A conductor of length 2 m carrying current 2 A is held parallel to an infinitely long conductor carrying current of 12 A at a distance of 100 mm, the force on small conductor is

Options

A. `8.6 xx 10^(-5)N`
B. `6.6 xx 10^(-5)N`
C. `7.6 xx 10^(-5)N`
D. `9.6 xx 10^(-5)N`

Explanation

To solve the problem, we will use the formula for the force between two parallel conductors carrying current. The force \( F \) on a conductor of length \( L \) carrying current \( I_1 \) in the presence of a magnetic field created by another conductor carrying current \( I_2 \) at a distance \( R \) is given by:

\[ F = \frac{\mu_0}{4\pi} \cdot \frac{2 I_1 I_2}{R} \cdot L \]

Where:
- \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( I_1 = 2 \, \text{A} \) (current in the small conductor)
- \( I_2 = 12 \, \text{A} \) (current in the infinitely long conductor)
- \( R = 100 \, \text{mm} = 0.1 \, \text{m} \) (distance between the conductors)
- \( L = 2 \, \text{m} \) (length of the small conductor)

### Step-by-step Solution:

1. **Convert units where necessary**:
- The distance \( R \) is given as 100 mm, which we convert to meters:
\[ R = 100 \, \text{mm} = 0.1 \, \text{m} \]

2. **Substitute the known values into the formula**:
\[ F = \frac{\mu_0}{4\pi} \cdot \frac{2 I_1 I_2}{R} \cdot L \]
Substituting the values:
\[ F = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{2 \cdot 2 \cdot 12}{0.1} \cdot 2 \]

3. **Simplify the expression**:
- The \( 4\pi \) cancels out:
\[ F = 10^{-7} \cdot \frac{2 \cdot 2 \cdot 12}{0.1} \cdot 2 \]
- Calculate \( 2 \cdot 2 \cdot 12 = 48 \):
\[ F = 10^{-7} \cdot \frac{48}{0.1} \cdot 2 \]
- This simplifies to:
\[ F = 10^{-7} \cdot 480 \cdot 2 = 10^{-7} \cdot 960 \]

4. **Calculate the final force**:
\[ F = 960 \times 10^{-7} \, \text{N} = 9.6 \times 10^{-5} \, \text{N} \]

5. **Final answer**:
The force on the small conductor is:
\[ F = 9.6 \times 10^{-5} \, \text{N} \]