A short conducting rod P of length `3.0 cm` is placed parallel to an near the centre of a long conducting rod Q of length 3.0 m. Conductors P and carry currents of 3.0 A and 4.0 A respectively in the same direction. The two conductors are separated by a distance of `2.0 cm` in air. What is the force experienced by the long conductor Q ?

Options

A. `1.6 xx 10^(-6)N`
B. `2.6 xx 10^(-6)N`
C. `3.6 xx 10^(-6)N`
D. `4.6 xx 10^(-6)N`

Answer

To find the force experienced by the long conductor Q due to the short conductor P, we can use the formula for the magnetic force between two parallel current-carrying conductors. The formula is given by:

\[ F = \frac{{\mu_0}}{2\pi} \cdot \frac{{I_1 \cdot I_2 \cdot L}}{{d}} \]

Where:
- \( F \) is the force between the conductors,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( I_1 \) and \( I_2 \) are the currents in the conductors,
- \( L \) is the length of the shorter conductor,
- \( d \) is the distance between the conductors.

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Length of conductor P, \( L = 3.0 \, \text{cm} = 0.03 \, \text{m} \)
- Current in conductor P, \( I_1 = 3.0 \, \text{A} \)
- Current in conductor Q, \( I_2 = 4.0 \, \text{A} \)
- Distance between the conductors, \( d = 2.0 \, \text{cm} = 0.02 \, \text{m} \)

**Step 2: Substitute the values into the formula.**
Using the formula for the force:
\[ F = \frac{{\mu_0}}{2\pi} \cdot \frac{{I_1 \cdot I_2 \cdot L}}{{d}} \]
Substituting the known values:
\[ F = \frac{{4\pi \times 10^{-7}}}{2\pi} \cdot \frac{{3.0 \cdot 4.0 \cdot 0.03}}{{0.02}} \]

**Step 3: Simplify the expression.**
The \( \pi \) terms cancel out:
\[ F = \frac{{4 \times 10^{-7}}}{2} \cdot \frac{{3.0 \cdot 4.0 \cdot 0.03}}{{0.02}} \]
Calculating \( \frac{4 \times 10^{-7}}{2} = 2 \times 10^{-7} \):
\[ F = 2 \times 10^{-7} \cdot \frac{{3.0 \cdot 4.0 \cdot 0.03}}{{0.02}} \]

**Step 4: Calculate the remaining expression.**
Calculating the numerator:
\[ 3.0 \cdot 4.0 \cdot 0.03 = 0.36 \]
Now, substituting back:
\[ F = 2 \times 10^{-7} \cdot \frac{0.36}{0.02} = 2 \times 10^{-7} \cdot 18 = 36 \times 10^{-7} \, \text{N} \]

**Step 5: Convert to standard form.**
\[ F = 3.6 \times 10^{-6} \, \text{N} \]

### Final Answer:
The force experienced by the long conductor Q is \( 3.6 \times 10^{-6} \, \text{N} \).