Two electrons are moving with speed of `5xx 10^(5)` m/s parallel to each other, then the electrostatic and magnetic force between them is

Options

A. `5xx 10^(5)`
B. `2.5 xx 10^(4)`
C. `3.6 xx 10^(5)`
D. `4.4 xx 10^(3)`

Explanation

To solve the problem of finding the electrostatic and magnetic forces between two electrons moving parallel to each other with a speed of \(5 \times 10^5\) m/s, we will follow these steps:

### Step 1: Understand the Forces Involved
We need to consider both the electrostatic force and the magnetic force between the two electrons.

### Step 2: Calculate the Electrostatic Force
The electrostatic force \(F_e\) between two charges is given by Coulomb's law:
\[ F_e = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]
where:
- \(k\) is Coulomb's constant, approximately \(9 \times 10^9 \, \text{N m}^2/\text{C}^2\),
- \(q_1\) and \(q_2\) are the charges of the electrons, which is the elementary charge \(e \approx 1.6 \times 10^{-19} \, \text{C}\),
- \(r\) is the distance between the two electrons.

Substituting the values:
\[ F_e = \frac{9 \times 10^9 \cdot (1.6 \times 10^{-19})^2}{r^2} \]

### Step 3: Calculate the Magnetic Force
The magnetic force \(F_m\) on a charge moving in a magnetic field is given by:
\[ F_m = q \cdot v \cdot B \]
where:
- \(B\) is the magnetic field created by one of the moving electrons at the location of the other,
- \(v\) is the speed of the electrons.

To find \(B\), we use the formula for the magnetic field due to a moving charge:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{q \cdot v}{r^2} \]
where \(\mu_0\) is the permeability of free space, approximately \(4\pi \times 10^{-7} \, \text{T m/A}\).

Substituting \(B\) into the magnetic force equation:
\[ F_m = q \cdot v \cdot \left(\frac{\mu_0}{4\pi} \cdot \frac{q \cdot v}{r^2}\right) \]
\[ F_m = \frac{\mu_0 \cdot q^2 \cdot v^2}{4\pi r^2} \]

### Step 4: Find the Ratio of Electrostatic to Magnetic Force
Now we find the ratio \( \frac{F_e}{F_m} \):
\[ \frac{F_e}{F_m} = \frac{\frac{k \cdot q^2}{r^2}}{\frac{\mu_0 \cdot q^2 \cdot v^2}{4\pi r^2}} = \frac{k \cdot 4\pi}{\mu_0 \cdot v^2} \]

### Step 5: Substitute Known Values
Substituting \(k = 9 \times 10^9\), \(\mu_0 = 4\pi \times 10^{-7}\), and \(v = 5 \times 10^5\):
\[ \frac{F_e}{F_m} = \frac{9 \times 10^9 \cdot 4\pi}{4\pi \times 10^{-7} \cdot (5 \times 10^5)^2} \]
\[ = \frac{9 \times 10^9}{10^{-7} \cdot 25 \times 10^{10}} = \frac{9 \times 10^9}{25 \times 10^3} = \frac{9}{25} \times 10^6 \]

### Step 6: Final Calculation
Calculating this gives:
\[ \frac{F_e}{F_m} = 0.36 \times 10^6 = 3.6 \times 10^5 \]

### Conclusion
Thus, the ratio of the electrostatic force to the magnetic force between the two electrons is \(3.6 \times 10^5\).

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