# How Is Current Density Calculated for Protons in the Solar Wind?

• Sanjay101
In summary, the density of protons in the solar wind near Earth is 8.90 cm-3 with a speed of 490 km/s. To find the current density of these protons, the formula J = nqv is used, where n represents the number density of protons per unit of volume. The number density is already given, so it can be plugged into the formula along with the charge of a proton (q) and the speed (v). For part b, the total current Earth would receive without the deflection of the protons by its magnetic field would be the product of the current density found in part a and the cross sectional area. However, for part a, it is important to note that the number
Sanjay101
Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 8.90 cm-3 and their speed is about 490 km/s.

(a) Find the current density of these protons.(b) If Earth's magnetic field did not deflect the protons, what total current would Earth receive?For part a I used Current density (J) = nqv, we know q and v. I tried finding n(number of protons) by finding the volume by m/density. Then using Avogadro's number/volume to get n.

For part b I said that the Current would just me Part a answer multiplied by the cross sectional area, but i did not get part a right.

Last edited:
Sanjay101 said:
Near Earth, the density of protons in the solar wind (a stream of particles from the Sun) is 8.90 cm-3 and their speed is about 490 km/s.

For part a I used Current density (J) = nqv, we know q and v. I tried finding n(number of protons) by finding the volume by m/density. Then using Avogadro's number/volume to get n.

In n.q.v, n represents the NUMBER density (that means, the number of charges with charge q per unit of volume). In this case, this would mean the number of protons per m^3, or per cm^3 (depending on the unit system in which you work). But that's given! The solar wind consists of 8.9 PROTONS per cubic cm (and not, as you seem to think, 8.9 GRAM of protons per cubic cm - which would be a terribly dense charge density!)

o, i cannot believe i mistook that info. Thanks.

im having trouoble with part a myself i am using vne=J
i have to convert km/s to m/s cm^-3 to m^-3 and e is 1.6e-19

i plug
500000*840*1.6x10^-19 but its wrong can i get help U_U ?

## 1. What is current density of protons?

The current density of protons refers to the amount of electric current carried by a group of protons per unit area. It is a measure of the concentration of protons in a given space.

## 2. How is current density of protons related to electric current?

The current density of protons is directly proportional to the electric current. This means that as the current density increases, the electric current also increases.

## 3. What is the unit of measurement for current density of protons?

The unit of measurement for current density of protons is amperes per square meter (A/m2). This unit is commonly used in scientific and engineering calculations.

## 4. What factors affect the current density of protons?

The current density of protons can be affected by several factors, including the number of protons present, the strength of the electric field, and the material through which the protons are passing.

## 5. How is current density of protons measured or calculated?

The current density of protons can be measured directly using specialized instruments such as a current density meter. It can also be calculated by dividing the total electric current by the cross-sectional area through which the protons are passing.

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