Two straight wires A and B of lengths 10 m and 12 m carrying currents of 4.0 A and 6.0 A respectively in opposite direction, lie parallel to each other at a distance of 3.0 cm. The force on a 15 cm section of the wire B near its centre is

Options

A. `2.4 xx 10^(-5)N,"attractive"`
B. `2.4 xx 10^(-5)N, "repulsive"`
C. `1.2 xx 10^(-5)N, "attractive"`
D. `1.2 xx 10^(-5)N,"attractive"`

Explanation

To solve the problem of finding the force on a 15 cm section of wire B, we can follow these steps:

### Step 1: Understand the Given Data
- Length of wire A, \( L_A = 10 \, \text{m} \)
- Length of wire B, \( L_B = 12 \, \text{m} \)
- Current in wire A, \( I_A = 4.0 \, \text{A} \)
- Current in wire B, \( I_B = 6.0 \, \text{A} \)
- Distance between the wires, \( D = 3.0 \, \text{cm} = 3 \times 10^{-2} \, \text{m} \)
- Length of the section of wire B we are considering, \( L = 15 \, \text{cm} = 15 \times 10^{-2} \, \text{m} \)

### Step 2: Use the Formula for Force Between Parallel Currents
The force per unit length between two parallel wires carrying currents \( I_1 \) and \( I_2 \) is given by the formula:
\[ F = \frac{\mu_0}{2\pi} \cdot \frac{I_1 I_2 L}{D} \]
where:
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \) (permeability of free space)
- \( D \) is the distance between the wires.

### Step 3: Substitute the Values into the Formula
Substituting the values into the formula:
\[ F = \frac{4\pi \times 10^{-7}}{2\pi} \cdot \frac{(4 \, \text{A})(6 \, \text{A})(15 \times 10^{-2} \, \text{m})}{3 \times 10^{-2} \, \text{m}} \]

### Step 4: Simplify the Expression
First, simplify the constants:
\[ F = \frac{4 \times 10^{-7}}{2} \cdot \frac{(4)(6)(15 \times 10^{-2})}{3 \times 10^{-2}} \]
\[ = 2 \times 10^{-7} \cdot \frac{(4)(6)(15)}{3} \]

Calculating the numerical part:
\[ = 2 \times 10^{-7} \cdot \frac{360}{3} = 2 \times 10^{-7} \cdot 120 = 2.4 \times 10^{-5} \, \text{N} \]

### Step 5: Determine the Direction of the Force
Since the currents in the wires are in opposite directions, the force between them will be repulsive.

### Final Answer
The force on the 15 cm section of wire B is:
\[ F = 2.4 \times 10^{-5} \, \text{N} \quad \text{(repulsive)} \]

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