially is at rest. This energy increase is converted into kinetic energy as the electron speeds up.
In our example, the potential difference of 350 volts accelerates the electron, giving it energy. This process is described by the equation:
  • \( KE = e \cdot V \)
Here, \( e \) represents the charge of an electron, which is a small but significant value of \( 1.6 \times 10^{-19} \) Coulombs. Hence, the kinetic energy attained by the electron translates to \( 5.6 \times 10^{-17} \) Joules, a crucial first step in understanding its acquired speed.
Kinetic Energy
Kinetic energy is the energy an object has due to its motion. The faster an object moves, the more kinetic energy it possesses. For an electron, as it is accelerated by the electric potential, it gains kinetic energy derived from its increased velocity.
The connection between kinetic energy and velocity is given by the formula:
  • \( KE = \frac{1}{2} m v^2 \)
where \( m \) stands for the electron's mass, specifically \( 9.11 \times 10^{-31} \) kg.
By rearranging this equation, we can solve for the speed \( v \) of the electron, allowing us to find that it reaches a significant velocity of approximately \( 1.1 \times 10^7 \) meters per second. This speed reflects the transformed electrical potential energy into kinetic energy.
Magnetic Force
Once in motion, an electron can experience magnetic forces, especially when it encounters a magnetic field perpendicularly. This force, also known as the Lorentz force, acts as a centripetal force that causes the electron to move in a circular path.
If an electron enters a magnetic field perpendicular to its velocity, the force it experiences can be calculated by the equation:
  • \( F_{mag} = e v B \)
where \( e \) is the electron's charge, \( v \) is its velocity, and \( B \) is the magnetic field's strength. In our scenario, with \( B \) being 200 milliTesla or 0.2 Tesla, the Lorentz force directs the electron into a curved trajectory, maintaining circular motion.
Circular Motion
In physics, when an object moves in a circle, it experiences centripetal force that keeps it moving along its circular path. This is exactly what happens to an electron in a magnetic field; the magnetic force acts as the centripetal force.
The electron鈥檚 trajectory can be calculated using:
  • \( F_{centripetal} = \frac{m v^2}{r} \)
where \( r \) represents the radius of the circular path. For our accelerated electron, we equate the centripetal force with the magnetic force.
Solving for \( r \), the formula:
  • \( r = \frac{m v}{e B} \)
lets us determine the radius to be approximately 3.13 mm. This reveals how the interplay of inertia and magnetic forces crafts the electron's graceful circular dance within the magnetic field.

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Most popular questions from this chapter

Two concentric, circu- endlar wire loops, of radii \(r_{1}=20.0 \mathrm{~cm}\) and \(r_{2}=30.0 \mathrm{~cm}\), are located in an \(x y\) plane; each carries a clockwise current of \(7.00 \mathrm{~A}\) (Fig. \(28-47\) ). (a) Find the magnitude of the net magnetic dipole moment of the system. (b) Repeat for reversed current in the inner loop.

A proton, a deuteron \((q=+e, m=2.0 \mathrm{u})\), and an alpha particle \((q=+2 e, m=4.0 \mathrm{u})\) all having the same kinetic energy enter a region of uniform magnetic field \(\vec{B}\), moving perpendicular to \(\vec{B}\). What is the ratio of (a) the radius \(r_{d}\) of the deuteron path to the radius \(r_{p}\) of the proton path and (b) the radius \(r_{\alpha}\) of the alpha particle path to \(r_{2} ?\)

The magnetic dipole moment of Earth has magnitude \(8.00 \times\) \(10^{22} \mathrm{~J} / \mathrm{T}\). Assume that this is produced by charges flowing in Earth's molten outer core. If the radius of their circular path is \(3500 \mathrm{~km}\), calculate the current they produce.

A circular wire loop of radius \(15.0 \mathrm{~cm}\) carries a current of \(2.60 \mathrm{~A}\). It is placed so that the normal to its plane makes an angle of \(41.0^{\circ}\) with a uniform magnetic field of magnitude \(12.0 \mathrm{~T}\). (a) Calculate the magnitude of the magnetic dipole moment of the loop. (b) What is the magnitude of the torque acting on the loop?

A horizontal power line carries a current of \(5000 \mathrm{~A}\) from south to north. Earth's magnetic field \((60.0 \mu \mathrm{T})\) is directed toward the north and inclined downward at \(70.0^{\circ}\) to the horizontal. Find the (a) magnitude and (b) direction of the magnetic force on \(100 \mathrm{~m}\) of the line due to Earth's field.
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