Currents of equal magnitude pass through two long parallel wires having separation of 1.35 cm. If the force per unit length on each wire is `4.76xx 10^(-2)` N/m, what is I ?

Answer

To solve the problem, we will use the formula for the force per unit length between two parallel wires carrying currents. The force per unit length \( F/L \) between two long parallel wires is given by the formula:

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

Where:
- \( F \) is the force between the wires,
- \( L \) is the length of the wires,
- \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \, \text{T m/A} \)),
- \( I_1 \) and \( I_2 \) are the currents through the wires,
- \( D \) is the separation between the wires.

Given:
- \( D = 1.35 \, \text{cm} = 1.35 \times 10^{-2} \, \text{m} \)
- \( \frac{F}{L} = 4.76 \times 10^{-2} \, \text{N/m} \)
- Since the currents are equal, we can set \( I_1 = I_2 = I \).

### Step 1: Substitute the known values into the formula

We can rewrite the formula as:

\[ \frac{F}{L} = \frac{\mu_0}{4\pi} \cdot \frac{2I^2}{D} \]

Substituting the known values:

\[ 4.76 \times 10^{-2} = \frac{(4\pi \times 10^{-7})}{4\pi} \cdot \frac{2I^2}{1.35 \times 10^{-2}} \]

### Step 2: Simplify the equation

The \( 4\pi \) cancels out:

\[ 4.76 \times 10^{-2} = \frac{10^{-7}}{2} \cdot \frac{2I^2}{1.35 \times 10^{-2}} \]

This simplifies to:

\[ 4.76 \times 10^{-2} = \frac{10^{-7} I^2}{1.35 \times 10^{-2}} \]

### Step 3: Rearranging to solve for \( I^2 \)

Multiply both sides by \( 1.35 \times 10^{-2} \):

\[ (4.76 \times 10^{-2}) \cdot (1.35 \times 10^{-2}) = 10^{-7} I^2 \]

Calculating the left side:

\[ (4.76 \times 1.35) \times 10^{-4} = 10^{-7} I^2 \]

\[ 6.426 \times 10^{-4} = 10^{-7} I^2 \]

### Step 4: Solve for \( I^2 \)

Now, divide both sides by \( 10^{-7} \):

\[ I^2 = \frac{6.426 \times 10^{-4}}{10^{-7}} = 6.426 \times 10^{3} \]

### Step 5: Take the square root to find \( I \)

\[ I = \sqrt{6.426 \times 10^{3}} \approx 80.1 \, \text{A} \]

### Final Answer

The current \( I \) is approximately \( 80.1 \, \text{A} \).

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