91Ó°ÊÓ

An electron traveling from the sun as part of the solar wind strikes the earth's magnetosphere at latitude \(80.0^{\circ} \mathrm{N}\) in a region where the magnetic field has a strength of \(15.0 \mu \mathrm{T}\) and is directed toward the earth's center. The electron has a speed of \(400 \mathrm{~km} / \mathrm{s}\) and is directed toward the earth's axis parallel to the equator. Magnetic forces send the electron on a helical trajectory. (a) What is the radius of this helix? (b) With what speed does the electron approach the surface of the earth? (c) If you are looking downward toward the earth from space, is the electron's motion clockwise or counterclockwise? (d) What is the frequency of the motion? (e) This electron strikes the ionosphere, where it is further accelerated by an electric field with strength \(20.0 \mathrm{mV} / \mathrm{m}\) directed northward parallel to the earth's surface. What is the electron's new speed after it has been deflected \(100 \mathrm{~km}\) southward by this field? (f) By what factor has its kinetic energy been increased by the electric field?

Step by step solution

01

Calculate the Radius of the Helix

Firstly, when an electron moves in a magnetic field it experiences a force that sends it in a helical path. We can use the following equation to calculate the radius \( r \) of the helix: \( r = \frac{{mv}}{{qB}} \), where \( m \) is the mass of the electron (9.11 x \( 10^{-31} \) kg), \( v \) is the speed of the electron (convert 400 km/s to m/s), \( q \) is the charge of the electron (1.60 x \( 10^{-19} \) C), and \( B \) is the magnetic field strength, converting \(15 \mu T\) to Teslas! (1.5 x \( 10^{-5} \) T)

02

Determine the Speed the Electron Approaches the Earth

The electron moves in a helical path with a speed \( v \) that is constant. The component of the speed towards the Earth’s surface is the same as the initial speed, 400 km/s.

03

Determining the Direction of the Electron’s Motion

Here, the right-hand rule is used. If the fingers of the right hand are curled in the direction of the electron’s motion, the thumb points in the direction of the magnetic field. Hence, the electron’s motion is counterclockwise viewed from the north pole.

04

Frequency of the Motion

The frequency \( f \) of the motion can be found using the formula \( f = \frac{{qB}}{{2\pi m}} \), which is derived from the period of circular motion \( T = \frac{{2\pi r}}{v} \), where \( r \) and \( v \) are the helix radius and velocity calculated in step 1 and 2 respectively. In this case, substitute the values of \( q \), \( B \), and \( m \) calculated before into the formula to find \( f \).

05

Calculate the Electron's New Speed after Deflection

The electric field \( E \) does work on the electron, increasing its speed. The work \( W \) done by the field can be found using \( W = qEd \), where \( d \) is the distance the electron travels before reaching the surface. This work is added to the initial kinetic energy, and the result is the new speed. Solve the equation \( \frac{1}{2}mv^2 = \frac{1}{2}mv_0^2 + qEd \) for \( v \), where \( v_0 \) is the initial speed, \( m \) is the mass of the electron, and the other variables are as defined previously.

06

Calculate the Factor by which Kinetic Energy has Increased

The final kinetic energy \( K_f \) is \( \frac{1}{2}mv^2 \), and the initial kinetic energy \( K_i \) is \( \frac{1}{2}mv_0^2 \). The factor by which kinetic energy has increased is simply \( K_f / K_i \).

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Key Concepts

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

Helical Trajectory of Charged Particles

When charged particles such as electrons move through a magnetic field, they typically do not follow a straight path. Instead, these particles tend to move in a spiral or helical pattern. This movement is caused by the force exerted on the moving charge by the magnetic field, known as the Lorenz force. The path's precise shape is determined by several factors, including the particle's charge, its velocity, and the strength of the magnetic field.

The formula to calculate the radius of this helix, as seen in the given problem, is given by: \( r = \frac{mv}{qB} \) where \( m \) is the mass of the electron, \( v \) is the velocity, \( q \) is the charge of the electron, and \( B \) is the magnetic field strength. The ability to calculate this helix radius is quintessential for understanding not just the path of the particle, but also the energy dynamics as it negotiates this force.

Lorentz Force

The Lorenz force is the combined electric and magnetic force on a point charge due to electromagnetic fields. A charged particle, like an electron, in the presence of an electric and a magnetic field, experiences this force. The Lorenz force is given by the formula \( \mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B}) \), where \( \mathbf{F} \) is the force experienced, \( q \) is the charge of the particle, \( \mathbf{E} \) is the electric field, \( \mathbf{v} \) is the velocity of the particle, and \( \mathbf{B} \) is the magnetic field.

In situations where only a magnetic field is present, the effective Lorenz force simplifies to \( q(\mathbf{v} \times \mathbf{B}) \), which is perpendicular to both the magnetic field and the velocity of the particle. This force is what causes the electron in the exercise to follow a helical trajectory around the direction of the magnetic field lines.

Magnetic Field Strength Calculation

Calculating the strength of a magnetic field is crucial when analyzing the motion of charged particles. The magnetic field strength, denoted as \( B \), has a significant impact on the trajectory and dynamics of the charge's movement. Units of magnetic field strength are Tesla (T) or, on a smaller scale, microtesla \( (\mu T) \).

In the given problem, we are provided with the magnetic field strength as \( 15.0 \mu T \), which must be converted to Tesla for the purpose of calculations (\( 1 \mu T = 10^{-6} T \)). Once we have determined the correct value of the magnetic field strength, it can be applied in formulas, like the one used to calculate the helix radius or the frequency of the motion. Understanding how to accurately calculate the magnetic field strength is essential in predicting the effects of the field on charged particles, such as the electron described in the exercise.