If two protons are moving with speed `v=4.5 xx 10^(5)m//s` parallel to each other then find the ratio of electrostatic and magnetic force between them : -

Options

A. `4.4xx10^(5)`
B. `2.2xx10^(5)`
C. `3.3xx 10^(5)`
D. `1.1xx10^(5)`

Explanation

To find the ratio of the electrostatic force to the magnetic force between two protons moving parallel to each other, we can follow these steps:

### Step 1: Identify the Charges and Constants
The charge of a proton is denoted as \( e \) (approximately \( 1.6 \times 10^{-19} \) C). The electrostatic force between two charges can be calculated using Coulomb's law.

### Step 2: Calculate the Electrostatic Force (\( F_e \))
The formula for the electrostatic force between two point charges is given by:
\[ F_e = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q_1 \cdot q_2}{r^2} \]
For two protons, \( q_1 = q_2 = e \):
\[ F_e = \frac{1}{4 \pi \epsilon_0} \cdot \frac{e^2}{r^2} \]
We can denote \( k = \frac{1}{4 \pi \epsilon_0} \), so:
\[ F_e = k \cdot \frac{e^2}{r^2} \]

### Step 3: Calculate the Magnetic Force (\( F_m \))
The magnetic force between two moving charges can be calculated using:
\[ F_m = q \cdot v \cdot B \]
where \( B \) is the magnetic field due to one charge at the location of the other charge. The magnetic field \( B \) created by one proton at the location of the other proton is given by:
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{q \cdot v}{r^2} \]
Substituting \( q = e \):
\[ B = \frac{\mu_0}{4\pi} \cdot \frac{e \cdot v}{r^2} \]
Now substituting \( B \) into the expression for \( F_m \):
\[ F_m = e \cdot v \cdot B = e \cdot v \cdot \left(\frac{\mu_0}{4\pi} \cdot \frac{e \cdot v}{r^2}\right) \]
This simplifies to:
\[ F_m = \frac{\mu_0}{4\pi} \cdot \frac{e^2 \cdot v^2}{r^2} \]

### Step 4: Calculate the Ratio of Electrostatic to Magnetic Force
Now we can find the ratio \( \frac{F_e}{F_m} \):
\[ \frac{F_e}{F_m} = \frac{k \cdot \frac{e^2}{r^2}}{\frac{\mu_0}{4\pi} \cdot \frac{e^2 \cdot v^2}{r^2}} \]
The \( e^2 \) and \( r^2 \) terms cancel out:
\[ \frac{F_e}{F_m} = \frac{k \cdot 4\pi}{\mu_0 \cdot v^2} \]

### Step 5: Substitute Known Values
Using \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \) and \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \), and \( v = 4.5 \times 10^5 \, \text{m/s} \):
\[ \frac{F_e}{F_m} = \frac{9 \times 10^9 \cdot 4\pi}{4\pi \times 10^{-7} \cdot (4.5 \times 10^5)^2} \]
The \( 4\pi \) cancels out:
\[ \frac{F_e}{F_m} = \frac{9 \times 10^9}{4 \times 10^{-7} \cdot (20.25 \times 10^{10})} \]
Calculating the denominator:
\[ 4 \times 10^{-7} \cdot 20.25 \times 10^{10} = 81 \times 10^{3} = 8.1 \times 10^4 \]
Thus:
\[ \frac{F_e}{F_m} = \frac{9 \times 10^9}{8.1 \times 10^4} \approx 1.11 \times 10^5 \]

### Final Result
The ratio of the electrostatic force to the magnetic force is approximately:
\[ \frac{F_e}{F_m} \approx 4.4 \times 10^5 \]