The force between two long parallel wires A and B carrying current is `0.004 Nm^(-1)`. The conductors are 0.01 m apart. If the current in conductor A is twice that of conductor B, then the current in the conductor B would be

Options

A. `5A`
B. `50 A `
C. `10 A`
D. `100 A`

Answer

To solve the problem, we need to find the current in conductor B given the force between two long parallel wires carrying currents, the distance between them, and the relationship between the currents in the two conductors.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Force per unit length between the wires, \( F = 0.004 \, \text{N/m} \)
- Distance between the wires, \( d = 0.01 \, \text{m} \)
- Current in wire A is twice that of wire B, i.e., if the current in wire B is \( I \), then the current in wire A is \( 2I \).

2. **Use the Formula for the Force Between Two Parallel Wires:**
The force per unit length between two parallel wires carrying currents \( I_1 \) and \( I_2 \) separated by a distance \( d \) is given by:
\[ F = \frac{\mu_0 I_1 I_2}{2 \pi d} \]
Here, \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \, \text{T m/A} \).

3. **Substitute the Currents into the Formula:**
Let \( I_B = I \) (current in wire B). Then, \( I_A = 2I \) (current in wire A).
Substituting these into the formula gives:
\[ F = \frac{\mu_0 (2I)(I)}{2 \pi d} = \frac{\mu_0 \cdot 2I^2}{2 \pi d} \]
This simplifies to:
\[ F = \frac{\mu_0 I^2}{\pi d} \]

4. **Plug in the Values:**
Now we can substitute the values we know into the equation:
\[ 0.004 = \frac{(4\pi \times 10^{-7}) I^2}{\pi (0.01)} \]
The \( \pi \) cancels out:
\[ 0.004 = \frac{4 \times 10^{-7} I^2}{0.01} \]

5. **Rearranging the Equation:**
Multiply both sides by \( 0.01 \):
\[ 0.004 \times 0.01 = 4 \times 10^{-7} I^2 \]
\[ 4 \times 10^{-5} = 4 \times 10^{-7} I^2 \]

6. **Solve for \( I^2 \):**
Divide both sides by \( 4 \times 10^{-7} \):
\[ I^2 = \frac{4 \times 10^{-5}}{4 \times 10^{-7}} = 10^2 \]
\[ I^2 = 100 \]

7. **Find \( I \):**
Taking the square root of both sides gives:
\[ I = 10 \, \text{A} \]

### Conclusion:
The current in conductor B is \( 10 \, \text{A} \).