Magnetic field solar wind question

In summary: R = m|V2|/|F| = m(V2)/|F|In summary, the solar wind is a thin, hot gas emitted by the sun that can interact with the Earth's magnetic field, causing charged particles to experience a magnetic force. To find the radius of the circular path of a charged particle with a velocity of 8.20 106 m/s encountering the Earth's magnetic field at an altitude where the field has a magnitude of 1.10 10-7 T, you can use the equation R = m(V2)/|F|, where m is the mass of the particle and V is its velocity. This can be used for both an electron and a proton with a
  • #1
johnson.3131
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The solar wind is a thin, hot gas given off by the sun. Charged particles in this gas enter the magnetic field of the Earth and can experience a magnetic force. Suppose a charged particle traveling with a speed of 8.20 106 m/s encounters the Earth's magnetic field at an altitude where the field has a magnitude of 1.10 10-7 T. Assuming that the particle's velocity is perpendicular to the magnetic field, find the radius of the circular path on which the particle would move if it were each of the following.
(a) an electron


(b) a proton


Relevant Equations:

F=I*L*B

F=qvB

charge electron and proton= +/- 1.6E-19

ATTEMPT: i plugged in the charges into q then then velocity and the magnetic field and got the answer, now I'm confused on how to find the radius.
please help. thanks!
 
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  • #2
The Force vector is going to be qV X B

The cross product gives you a force perpendicular to it's motion.

If the B field is pointing out of your paper and the Velocity is up, then I get V X B as being a force vector right directed to the motion.
(It's not important that you see this to solve the problem, just that it may be useful to visualize.)

Given that you have found the |F| in the B field, then the curvature should be given by centripetal acceleration shouldn't it?

|F| = m|V2|/R
 
  • #3


I would like to clarify that the equations you have listed are not relevant for finding the radius of the circular path of a charged particle in a magnetic field. The correct equation to use is the Lorentz force equation, which is F = qvB, where F is the force experienced by the charged particle, q is the charge of the particle, v is the velocity of the particle, and B is the magnetic field strength.

To find the radius of the circular path, we can use the centripetal force equation, which is F = mv^2/r, where m is the mass of the particle and r is the radius of the circular path. By equating the Lorentz force and the centripetal force, we can solve for the radius:

F = qvB = mv^2/r

r = mv/qB

Now, let's plug in the given values for the electron and the proton:

(a) For an electron:

r = (9.11 x 10^-31 kg)(8.20 x 10^6 m/s) / (1.6 x 10^-19 C)(1.10 x 10^-7 T)

r = 0.0477 meters

Therefore, the radius of the circular path for an electron would be approximately 0.0477 meters.

(b) For a proton:

r = (1.67 x 10^-27 kg)(8.20 x 10^6 m/s) / (1.6 x 10^-19 C)(1.10 x 10^-7 T)

r = 2.19 meters

Therefore, the radius of the circular path for a proton would be approximately 2.19 meters.

It is important to note that these calculations assume that the charged particle's velocity is perpendicular to the magnetic field. If the angle between the velocity and the magnetic field is not 90 degrees, the radius of the circular path would be different and can be calculated using the same equations. I hope this helps clarify the calculation process.
 

Related to Magnetic field solar wind question

What is a magnetic field?

A magnetic field is a region in space where an object with magnetic properties experiences a force. It is created by the movement of electrically charged particles.

How is the magnetic field of the sun related to solar wind?

The magnetic field of the sun is closely connected to solar wind. Solar wind is a stream of charged particles that are constantly released from the sun's corona. This stream of particles carries the sun's magnetic field with it, creating a magnetic field in space known as the interplanetary magnetic field.

What is the impact of solar wind on Earth's magnetic field?

Solar wind has a significant impact on Earth's magnetic field. When the solar wind interacts with Earth's magnetic field, it can cause disturbances and fluctuations in the field. These disturbances can lead to phenomena such as auroras and disruptions in communication and navigation systems.

How does the strength of the magnetic field affect solar wind?

The strength of the magnetic field can greatly influence the behavior of solar wind. A stronger magnetic field can deflect or slow down the solar wind, while a weaker field allows the wind to penetrate deeper into space. This can have implications for the protection of Earth and other planets from the effects of solar wind.

How do scientists study the magnetic field and solar wind?

Scientists use a variety of instruments and techniques to study the magnetic field and solar wind. These include spacecraft, such as NASA's Solar Dynamics Observatory, as well as ground-based observatories and simulations. By studying the magnetic field and solar wind, scientists can better understand the sun and its impact on our solar system.

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