Near Earth the density of protons in the solar wind (a stream of particles from the Sun) can be `4.63cm^(-3)` and their suced can be 391 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?

Answer

To solve the problem step by step, we will break it down into two parts as given in the question.

### Part (a): Find the current density of the protons.

1. **Identify the formula for current density (J)**:
The current density \( J \) is given by the formula:
\[ J = n \cdot e \cdot v_d \]
where:
- \( n \) = number density of charge carriers (protons in this case)
- \( e \) = charge of a proton
- \( v_d \) = drift velocity of the protons

2. **Convert the number density from \( cm^{-3} \) to \( m^{-3} \)**:
Given that the density of protons \( n = 4.63 \, cm^{-3} \):
\[ n = 4.63 \, cm^{-3} = 4.63 \times 10^6 \, m^{-3} \]

3. **Use the charge of a proton**:
The charge of a proton \( e \) is approximately:
\[ e = 1.6 \times 10^{-19} \, C \]

4. **Convert the drift velocity from \( km/s \) to \( m/s \)**:
Given that the drift velocity \( v_d = 391 \, km/s \):
\[ v_d = 391 \times 10^3 \, m/s \]

5. **Substitute the values into the current density formula**:
Now we can substitute \( n \), \( e \), and \( v_d \) into the formula:
\[ J = (4.63 \times 10^6 \, m^{-3}) \cdot (1.6 \times 10^{-19} \, C) \cdot (391 \times 10^3 \, m/s) \]

6. **Calculate the current density**:
Performing the multiplication:
\[ J = 4.63 \times 10^6 \cdot 1.6 \times 10^{-19} \cdot 391 \times 10^3 \]
\[ J \approx 2.90 \times 10^{-7} \, A/m^2 \]

### Part (b): Find the total current that Earth would receive.

1. **Identify the relationship between current density and total current**:
The total current \( I \) can be calculated using the formula:
\[ J = \frac{I}{A} \]
Rearranging gives:
\[ I = J \cdot A \]

2. **Calculate the cross-sectional area (A)**:
Assuming the protons are hitting the Earth, the area \( A \) can be represented as:
\[ A = \pi r^2 \]
where \( r \) is the radius of the Earth. Given \( r \approx 6.37 \times 10^6 \, m \):
\[ A = \pi (6.37 \times 10^6)^2 \]

3. **Calculate the area**:
\[ A \approx 3.14 \times (6.37 \times 10^6)^2 \approx 1.27 \times 10^{14} \, m^2 \]

4. **Substitute \( J \) and \( A \) into the current formula**:
\[ I = (2.90 \times 10^{-7} \, A/m^2) \cdot (1.27 \times 10^{14} \, m^2) \]

5. **Calculate the total current**:
\[ I \approx 3.69 \times 10^{7} \, A \]

### Final Answers:
- (a) Current density \( J \approx 2.90 \times 10^{-7} \, A/m^2 \)
- (b) Total current \( I \approx 3.69 \times 10^{7} \, A \)